Continuity and Differentiability: Class 12 Maths Chapter 5 Hinglish mein

Mathematics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

145 min
45 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

Kisi point par function continuous hai ya nahi, yeh uske one-sided limits ko uski value se compare karke tay karna, aur piecewise function ke saare points of discontinuity dhoondhna.
Continuous functions ke algebra aur composition se dikhana ki polynomial, rational, trigonometric aur modulus functions apne domains par continuous hain.
Samjhana ki differentiability se continuity kyun milti hai, aur one-sided derivatives se dikhana ki |x − a| jaisa corner differentiable nahi hota.
Chain rule se composite, implicit aur inverse trigonometric functions ko differentiate karna.
Exponential aur logarithmic functions aur unke laws samjhana, aur eˣ, log x, aˣ aur unke composites ko differentiate karna.
Variable powers aur lambe products ke liye logarithmic differentiation use karna, aur parametric form mein diye curves ke liye dy/dx nikaalna.
Second order derivatives nikaalna aur function aur uske derivatives ke beech ke relations verify karna.

NCERT Mathematics Class 12, Chapter 5, pp. 104–146 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

5.1 Introduction

Ek aam din ke do graphs socho. Jaipur ka temperature subah bhar smoothly badhta hai, kabhi achaanak chhalaang nahi lagata. Car-park ka charge kuch der ₹40 par tika rehta hai, phir jaise hi ek aur ghanta poora hota hai, turant ₹60 par kood jaata hai.

Yeh chapter is roz ke fark ko mathematics mein badalta hai. Jis function ke graph mein kisi point par koi break nahi hota, woh wahan continuous (संतत) hai. Jis function ke graph ka kisi point par ek hi, well-defined slope bhi ho, woh wahan differentiable (अवकलनीय) hai. Hum dekhenge ki yeh dono ideas aapas mein kaise jude hain.

Phir hum derivatives ke liye tumhara Class XI wala toolkit badhaenge: composite functions, aise relations jo kabhi $y$ ke liye solve nahi kiye gaye, inverse trigonometric functions, exponentials aur logarithms, teesre variable ke through diye curves, aur derivatives ke derivatives.

5.2 Continuity

Ek pencil ko graph par left se right khiskao aur dekho ki kisi khaas point $x = c$ ke paas se guzarte waqt kya hota hai. Agar $c$ ke paas pencil kabhi paper se nahi uthti, to graph wahan unbroken hai. Fig. 5.1 dikhata hai ki graph $c$ par kin tareekon se toot sakta hai. Har break ek alag failure hai, aur hamari definition ko in sabko pakadna hoga.

3/5Left side aur right side alag-alag heights ki taraf jaati hain.

Fig. 5.1

x = c

par graph todne ke chaar tareeke, aur unbroken case.NCERT Figs 5.1–5.2 · p. 104

Axes, jin par x = c par ek dashed vertical guide hai. Step 1 mein ek unbroken rising curve hai jo x = c par ek filled dot se guzarta hai. Step 2 mein wahi curve hai, x = c par ek open circle ke saath, aur c ke upar kahin koi dot nahi, yaani f(c) defined nahi hai. Step 3 curve ko c par kaat deta hai: left piece ek filled dot par khatam hota hai aur right piece upar ek open circle se shuru hota hai, yaani dono sides alag heights ki taraf jaati hain. Step 4 mein curve phir se hai, c par uspar ek open circle hai, jabki f(c) dene wala filled dot curve se kaafi upar hai. Step 5 mein ek curve hai jo c ke right mein upar aur c ke left mein neeche bhaagta hai, dashed guide se chipka hua, aur approach karne ke liye koi finite height nahi hai.

Class XI se yaad karo ki $\lim_{x \to c^{-}} f (x)$ , yaani left-hand limit, woh value hai jiski taraf $f (x)$ jaata hai jab $x$ , $c$ se chhote numbers se hote hue $c$ ki taraf aata hai. Right-hand limit $\lim_{x \to c^{+}} f (x)$ mein $c$ se bade numbers use hote hain. Jab dono exist karte hain aur barabar hote hain, to unki common value limit $\lim_{x \to c} f (x)$ hoti hai. Yahi one-sided limits woh tools hain jo figure ke cases ko alag karte hain.

Definition

Ek point par continuity

Ek real function $f$ aur ek number $c$ lo jahan $f$ defined hai. Hum kehte hain ki $f$ , $c$ par continuous hai jab $\lim_{x \to c} f (x) = f (c)$ ho.

Example $f (x) = x^{2}$ , $c = 3$ par continuous hai: jab $x \to 3$ , to $x^{2} \to 9 = f (3)$ .

c par continuity, poori tarah

\lim_{x \to c^{-}} f (x) = \lim_{x \to c^{+}} f (x) = f (c)

Poora likhen to $c$ par continuity teen sawaal poochti hai, aur teeno ka jawab haan hona chahiye:

Kya $f (c)$ defined hai? (Hole yahan fail hota hai.)
Kya left-hand aur right-hand limits exist karte hain aur barabar hain? (Jump ya blow-up yahan fail hota hai.)
Kya woh common value $f (c)$ ke barabar hai? (Misplaced point yahan fail hota hai.)

Jab $f$ apne domain ke kisi point $c$ par yeh test fail karta hai, to hum $c$ ko point of discontinuity kehte hain aur kehte hain ki $f$ wahan discontinuous hai. Apne domain ke shabd zaroori hain: jis number par $f$ ki koi value hi nahi hai, use is chapter mein kabhi point of discontinuity nahi kaha jaata.

Example 1NCERT Example 1 · p. 106

Ek point par seedhi line

Kya $f (x) = 4 - 3 x$ , $x = 2$ par continuous hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 106

Negative point par ek cubic

Check karo ki $p (x) = x^{3} - 2 x$ , $x = - 1$ par continuous hai ya nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 106

Apne kink par ek modulus

$x = - 3$ par $f (x) = ∣ x + 3 ∣$ ki continuity discuss karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 106

Curve se utha hua ek point

Maan lo $x \neq 1$ ke liye $f (x) = x^{2} - 1$ aur $f (1) = 2$ . Dikhao ki $f$ , $x = 1$ par continuous nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Neeche diye probe se teeno checks khud chalao. Yeh $x$ ko dono sides se point ki taraf le jaata hai, har press par das guna paas, aur jo heights milti hain unki list banata hai.

InteractiveContinuity probe

x = −1 se 5 tak ka graph, c = 2 par dashed guide ke saath, aur chhe function buttons: jude pieces, ek jump, ek hole, ek misplaced point, ek blow-up aur ek corner. Graph par do probe dots 2 − h aur 2 + h par hain. Move closer ka har press h ko 10 se divide karta hai, 1 se 0.001 tak, aur f(2 − h) aur f(2 + h) ki table mein ek row jodta hai; ek line f(2) dikhati hai ya batati hai ki woh undefined hai. Aakhri press ke baad teen checks par tick ya cross lagta hai (f(2) defined hai, dono sides ek number par settle hoti hain, woh number f(2) ke barabar hai) aur ek verdict batata hai ki function 2 par continuous hai ya nahi. Reset se phir shuru hota hai.

Fig. 5.2Ek function chuno, Move closer dabakar

x

ko dono sides se

c = 2

ki taraf lao, phir verdict padho.

Example 5NCERT Example 5 · p. 107

Ek constant function

Dikhao ki constant function $f (x) = - 4$ har real number par continuous hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6NCERT Example 6 · p. 107

Identity function

Dikhao ki identity function $f (x) = x$ ka koi point of discontinuity nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ab tak humne ek baar mein ek point test kiya hai. Aksar hum domain ke har point ke baare mein ek saath jaanna chahte hain.

Definition

Continuous function

Aisa real function jo apne domain ke har point par continuous ho.

Example $f (x) = x^{2}$ ek continuous function hai: yeh har real number par continuous hai.

Do edge cases par ek pal rukna chahiye. Agar $f$ ka domain sirf ek point hai, to $f$ test apne aap pass kar leta hai, kyunki $f (c)$ se asehmat hone ke liye koi paas ki values hi nahi hain. Agar $f$ ek closed interval $[a, b]$ par hai, to har end ke padosi sirf ek side par hote hain, isliye wahan hum one-sided limits use karte hain: left end par $\lim_{x \to a^{+}} f (x) = f (a)$ aur right end par $\lim_{x \to b^{-}} f (x) = f (b)$ . $a$ par left-hand limit maangne ka matlab hoga aisi values maangna jo $f$ ke paas hain hi nahi.

Example 7NCERT Example 7 · p. 108

Har jagah modulus

Kya $f (x) = ∣ x + 3 ∣$ ek continuous function hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 108

Degree four ka ek polynomial

$f (x) = 2 x^{4} - 5 x + 1$ ki continuity discuss karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 108

Apne domain par ek reciprocal

$f (x) = \frac{1}{x - 2}$ , $x \neq 2$ ki continuity discuss karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

To $x = 2$ ke paas aakhir hota kya hai? Table 5.1 ki tarah, har side se 2 ke aur paas ki values try karo. Table do bekaabu trends dikhati hai.

Table 5.1Jab

x

, 2 ke paas aata hai to

f (x) = 1 / (x - 2)

ki values

$x$ (2 ke right)	$f (x)$	$x$ (2 ke left)	$f (x)$
3	1	1	−1
2.5	2	1.5	−2
2.1	10	1.9	−10
2.01	100	1.99	−100
2.001	1000	1.999	−1000

2 ke right mein values har bound ke paar chali jaati hain: koi number bolo, maan lo ek million, aur jaise hi $x$ , 2 se ek millionth ke andar aata hai, $f (x)$ use paar kar leta hai. Iska shorthand hai $\lim_{x \to 2^{+}} f (x) = + \infty$ . Left mein values isi tarah har bound se neeche gir jaati hain, jise $\lim_{x \to 2^{-}} f (x) = - \infty$ likhte hain. Symbol ke saath dhyaan rakho: $+ \infty$ aur $- \infty$ real numbers nahi hain. Yeh equations kehti hain ki one-sided limits real numbers ke roop mein exist nahi karte, aur batati hain ki woh kaise fail hote hain.

2/2Jab $x \to 2^{-}$ , dots bina bound ke neeche girte hain: $f (x) \to - \infty$ .

Fig. 5.3

x = 2

ke paas

y = \frac{1}{x - 2}

ka graph.NCERT Fig 5.3 · p. 109

Axes, jin par x = 2 par ek dashed vertical guide hai. Pehle step mein right branch highlighted hai, jismein x = 3, 2.5, 2.25 aur 2.2 par dots hain jinki heights 1, 2, 4 aur 5 frame ke top ki taraf chadhti hain, jahan plus infinity likha hai. Doosre step mein left branch highlighted hai, jismein x = 1, 1.5, 1.75 aur 1.8 par dots hain jinki heights −1, −2, −4 aur −5 bottom ki taraf girti hain, jahan minus infinity likha hai. Dono branches kabhi guide ko nahi chhooti.

Example 10NCERT Example 10 · p. 110

Do pieces jo nahi milte

$f (x) = {\begin{cases} 2 x - 1, & x \leq 3 \\ x + 4, & x > 3 \end{cases}$ ki continuity discuss karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 110

Teen hisson wala rule

$f (x) = {\begin{cases} x^{2}, & x < 2 \\ 1, & x = 2 \\ 6 - x, & x > 2 \end{cases}$ ke saare points of discontinuity dhoondho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 111

Domain mein gap, break nahi hota

$f (x) = {\begin{cases} 3 - x, & x < 0 \\ x^{2} + 1, & x > 0 \end{cases}$ ki continuity discuss karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 111

Pieces jo milte hain

Dikhao ki $f (x) = {\begin{cases} x^{2} + 2, & x \leq 1 \\ 4 x - 1, & x > 1 \end{cases}$ ek continuous function hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 112

Har polynomial continuous hai

Prove karo ki har polynomial $p (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ (real $a_{i}$ ke saath) ek continuous function hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Greatest integer function $[x]$ woh sabse bada integer deta hai jo $x$ se chhota ya uske barabar ho. Jaise $[2.7] = 2$ , $[5] = 5$ aur $[- 1.2] = - 2$ ( $- 1$ nahi, kyunki $- 1$ , $- 1.2$ se bada hai). Do paas-paas ke integers ke beech $[x]$ constant rehta hai, aur har integer par 1 se upar chadhta hai. Steps mein badhne wale charges bhi aise hi behave karte hain, jaisa Fig. 5.4 dikhata hai.

2/3 $x = 2$ par left side ₹40 ki taraf jaati hai lekin fee ₹60 hai: ek jump.

Fig. 5.4

x

ghanton ke liye fee

F (x) = 20 + 20 [x]

rupees,

0 \leq x < 4

.NCERT Fig 5.8 · p. 112

Axes par 0 se 4 tak park kiye ghante aur 0 se 100 tak fee rupees mein hai. Graph mein 20, 40, 60 aur 80 rupees par chaar flat steps hain; har step left mein filled dot se shuru hota hai aur right mein open circle par khatam hota hai. Doosre step mein poora ghanta x = 2 highlighted hai: ek arrow dikhata hai ki left step ₹40 ki taraf jaata hai, jabki theek 2 ghante ki fee, ₹60, upar wale step par mark hai. Teesre step mein x = 2.5 highlighted hai, ek flat step ke beech mein, jahan graph dono sides par level hai aur kuch nahi tootta.

Example 15NCERT Example 15 · p. 112

Parking fee kahan jump karti hai?

Ek car park entry par ₹20 leta hai aur har poore hue ghante ke end par ₹20 aur, 4 ghante tak. To $x$ ghanton ki fee $F (x) = 20 + 20 [x]$ , $0 \leq x < 4$ hai. Woh points dhoondho jahan $F$ discontinuous hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.2.1 Continuous functions ka algebra

Har function ko shuru se test karna slow hoga. Kyunki continuity limits ke baare mein ek statement hai, Class XI ke limit laws seedhe yahan bhi lagte hain: continuous functions ko $+$ , $-$ , $\times$ ya $\div$ se jodo aur result bhi continuous hoga.

Theorem 1Continuous functions ka algebra

Agar do real functions $f$ aur $g$ dono $c$ par continuous hain, to unka sum $f + g$ , difference $f - g$ aur product $f \cdot g$ bhi continuous hain. Unka quotient $\frac{f}{g}$ bhi $c$ par continuous hai, jab tak $g (c) \neq 0$ ho.

Hum product wala part prove karte hain; baaki parts matching limit law ke saath isi pattern par chalte hain. $f$ aur $g$ dono $c$ par defined hain, isliye $f \cdot g$ bhi hai.

\lim_{x \to c} (f \cdot g) (x) = \lim_{x \to c} f (x) \cdot \lim_{x \to c} g (x) = f (c) g (c) = (f \cdot g) (c)

Pehla step limits ka product law hai, aur doosra $c$ par $f$ aur $g$ ki continuity use karta hai. Quotient ke liye condition $g (c) \neq 0$ theek wahi hai jo limits ke quotient law ko chahiye.

Product part mein $f$ ko ek constant $λ$ lo: agar $g$ continuous hai, to $λ g$ bhi hai. $λ = - 1$ ke saath, $g$ ki continuity se $- g$ ki continuity milti hai.
Quotient part mein $f = λ$ lo: $\frac{λ}{g}$ har us jagah continuous hai jahan $g (x) \neq 0$ . Khaas taur par $\frac{1}{g}$ bhi.

Example 16NCERT Example 16 · p. 114

Rational functions

Dikhao ki $r (x) = \frac{x^{2} + 1}{x - 3}$ continuous hai, aur samjhao ki yahi baat har rational function ke liye kyun sach hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 114

Cosine function

Prove karo ki $\cos x$ continuous hai. Do facts use karo jo 0 ke paas sine aur cosine ke graphs se saaf dikhte hain: $\lim_{h \to 0} \sin h = 0$ aur $\lim_{h \to 0} \cos h = 1$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 115

Secant function

Dikhao ki $\sec x$ jahan bhi defined hai, wahan continuous hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Functions banane ka doosra tareeka composition hai: $g$ ka output $f$ mein daalo to $(f \circ g) (x) = f (g (x))$ milta hai, jiske liye $g$ ka har output $f$ ka allowed input hona chahiye. Fig. 5.5 dikhata hai ki continuity isme bhi kyun bachi rehti hai.

3/3 $f$ , $g (c)$ par continuous hai, isliye $f (g (x))$ , $f (g (c))$ ke paas hi rehta hai.

Fig. 5.5Har continuous stage se guzarkar chhote changes chhote hi rehte hain.NCERT p. 115

Teen horizontal number lines ek ke neeche ek hain, jin par x, u = g(x) aur y = f(u) likha hai. Step 1 mein x line par point c ke around ek chhota highlighted interval hai. Step 2 mein g likhe curved arrows use beech wali line par g(c) ke around ek chhote highlighted interval tak le jaate hain. Step 3 mein f likhe arrows use bottom line par f(g(c)) ke around ek chhote highlighted interval tak le jaate hain, yaani shuru mein chhota change end mein bhi sirf chhota change laata hai.

Theorem 2Composite ki continuity

Maan lo $f \circ g$ , $c$ par defined hai. $c$ par $g$ ki continuity, aur point $g (c)$ par $f$ ki continuity, milkar $f \circ g$ ko $c$ par continuous banati hain.

Yeh theorem hum bina proof ke use karte hain; idea figure mein hai.

Example 19NCERT Example 19 · p. 115

Polynomial ka sine

Dikhao ki $h (x) = \sin (2 x^{3} - x)$ ek continuous function hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 115

Modulus ke andar modulus

Dikhao ki $f (x) = ∣ ∣ x ∣ - 2 ∣$ ek continuous function hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 12 questions

Practice 5.1: answers ke saath

NCERT Exercise 5.1 ki skills par naye questions.

1
Short answerEasyNCERT Exercise 5.1, Q1–Q2
Continuity ki definition use karo.
1. Dikhao ki $f (x) = 7 - 2 x$ , $x = - 1$ , $x = 0$ aur $x = 4$ par continuous hai.
  Solution dikhao
  AnswerTeeno points par continuous.
  $f$ ek polynomial hai, isliye har $c$ ke liye $\lim_{x \to c} f (x) = 7 - 2 c = f (c)$ . Khaas taur par $f (- 1) = 9$ , $f (0) = 7$ aur $f (4) = - 1$ wahan ki limits ke barabar hain.
2. $x = - 2$ par $g (x) = 3 x^{2} + x - 4$ ki continuity examine karo.
  Solution dikhao
  Answer $x = - 2$ par continuous.
  $g (- 2) = 12 - 2 - 4 = 6$ , aur $\lim_{x \to - 2} (3 x^{2} + x - 4) = 6$ . Limit value ke barabar hai.
2
Short answerEasyNCERT Exercise 5.1, Q3
Har function ki continuity examine karo.
1. $h (x) = 4 x + 9$
  Solution dikhao
  AnswerHar jagah continuous.
  Polynomial har real number par continuous hota hai.
2. $h (x) = \frac{1}{x + 3}$ , $x \neq - 3$
  Solution dikhao
  AnswerApne domain par continuous.
  $c \neq - 3$ ke liye $\lim_{x \to c} \frac{1}{x + 3} = \frac{1}{c + 3} = h (c)$ . Point $- 3$ domain mein nahi hai, isliye $h$ ek continuous function hai.
3. $h (x) = \frac{x^{2} - 9}{x - 3}$ , $x \neq 3$
  Solution dikhao
  AnswerApne domain par continuous.
  $x \neq 3$ ke liye $h (x) = x + 3$ . Har $c \neq 3$ par $\lim_{x \to c} h (x) = c + 3 = h (c)$ .
4. $h (x) = ∣ x + 1 ∣$
  Solution dikhao
  AnswerHar jagah continuous.
  $c < - 1$ ke liye $c$ ke paas $h = - (x + 1)$ ; $c > - 1$ ke liye $c$ ke paas $h = x + 1$ ; dono continuous hain. $- 1$ par dono one-sided limits 0 hain, jo $h (- 1)$ ke barabar hai.
Short answerEasyNCERT Exercise 5.1, Q4
Prove karo ki har positive integer $m$ ke liye $f (x) = x^{m}$ , $x = 2$ par continuous hai.
Solution dikhao
Answer $\lim_{x \to 2} x^{m} = 2^{m} = f (2)$ .
$x^{m}$ , $x$ ki $m$ copies ka product hai, aur $\lim_{x \to 2} x = 2$ . Product law se $\lim_{x \to 2} x^{m} = 2^{m}$ , jo $f (2)$ hai. Isliye $f$ , 2 par continuous hai.
Short answerMediumNCERT Exercise 5.1, Q5
Maan lo $x \leq 2$ ke liye $f (x) = x + 1$ aur $x > 2$ ke liye $f (x) = 4$ . Kya $f$ , $x = 0$ par continuous hai? $x = 2$ par? $x = 3$ par?
Solution dikhao
Answer0 par haan, 2 par nahi, 3 par haan.
0 aur 3 par, point ke paas $f$ ek polynomial ya constant jaisa hai, isliye wahan continuous hai.
2 par: $f (2) = 3$ , $\lim_{x \to 2^{-}} (x + 1) = 3$ lekin $\lim_{x \to 2^{+}} 4 = 4$ . Sides barabar nahi hain, isliye $f$ , 2 par discontinuous hai.
5
Short answerMediumNCERT Exercise 5.1, Q6–Q12
Har function ke saare points of discontinuity dhoondho.
1. $x \leq 1$ ke liye $f (x) = 3 x - 1$ ; $x > 1$ ke liye $f (x) = 3 x + 2$
  Solution dikhao
  AnswerSirf $x = 1$ .
  1 par: left-hand limit $2$ , right-hand limit $5$ . Yeh alag hain, isliye 1 ek point of discontinuity hai; baaki jagah $f$ continuous hai.
2. $x \leq - 1$ ke liye $f (x) = ∣ x ∣ + 2$ ; $- 1 < x < 2$ ke liye $x^{2} + 2$ ; $x \geq 2$ ke liye $3 x + 1$
  Solution dikhao
  AnswerSirf $x = 2$ .
  $- 1$ par: $f (- 1) = 3$ , left-hand limit $3$ , right-hand limit $1 + 2 = 3$ , isliye continuous. 2 par: left-hand limit $4 + 2 = 6$ , right-hand limit $f (2) = 7$ , isliye discontinuous.
3. $x \neq 2$ ke liye $f (x) = \frac{∣ x - 2 ∣}{x - 2}$ ; $f (2) = 1$
  Solution dikhao
  AnswerSirf $x = 2$ .
  $x > 2$ ke liye $f (x) = 1$ ; $x < 2$ ke liye $f (x) = - 1$ . 2 par one-sided limits $- 1$ aur $1$ hain, isliye $f$ wahan discontinuous hai. Baaki jagah $f$ har point ke paas constant hai.
4. $x > 0$ ke liye $f (x) = \frac{∣ x ∣}{x}$ ; $x \leq 0$ ke liye $f (x) = 1$
  Solution dikhao
  AnswerKoi nahi.
  $x > 0$ ke liye $∣ x ∣ / x = x / x = 1$ . Isliye har real $x$ ke liye $f (x) = 1$ , ek constant function, jo har jagah continuous hai.
5. $x \geq 2$ ke liye $f (x) = 2 x + 1$ ; $x < 2$ ke liye $f (x) = x^{2} + 1$
  Solution dikhao
  AnswerKoi nahi.
  2 par: $f (2) = 5$ , left-hand limit $4 + 1 = 5$ , right-hand limit 5. Sab barabar hain, isliye $f$ har jagah continuous hai.
6. $x \leq 1$ ke liye $f (x) = x^{3} + 1$ ; $x > 1$ ke liye $f (x) = 3 x - 1$
  Solution dikhao
  AnswerKoi nahi.
  1 par: $f (1) = 2$ , left-hand limit 2, right-hand limit $3 - 1 = 2$ . Har jagah continuous.
7. $x \leq - 1$ ke liye $f (x) = x^{5} + 1$ ; $x > - 1$ ke liye $f (x) = x^{2}$
  Solution dikhao
  AnswerSirf $x = - 1$ .
  $- 1$ par: $f (- 1) = - 1 + 1 = 0$ , lekin right-hand limit $(- 1)^{2} = 1$ hai. Sirf $- 1$ par discontinuous.
6
Short answerMediumNCERT Exercise 5.1, Q13–Q16
Har function ki continuity discuss karo.
1. $x \leq 0$ ke liye $f (x) = 4 - x$ ; $x > 0$ ke liye $f (x) = x + 4$ . Kya $f$ ek continuous function hai?
  Solution dikhao
  AnswerHaan.
  0 par: $f (0) = 4$ aur dono one-sided limits 4 hain. Baaki jagah har point ke paas $f$ ek polynomial hai. Isliye $f$ continuous hai.
2. Ek courier 2 kg tak ke parcels ke ₹10, 2 kg se zyada aur 5 kg tak ke ₹15, aur 5 kg se zyada aur 8 kg tak ke ₹25 leta hai. Charge $C (w)$ , $0 \leq w \leq 8$ ki continuity discuss karo.
  Solution dikhao
  AnswerSirf $w = 2$ aur $w = 5$ par discontinuous.
  Har interval par $C$ constant hai. 2 par: $C (2) = 10$ , left-hand limit 10, right-hand limit 15, isliye discontinuous. 5 par: $C (5) = 15$ , right-hand limit 25, isliye discontinuous. End points 0 aur 8 par one-sided limits values se match karti hain.
3. $x < 0$ ke liye $f (x) = - 3 x$ ; $0 \leq x \leq 2$ ke liye $0$ ; $x > 2$ ke liye $2 x - 3$
  Solution dikhao
  AnswerSirf $x = 2$ par discontinuous.
  0 par: left-hand limit 0, $f (0) = 0$ , right-hand limit 0, isliye continuous. 2 par: $f (2) = 0$ , lekin right-hand limit $4 - 3 = 1$ hai, isliye discontinuous.
4. $x \leq 0$ ke liye $f (x) = 1$ ; $0 < x \leq π$ ke liye $\cos x$ ; $x > π$ ke liye $- 1$
  Solution dikhao
  AnswerHar jagah continuous.
  0 par: $f (0) = 1$ aur $\lim_{x \to 0^{+}} \cos x = 1$ . $π$ par: $f (π) = \cos π = - 1$ aur right-hand limit $- 1$ hai. Har piece continuous hai, isliye $f$ har jagah continuous hai.
7
Short answerMediumNCERT Exercise 5.1, Q17–Q18
Unknown constants ke saath pieces jodna.
1. $a$ aur $b$ ke beech woh relation dhoondho jiske liye $f (x) = a x - 2$ ( $x \leq 2$ ), $f (x) = b x^{2} + 1$ ( $x > 2$ ), $x = 2$ par continuous ho.
  Solution dikhao
  Answer $2 a - 4 b = 3$ .
  $f (2) = 2 a - 2$ aur right-hand limit $4 b + 1$ hai. Continuity ke liye $2 a - 2 = 4 b + 1$ chahiye, yaani $2 a - 4 b = 3$ .
2. Kis $λ$ ke liye $f (x) = λ (x^{2} + 3)$ ( $x \leq 1$ ), $f (x) = 2 x + 6$ ( $x > 1$ ), $x = 1$ par continuous hai? Kya har $λ$ ke liye $f$ , $x = 0$ par continuous hai?
  Solution dikhao
  Answer $λ = 2$ ; haan, har $λ$ ke liye $f$ , 0 par continuous hai.
  1 par: $f (1) = 4 λ$ aur right-hand limit 8 hai, isliye $λ = 2$ . 0 ke paas dono sides par saare $x$ ke liye $f (x) = λ (x^{2} + 3)$ hai, jo ek polynomial hai, isliye $λ$ kuch bhi ho, $f$ , 0 par continuous hai.
Short answerHardNCERT Exercise 5.1, Q19
Dikhao ki $h (x) = [x] + [- x]$ har integer par discontinuous hai.
Solution dikhao
AnswerIntegers par $h = 0$ aur baaki jagah $h = - 1$ , isliye $\lim_{x \to n} h (x) = - 1 \neq 0 = h (n)$ .
Agar $x$ integer nahi hai, to $[x] = n$ likho jahan $n < x < n + 1$ . Tab $- n - 1 < - x < - n$ , isliye $[- x] = - n - 1$ aur $h (x) = - 1$ .
Agar $x = n$ ek integer hai, to $h (n) = n + (- n) = 0$ .
Kisi bhi integer $n$ ke paas (lekin us par nahi) $h$ , $- 1$ ke barabar hai, isliye $\lim_{x \to n} h (x) = - 1 \neq h (n)$ . Har integer ek point of discontinuity hai.
9
Short answerMediumNCERT Exercise 5.1, Q20–Q22
Trigonometric combinations.
1. Kya $f (x) = 2 \cos x + x^{3}$ , $x = \frac{π}{2}$ par continuous hai?
  Solution dikhao
  AnswerHaan.
  Yeh continuous functions ka sum hai, isliye har jagah continuous hai, $\frac{π}{2}$ par bhi, jahan iski value $\frac{π^{3}}{8}$ hai.
2. $3 \sin x - x^{2}$ , $x \sin x$ aur $\frac{\sin x}{2 + \cos x}$ ki continuity discuss karo.
  Solution dikhao
  AnswerTeeno har jagah continuous hain.
  Pehla continuous functions ka difference hai aur doosra unka product. Teesre mein $2 + \cos x \geq 1$ , isliye denominator kabhi zero nahi hota aur quotient har real number par continuous hai.
3. $\sin x$ , $\tan x$ , $cosec x$ aur $\cot x$ ki continuity discuss karo.
  Solution dikhao
  AnswerHar ek apne domain par continuous hai: $\sin$ har jagah, $\tan$ , $x \neq (2 n + 1) \frac{π}{2}$ ke liye, $cosec$ aur $\cot$ , $x \neq n π$ ke liye.
  Sine ke liye $\sin (c + h) = \sin c \cos h + \cos c \sin h \to \sin c$ use karo. Phir $\tan = \frac{\sin}{}$ , $cosec = \frac{1}{\sin}$ aur $\cot = \frac{\cos}{}$ continuous functions ke quotients hain, jo har us jagah continuous hain jahan denominator non-zero ho.
10
Short answerHardNCERT Exercise 5.1, Q23–Q25
Sine wali limits.
1. $x < 0$ ke liye $f (x) = \frac{\sin 2 x}{x}$ , $x \geq 0$ ke liye $f (x) = x + 2$ ke points of discontinuity dhoondho.
  Solution dikhao
  AnswerKoi nahi.
  $\lim_{x \to 0^{-}} \frac{\sin 2 x}{x} = \lim 2 \cdot \frac{\sin 2 x}{2 x} = 2$ , right-hand limit 2 hai aur $f (0) = 2$ . Har piece apne interval par continuous hai, isliye $f$ har jagah continuous hai.
2. Kya $f (x) = x \sin \frac{1}{x}$ ( $x \neq 0$ ), $f (0) = 0$ ek continuous function hai?
  Solution dikhao
  AnswerHaan.
  $x \neq 0$ ke liye $∣ x \sin \frac{1}{x} ∣ \leq ∣ x ∣$ , jo 0 ki taraf jaata hai. Isliye $\lim_{x \to 0} f (x) = 0 = f (0)$ . 0 se door, $f$ continuous functions ka product aur composite hai.
3. $f (x) = \cos x + \sin x$ ( $x \neq 0$ ), $f (0) = 2$ ki continuity examine karo.
  Solution dikhao
  AnswerSirf 0 par discontinuous.
  $\lim_{x \to 0} (\cos x + \sin x) = 1 + 0 = 1$ , lekin $f (0) = 2$ . Baaki jagah $f$ continuous functions ka sum hai.
11
Short answerMediumNCERT Exercise 5.1, Q26–Q30
Woh constants dhoondho jo har function ko diye gaye point par continuous banate hain.
1. $x \neq π$ ke liye $f (x) = \frac{k \sin x}{π - x}$ , $f (π) = 5$ , $x = π$ par. $k$ nikaalo.
  Solution dikhao
  $t = π - x$ rakho. Tab $\sin x = \sin (π - t) = \sin t$ , aur $t \to 0$ par $\frac{\sin t}{t} \to 1$ . Limit $k$ hai, isliye $k = f (π) = 5$ .
2. $x \leq 3$ ke liye $f (x) = k x^{2} + 1$ , $x > 3$ ke liye $f (x) = 4$ , $x = 3$ par. $k$ nikaalo.
  Solution dikhao
  $f (3) = 9 k + 1$ ko right-hand limit 4 ke barabar hona chahiye, isliye $k = \frac{1}{3}$ .
3. $x \leq \frac{π}{2}$ ke liye $f (x) = k x - 1$ , $x > \frac{π}{2}$ ke liye $f (x) = \sin x$ , $x = \frac{π}{2}$ par. $k$ nikaalo.
  Solution dikhao
  $f (\frac{π}{2}) = \frac{k π}{2} - 1$ ko $\sin \frac{π}{2} = 1$ ke barabar hona chahiye, isliye $k = \frac{4}{π}$ .
4. $x \leq 4$ ke liye $f (x) = k x + 3$ , $x > 4$ ke liye $f (x) = 2 x - 1$ , $x = 4$ par. $k$ nikaalo.
  Solution dikhao
  $4 k + 3 = 8 - 1 = 7$ , isliye $k = 1$ .
5. $a$ aur $b$ aise nikaalo ki $f (x) = 2$ ( $x \leq 1$ ), $a x + b$ ( $1 < x < 4$ ), $14$ ( $x \geq 4$ ) ek continuous function ho.
  Solution dikhao
  Answer $a = 4$ , $b = - 2$ .
  1 par: $a + b = 2$ . 4 par: $4 a + b = 14$ . Ghataane par $3 a = 12$ , isliye $a = 4$ aur $b = - 2$ .
12
Short answerMediumNCERT Exercise 5.1, Q31–Q34
Composites aur moduli.
1. Dikhao ki $f (x) = \cos (\frac{1}{1 + x^{2}})$ continuous hai.
  Solution dikhao
  AnswerHar jagah continuous.
  $u (x) = \frac{1}{1 + x^{2}}$ rational hai aur iska denominator kabhi zero nahi hota, isliye yeh har jagah continuous hai; cosine continuous hai; Theorem 2 se composite continuous hai.
2. Dikhao ki $f (x) = ∣ \sin 2 x ∣$ continuous hai.
  Solution dikhao
  AnswerHar jagah continuous.
  $\sin 2 x$ , $2 x$ aur sine ka composite hai, isliye continuous hai; continuous function ka modulus continuous hota hai ( $m (u) = ∣ u ∣$ ke saath Theorem 2).
3. Dikhao ki $f (x) = \sin (∣ x ∣ - 2)$ continuous hai.
  Solution dikhao
  AnswerHar jagah continuous.
  $∣ x ∣ - 2$ continuous hai aur sine continuous hai, isliye composite continuous hai.
4. $f (x) = ∣ x - 1 ∣ + ∣ x + 2 ∣$ ke saare points of discontinuity dhoondho.
  Solution dikhao
  AnswerKoi nahi.
  Har modulus ek polynomial aur $∣ u ∣$ ka composite hai, isliye continuous hai; unka sum har jagah continuous hai.

5.3 Differentiability

Class XI mein tumne real function $f$ ka uske domain ke point $c$ par derivative dekha tha:

c par derivative

f^{'} (c) = \lim_{h \to 0} \frac{f (c + h) - f (c)}{h}

jahan

$h$: $c$ se ek chhota step, positive ya negative

Geometrically, $f^{'} (c)$ woh slope hai jis par chords ke chhote hote jaane par chord slopes settle hote hain. Har us $x$ par, jahan limit exist karti hai, yeh slopes ikattha karne se ek naya function banta hai, derivative $f^{'}$ . Is chapter mein tum iske saare naam dekhoge: $f^{'} (x)$ , $\frac{d}{d x} f (x)$ , aur jab $y = f (x)$ ho to $\frac{d y}{d x}$ ya $y^{'}$ ; poora process differentiation kehlata hai. $x$ ke functions $u$ aur $v$ ke liye, Class XI ke teen rules zyadatar kaam karte hain:

$(u \pm v)^{'} = u^{'} \pm v^{'}$
$(u v)^{'} = u^{'} v + u v^{'}$ (product rule, jise Leibniz's rule bhi kehte hain)
${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ jahan bhi $v \neq 0$ (quotient rule)

Table 5.2Class XI ke standard derivatives

$f (x)$	$f^{'} (x)$
$x^{n}$	$n x^{n - 1}$
$\sin x$	$\cos x$
$\cos x$	$- \sin x$
$\tan x$	$\sec^{2} x$

$f^{'} (c)$ ka formula tabhi kaam karta hai jab uski limit exist kare, aur kuch points par woh nahi karti. Wahan $f$ ka koi derivative hi nahi hota: woh $c$ par differentiable nahi hai. Limit kis taraf se fail hoti hai, yeh dekhne ke liye $h$ ko har side se alag-alag 0 ki taraf le jao. Isse ek left-hand derivative ( $h < 0$ se) aur ek right-hand derivative ( $h > 0$ se) milta hai.

Left-hand aur right-hand derivatives

L f^{'} (c) = \lim_{h \to 0^{-}} \frac{f (c + h) - f (c)}{h}, R f^{'} (c) = \lim_{h \to 0^{+}} \frac{f (c + h) - f (c)}{h}

Definition

Ek point par differentiable

Hum $f$ ko uske domain ke point $c$ par differentiable kehte hain jab $c$ par uske left-hand aur right-hand derivatives dono finite aur barabar hon; unki common value $f^{'} (c)$ hai.

Example $c = 1$ par $f (x) = x^{2}$ : dono one-sided difference quotients $2 + h$ ke barabar hain, jo 2 ki taraf jaata hai, isliye $f^{'} (1) = 2$ .

Open interval $(a, b)$ par differentiability ka matlab hai uske har point par differentiability. Closed interval $[a, b]$ par hamein ends par one-sided derivatives bhi chahiye: $a$ par right-hand derivative aur $b$ par left-hand derivative, bilkul waise hi jaise end point par continuity sirf ek side use karti hai.

Theorem 3Differentiable matlab continuous

Kisi point par differentiability wahan continuity ko zaroori bana deti hai: agar $f^{'} (c)$ exist karta hai, to $f$ , $c$ par continuous hai.

Corollary 1Differentiable functions

Har differentiable function continuous hota hai.

Iska converse galat hai: graph kisi point par unbroken hokar bhi wahan ek single slope ke bina ho sakta hai. Sabse simple culprit ek corner hai.

ExampleNCERT p. 120

x = 3 par ek corner

Dikhao ki $f (x) = ∣ x - 3 ∣$ , $x = 3$ par continuous hai lekin differentiable nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Corner par dono sides ke chords do alag slopes par settle hote hain. Fig. 5.6 ise ek curved corner ke liye dikhata hai, jahan chords chhote hote hue saaf-saaf mudte hain.

4/4 $- 2 \neq 2$ : koi single tangent nahi, isliye $f$ , 1 par differentiable nahi hai.

Fig. 5.6

x = 1

par

y = ∣ x^{2} - 1 ∣

ke corner tak chords.NCERT p. 120

x lagbhag −0.2 se 2 tak y = |x² − 1| ke graph mein point (1, 0) par ek sharp corner hai. Step 1 mein sirf curve aur corner hai. Step 2 corner se left branch par x = 0.2, 0.6 aur 0.85 ke points tak teen chords banata hai; yeh slope −2 ki taraf steep hote jaate hain, aur slope −2 ki ek dashed line labelled hai. Step 3 right branch par x = 1.8, 1.4 aur 1.15 ke points tak teen chords banata hai, jo slope 2 ki dashed line ki taraf settle hote hain. Step 4 dono dashed limiting lines saath dikhata hai: do alag slopes, isliye koi single tangent nahi.

Neeche ki zoom lab yahi baat doosre tareeke se dikhati hai. Smooth graph ke kisi point par zoom in karo to graph jaldi hi ek seedhi line jaisa dikhta hai, jiska slope derivative hai. Corner par zoom in karo to woh kitna bhi aage jao, corner hi rehta hai.

InteractiveZoom lab

Ek marked point par centred graph window, paanch function buttons ke saath: 1 par x squared, 1 par smooth join, 1 par kinked join, 1 par |x − 1| aur 1 par x − 1 ka cube root. Zoom in aur Zoom out buttons window ko aadha ya double karte hain; ek readout window ki width batata hai. Do chords point ko uske left aur right mein h door ke points se jodte hain, jahan h window width se juda hai, aur ek table h = 0.1, 0.01 aur 0.001 ke liye left aur right chord slopes ki list deti hai. Ek verdict line batati hai ki dono slopes ek number par settle hote hain (derivative), do alag numbers par (corner), ya bina bound ke badhte hain (vertical tangent).

Fig. 5.7Ek function chuno, marked point par zoom in karo, aur left aur right se aane wale chords ke slopes compare karo.

5.3.1 Composite functions ke derivatives

Tum $y = (5 - 2 x)^{2}$ ko kaise differentiate karoge? Ek tareeka hai pehle expand karna: $y = 25 - 20 x + 4 x^{2}$ , isliye $\frac{d y}{d x} = - 20 + 8 x = - 4 (5 - 2 x)$ . Ab answer ko alag nazar se dekho. Bahar wala operation, squaring, $2 (5 - 2 x)$ deta hai; andar ka expression $5 - 2 x$ , rate $- 2$ se badalta hai; aur $2 (5 - 2 x) \times (- 2)$ theek $- 4 (5 - 2 x)$ hai.

Yeh pattern kaam ka hai, kyunki $(5 - 2 x)^{40}$ ko expand karna bekaar hai, jabki pattern turant $40 (5 - 2 x)^{39} \times (- 2) = - 80 (5 - 2 x)^{39}$ de deta hai. Iske peeche idea yeh hai ki chain ke saath rates multiply hote hain, jaisa Fig. 5.8 dikhata hai.

3/3 $y = u^{2}$ , 16 se 18.49 par jaata hai: $Δ y \approx 8 Δ u \approx 24 Δ x$ .

Fig. 5.8

x

ka change

u

se hokar

y

tak jaata hai, aur har link par multiply hota hai.NCERT p. 121

Teen horizontal rulers ek ke neeche ek hain: upar wala x ke liye, beech wala u = 3x + 1 ke liye aur neeche wala y = u squared ke liye. Step 1 mein x = 1 se 1.1 tak jaana mark hai, 0.1 ka change. Step 2 mein u, 4 se 4.3 tak jaata hai, 0.3 ka change, teen guna bada, aur du/dx = 3 likha hai. Step 3 mein y, 16 se 18.49 tak jaata hai, 2.49 ka change, jo u ke change ka lagbhag 8 guna hai kyunki dy/du = 2u = 8, isliye y, x se lagbhag 24 guna tez badalta hai.

Theorem 4Chain rule

Maan lo $f (x) = v (u (x))$ , aur andar ki value ke liye $t = u (x)$ likho. Jab bhi andar ka rate $\frac{d t}{d x}$ aur bahar ka rate $\frac{d v}{d t}$ dono exist karte hain, composite ka rate unka product hota hai: $\frac{d f}{d x} = \frac{d v}{d t} \cdot \frac{d t}{d x}$ .

Hum chain rule bina proof ke use karte hain. Figure mein $u = 4$ par $\frac{d u}{d x} = 3$ aur $\frac{d y}{d u} = 2 u = 8$ , isliye $\frac{d y}{d x} = 24$ ; 0.1 ke step ke liye asli change 2.49, $24 \times 0.1$ ke kaafi paas hai. Lambi chains bhi isi tarah kaam karti hain. Agar $x \to t = u (x) \to s = v (t) \to y = w (s)$ , to, agar teeno derivatives exist karte hain,

Teen links wala chain rule

\frac{d y}{d x} = \frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x}

Example 21NCERT Example 21 · p. 121

Cube ka cosine

$f (x) = \cos (x^{3})$ ko differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

ExampleNCERT p. 121

Teen links

$y = \sin^{3} (2 x + 1)$ ko differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 3 questions

Practice 5.2: answers ke saath

NCERT Exercise 5.2 ki skills par naye questions.

1
Short answerMediumNCERT Exercise 5.2, Q1–Q8
Har function ko $x$ ke respect mein differentiate karo.
1. $\cos (x^{3} - 2)$
  Solution dikhao
  Answer $- 3 x^{2} \sin (x^{3} - 2)$
  Bahar ka $\cos t$ , $- \sin t$ deta hai; andar ka $t = x^{3} - 2$ , $3 x^{2}$ deta hai. Isliye $- 3 x^{2} \sin (x^{3} - 2)$ .
2. $\tan (\sin x)$
  Solution dikhao
  Answer $\cos x \sec^{2} (\sin x)$
  $t = \sin x$ ke saath $\frac{d}{d x} \tan t = \sec^{2} t \cdot \frac{d t}{d x}$ , jo $\sec^{2} (\sin x) \cos x$ deta hai.
3. $\cos (4 - 3 x)$
  Solution dikhao
  Answer $3 \sin (4 - 3 x)$
  $- \sin (4 - 3 x) \cdot (- 3) = 3 \sin (4 - 3 x)$ .
4. $\cos (\tan 2 x)$
  Solution dikhao
  Answer $- 2 \sec^{2} 2 x \cdot \sin (\tan 2 x)$
  Teen links: $- \sin (\tan 2 x) \cdot \sec^{2} (2 x) \cdot 2$ .
5. $\frac{\cos 2 x}{\sin (x^{2})}$
  Solution dikhao
  Answer $- \frac{2 \sin 2 x}{\sin (x^{2})} - \frac{2 x \cos 2 x \cos (x^{2})}{\sin^{2} (x^{2})}$
  Quotient rule, $u = \cos 2 x$ , $u^{'} = - 2 \sin 2 x$ , $v = \sin (x^{2})$ aur $v^{'} = 2 x \cos (x^{2})$ ke saath, jahan bhi $\sin (x^{2}) \neq 0$ :
  $\frac{u^{'} v - u v^{'}}{v^{2}} = \frac{- 2 \sin 2 x \sin (x^{2}) - 2 x \cos 2 x \cos (x^{2})}{\sin^{2} (x^{2})}$
6. $\sin (x^{2}) \cos^{3} (2 x)$
  Solution dikhao
  Answer $2 x \cos (x^{2}) \cdot \cos^{3} 2 x - 6 \sin (x^{2}) \cdot \cos^{2} 2 x \sin 2 x$
  Product rule. $\frac{d}{d x} \sin (x^{2}) = 2 x \cos (x^{2})$ aur $\frac{d}{d x} \cos^{3} 2 x = 3 \cos^{2} 2 x \cdot (- \sin 2 x) \cdot 2 = - 6 \cos^{2} 2 x \sin 2 x$ .
7. $\sqrt{\tan 3 x}$ (jahan $\tan 3 x > 0$ )
  Solution dikhao
  Answer $\frac{3 \sec^{2} 3 x}{2 \sqrt{\tan 3 x}}$
  $t = \tan 3 x$ ke saath $\frac{1}{2 \sqrt{t}} \cdot \sec^{2} 3 x \cdot 3$ .
8. $\cos (\frac{1}{x})$ , $x \neq 0$
  Solution dikhao
  Answer $\frac{1}{x^{2}} \sin (\frac{1}{x})$
  $- \sin (\frac{1}{x}) \cdot (- \frac{1}{x^{2}}) = \frac{1}{x^{2}} \sin \frac{1}{x}$ .
Short answerMediumNCERT Exercise 5.2, Q9
Prove karo ki $f (x) = ∣ x + 2 ∣$ , $x = - 2$ par differentiable nahi hai.
Solution dikhao
AnswerLeft-hand derivative −1 hai aur right-hand derivative 1.
$\frac{f (- 2 + h) - f (- 2)}{h} = \frac{∣ h ∣ - 0}{h}$ . $h < 0$ ke liye yeh $- 1$ hai; $h > 0$ ke liye $1$ . One-sided derivatives alag hain, isliye $f^{'} (- 2)$ exist nahi karta.
Short answerHardNCERT Exercise 5.2, Q10
Maan lo $0 < x < 2$ ke liye $f (x) = [2 x]$ . Dikhao ki $f$ , $x = \frac{1}{2}$ , $x = 1$ aur $x = \frac{3}{2}$ par differentiable nahi hai.
Solution dikhao
AnswerIn har point par $f$ , 1 se jump karta hai, isliye wahan continuous nahi hai aur differentiable nahi ho sakta.
$[2 x]$ tab jump karta hai jab $2 x$ kisi integer ko paar karta hai, yaani $(0, 2)$ mein $x = \frac{1}{2}, 1, \frac{3}{2}$ par. Jaise $x = 1$ par: left-hand limit $[2^{-}] = 1$ hai lekin $f (1) = [2] = 2$ .
Jo function kisi point par continuous nahi hai, woh wahan differentiable nahi ho sakta (Theorem 3). Seedhe dekho to: $h < 0$ ke liye $\frac{f (1 + h) - f (1)}{h} = \frac{1 - 2}{h} = - \frac{1}{h}$ , jo $h \to 0^{-}$ par bina bound ke badhta hai.

5.3.2 Implicit functions ke derivatives

Kai curves $y$ ke formula ki jagah $x$ aur $y$ ko jodne wali equation ke roop mein aate hain. Relation $3 x + 2 y = 7$ aasaani se $y = \frac{7 - 3 x}{2}$ mein badla ja sakta hai: yahan $y$ , $x$ ka explicit function hai. Relation $y^{5} + y = x$ alag hai. Har $x$ ke liye theek ek $y$ hai, kyunki $y^{5} + y$ lagataar badhta hai, phir bhi algebra ka koi formula $y$ ko $x$ ke terms mein nahi likhta.

Definition

Implicit function

Jab $x$ aur $y$ ke beech ka relation $y$ ko $x$ ka function fix kar deta hai, lekin $y = f (x)$ form mein solve nahi kiya gaya (ya aasaani se solve nahi ho sakta), to hum kehte hain ki relation $y$ ko $x$ ka implicit function deta hai.

Example $y^{5} + y = x$ , $y$ ko implicitly define karta hai.

$\frac{d y}{d x}$ nikaalne ke liye hamein formula ki zaroorat nahi. Relation ki dono sides ko $x$ ke respect mein differentiate karo, yeh yaad rakhte hue ki $y$ , $x$ ka function hai, isliye chain rule se $y$ wala har term ek factor $\frac{d y}{d x}$ laata hai: $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ . Phir $\frac{d y}{d x}$ ke liye solve karo. Fig. 5.9 ek circle use karta hai, jo ek single function ka graph nahi hai, lekin uska har half hai.

3/3 $2 x + 2 y y^{'} = 0$ se $(3, 4)$ par $y^{'} = - \frac{x}{y} = - \frac{3}{4}$ milta hai.

Fig. 5.9

x^{2} + y^{2} = 25

ka implicit differentiation.NCERT p. 122

Square axes par origin par centred radius 5 ka circle. Step 1 mein poora circle hai aur ek vertical line x = 3 use do baar milti hai, (3, 4) aur (3, −4) par, isliye circle ek function ka graph nahi hai. Step 2 upper half ko highlight karta hai, y barabar 25 minus x squared ka square root, aur lower half ko halka karta hai. Step 3 point (3, 4) par slope −3/4 wali tangent line jodta hai, aur ek label dy/dx = −x/y deta hai.

Tangent ka slope doosre tareeke se check karo: upper half par $y = \sqrt{25 - x^{2}}$ se $x = 3$ par $y^{'} = \frac{- x}{\sqrt{25 - x^{2}}} = - \frac{3}{4}$ milta hai, pehle jaisa hi. Implicit route ne bina koi half chune yahi answer de diya.

Example 22NCERT Example 22 · p. 123

Do raaste, ek answer

Agar $4 x - 2 y = 9$ hai to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 23NCERT Example 23 · p. 123

Aisa relation jo solve nahi ho sakta

Agar $y^{2} + \sin y = x^{3}$ hai to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.3.3 Inverse trigonometric functions ke derivatives

Chapter 2 ke inverse trigonometric functions apne domains par continuous hain; yeh hum bina proof ke maante hain. Inhe differentiate karne ke liye har ek ko ulta karke trigonometric equation banao aur implicitly differentiate karo.

Example 24NCERT Example 24 · p. 124

tan⁻¹ x ka derivative

Yeh maan kar ki derivative exist karta hai, $y = \tan^{- 1} x$ ka derivative nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Derivationsin⁻¹ x aur cos⁻¹ x ke derivatives

Maan lo $y = \sin^{- 1} x$ , yaani $x = \sin y$ , jahan $- \frac{π}{2} \leq y \leq \frac{π}{2}$ . Differentiate karne par $1 = \cos y \frac{d y}{d x}$ , isliye $\frac{d y}{d x} = \frac{1}{\cos y}$ . Iske liye $\cos y \neq 0$ chahiye, jo $y = \pm \frac{π}{2}$ , yaani $x = \pm 1$ , ko rule out karta hai.

$- \frac{π}{2} < y < \frac{π}{2}$ ke liye cosine positive hai, isliye $\cos y = + \sqrt{1 - \sin^{2} y} = \sqrt{1 - x^{2}}$ . Isliye

\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}, - 1 < x < 1

$y = \cos^{- 1} x$ ke liye $x = \cos y$ , jahan $0 \leq y \leq π$ , aur $1 = - \sin y \frac{d y}{d x}$ . $0 < y < π$ par sine positive hai, isliye $\sin y = \sqrt{1 - x^{2}}$ :

\frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}, - 1 < x < 1

Ek check: $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ , ek constant, isliye dono derivatives ka sum 0 hona chahiye, aur hota bhi hai.

Table 5.3Inverse trigonometric functions ke derivatives

$f (x)$	$f^{'} (x)$	kab valid
$\sin^{- 1} x$	$\frac{1}{\sqrt{1 - x^{2}}}$	$- 1 < x < 1$
$\cos^{- 1} x$	$- \frac{1}{\sqrt{1 - x^{2}}}$	$- 1 < x < 1$
$\tan^{- 1} x$	$\frac{1}{1 + x^{2}}$	saare real $x$

Reciprocals kyun? $\sin^{- 1}$ ka graph line $y = x$ mein sine graph ke ek piece ki mirror image hai. Reflection rise aur run ko aapas mein badal deta hai, isliye slope $m$ wali tangent slope $\frac{1}{m}$ wali tangent ban jaati hai, jaisa Fig. 5.10 dikhata hai.

3/3Reflected tangent ka slope $2 = \frac{1}{\sqrt{1 - (\sqrt{3} / 2)^{2}}}$ hai, jo $\frac{1}{2}$ ka reciprocal hai.

Fig. 5.10Dashed line

y = x

mein mirror images ke matching points par slopes reciprocal hote hain.NCERT p. 124

Dashed mirror line y = x ke saath square axes. Step 1 mein −π/2 se π/2 tak y = sin x ka piece hai aur point (π/3, √3/2), lagbhag (1.05, 0.87), apni slope cos(π/3) = 1/2 wali tangent ke saath. Step 2 −1 se 1 tak reflected curve y = sin⁻¹ x jodta hai, aur ek dotted segment jo point ko uski mirror image (√3/2, π/3) se jodta hai. Step 3 (√3/2, π/3) par reflected curve ki tangent jodta hai, jiska slope 2 hai, yaani 1/2 ka reciprocal.

Quick check

4 mein se 1

\frac{d}{d x} \sin (5 x^{2}) =

\cos (5 x^{2})

10 x \cos (5 x^{2})

10 x \cos (10 x)

5 \cos (x^{2})

Answer dekho

Answer (B) $10 x \cos (5 x^{2})$

Bahar:

\cos (5 x^{2})

; andar:

\frac{d}{d x} 5 x^{2} = 10 x

; multiply karo. Option A andar ka rate bhool jaata hai, aur option C andar ko do baar differentiate kar deta hai.

Practice · 2 questions

Practice 5.3: answers ke saath

NCERT Exercise 5.3 ki skills par naye questions.

1
Short answerMediumNCERT Exercise 5.3, Q1–Q8
Har relation ke liye $\frac{d y}{d x}$ nikaalo.
1. $5 x - 2 y = \cos x$
  Solution dikhao
  Answer $\frac{5 + \sin x}{2}$
  $5 - 2 y^{'} = - \sin x$ , isliye $y^{'} = \frac{5 + \sin x}{2}$ .
2. $x + 4 y = \cos y$
  Solution dikhao
  Answer $- \frac{1}{4 + \sin y}$
  $1 + 4 y^{'} = - \sin y y^{'}$ , isliye $y^{'} (4 + \sin y) = - 1$ .
3. $p x^{2} + q y = \sin y$ ( $p$ , $q$ constants)
  Solution dikhao
  Answer $\frac{2 p x}{\cos y - q}$
  $2 p x + q y^{'} = \cos y y^{'}$ , isliye $y^{'} (\cos y - q) = 2 p x$ .
4. $x y^{2} + y = \sec x$
  Solution dikhao
  Answer $\frac{\sec x \tan x - y^{2}}{2 x y + 1}$
  $y^{2} + 2 x y y^{'} + y^{'} = \sec x \tan x$ , isliye $y^{'} (2 x y + 1) = \sec x \tan x - y^{2}$ .
5. $x^{2} - x y + 2 y^{2} = 50$
  Solution dikhao
  Answer $\frac{y - 2 x}{4 y - x}$
  $2 x - (y + x y^{'}) + 4 y y^{'} = 0$ , isliye $y^{'} (4 y - x) = y - 2 x$ .
6. $x^{3} + y^{3} = 3 x y + 7$
  Solution dikhao
  Answer $\frac{y - x^{2}}{y^{2} - x}$
  $3 x^{2} + 3 y^{2} y^{'} = 3 y + 3 x y^{'}$ , isliye $y^{'} (y^{2} - x) = y - x^{2}$ .
7. $\cos^{2} y + \sin (x y) = \frac{1}{2}$
  Solution dikhao
  Answer $\frac{y \cos (x y)}{\sin 2 y - x \cos (x y)}$
  $- 2 \cos y \sin y y^{'} + \cos (x y) (y + x y^{'}) = 0$ . $2 \sin y \cos y = \sin 2 y$ ke saath: $y^{'} (x \cos (x y) - \sin 2 y) = - y \cos (x y)$ .
8. $\tan^{2} x + \sec^{2} y = 4$
  Solution dikhao
  Answer $- \frac{\tan x \sec^{2} x}{\tan y \sec^{2} y}$
  $2 \tan x \sec^{2} x + 2 \sec y \cdot \sec y \tan y y^{'} = 0$ , isliye $y^{'} = - \frac{\tan x \sec^{2} x}{\tan y \sec^{2} y}$ .
2
Short answerHardNCERT Exercise 5.3, Q9–Q15
Substitution se simplify karo, phir $\frac{d y}{d x}$ nikaalo.
1. $y = \tan^{- 1} \frac{2 x}{1 - x^{2}}$ , $- 1 < x < 1$
  Solution dikhao
  Answer $\frac{2}{1 + x^{2}}$
  $x = \tan θ$ , $- \frac{π}{4} < θ < \frac{π}{4}$ ke saath: $\frac{2 \tan θ}{1 - \tan^{2} θ} = \tan 2 θ$ aur $2 θ \in (- \frac{π}{2}, \frac{π}{2})$ , isliye $y = 2 θ = 2 \tan^{- 1} x$ .
2. $y = \sin^{- 1} (3 x - 4 x^{3})$ , $- \frac{1}{2} < x < \frac{1}{2}$
  Solution dikhao
  Answer $\frac{3}{\sqrt{1 - x^{2}}}$
  $x = \sin θ$ , $- \frac{π}{6} < θ < \frac{π}{6}$ ke saath: $3 \sin θ - 4 \sin^{3} θ = \sin 3 θ$ aur $3 θ \in (- \frac{π}{2}, \frac{π}{2})$ , isliye $y = 3 \sin^{- 1} x$ .
3. $y = \sin^{- 1} \frac{1 - x^{2}}{1 + x^{2}}$ , $0 < x < 1$
  Solution dikhao
  Answer $- \frac{2}{1 + x^{2}}$
  $x = \tan θ$ , $0 < θ < \frac{π}{4}$ ke saath: fraction $\cos 2 θ = \sin (\frac{π}{2} - 2 θ)$ hai aur $\frac{π}{2} - 2 θ \in (0, \frac{π}{2})$ . Isliye $y = \frac{π}{2} - 2 \tan^{- 1} x$ .
4. $y = \cos^{- 1} (2 x \sqrt{1 - x^{2}})$ , $0 < x < \frac{1}{\sqrt{2}}$
  Solution dikhao
  Answer $- \frac{2}{\sqrt{1 - x^{2}}}$
  $x = \sin θ$ , $0 < θ < \frac{π}{4}$ ke saath: $2 \sin θ \cos θ = \sin 2 θ = \cos (\frac{π}{2} - 2 θ)$ , jahan $\frac{π}{2} - 2 θ \in (0, \frac{π}{2})$ . Isliye $y = \frac{π}{2} - 2 \sin^{- 1} x$ .
5. $y = \tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}$ , $x \neq 0$
  Solution dikhao
  Answer $\frac{1}{2 (1 + x^{2})}$
  $x = \tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $θ \neq 0$ ke saath: $\frac{\sec θ - 1}{\tan θ} = \frac{1 - \cos θ}{\sin θ} = \tan \frac{θ}{2}$ . Isliye $y = \frac{θ}{2} = \frac{1}{2} \tan^{- 1} x$ .
6. $y = \cos^{- 1} (4 x^{3} - 3 x)$ , $\frac{1}{2} < x < 1$
  Solution dikhao
  Answer $- \frac{3}{\sqrt{1 - x^{2}}}$
  $x = \cos θ$ , $0 < θ < \frac{π}{3}$ ke saath: $4 \cos^{3} θ - 3 \cos θ = \cos 3 θ$ aur $3 θ \in (0, π)$ . Isliye $y = 3 \cos^{- 1} x$ .
7. $y = \sec^{- 1} \frac{1}{1 - 2 x^{2}}$ , $0 < x < \frac{1}{\sqrt{2}}$
  Solution dikhao
  Answer $\frac{2}{\sqrt{1 - x^{2}}}$
  $x = \sin θ$ , $0 < θ < \frac{π}{4}$ ke saath: $1 - 2 \sin^{2} θ = \cos 2 θ$ , isliye argument $\sec 2 θ$ hai, jahan $2 θ \in (0, \frac{π}{2})$ . Isliye $y = 2 \sin^{- 1} x$ .

5.4 Exponential aur Logarithmic Functions

Interest kamaati savings, doodh mein badhte bacteria, bloodstream se nikalti medicine: yeh sab barabar time mein barabar factors se badalte hain, jise koi polynomial describe nahi kar sakta. Jo functions kar sakte hain woh exponentials hain, aur unke inverses logarithms hain. Hum inki main properties ko prove karne ki jagah unke liye argument denge; proofs ke liye aage ke courses ke tools chahiye.

Powers se shuru karo. Jab $x$ , 1 se 2 hota hai, to $x^{2}$ , 1 se 4 hota hai, $x^{5}$ , 1 se 32 aur $x^{10}$ , 1 se 1024. $x > 1$ ke liye power jitni badi, growth utni tez. Phir bhi ek alag tarah ka function, $2^{x}$ , aakhir mein har power ko peeche chhod deta hai. $x = 100$ par $x^{10}$ ko $2^{x}$ se compare karo: $100^{10} = 10^{20}$ , jabki $2^{100} \approx 1.27 \times 10^{30}$ , das billion guna se bhi zyada. Lagbhag $x = 59$ se aage $2^{x}$ hamesha ke liye $x^{10}$ se aage rehta hai, aur tum koi bhi power chuno, uske against aisa hi crossover hota hai. Fig. 5.11 $x^{3}$ ke against race dikhata hai, jahan crossover itni jaldi aata hai ki draw kiya ja sake.

3/3 $x = 12$ par: $4096$ vs $1728$ . 1 ka har step $2^{x}$ ko double karta hai, lekin $x^{3}$ ko kam se kam factor se multiply karta hai.

Fig. 5.11

y = x^{3}

y = 2^{x}

.NCERT Fig 5.9 · p. 126

Step 1 ek close-up hai, x 0 se 6 aur y 0 se 230 tak: cube curve exponential se kaafi upar hai, x = 5 ke upar 125 aur 32 par dots hain. Step 2 aur 3 zoom out karke x 0 se 12.6 aur y 0 se 4600 tak dikhate hain. Step 2 x = 9.94 ke paas crossing mark karta hai, jahan dono values lagbhag 980 hain. Step 3 x = 12 par values mark karta hai, jahan 2 to the power 12, 4096 hai lekin 12 cubed sirf 1728, aur exponential lagbhag vertical chadhta hai jabki cube dheere-dheere badhta hai.

Definition

Exponential function

Real number $b > 1$ ke liye, base $b$ wala exponential function $y = f (x) = b^{x}$ hai.

Example $y = 2^{x}$ , $y = 10^{x}$ (common exponential function) aur $y = e^{x}$ .

Ek hi axes par $y = 2^{x}$ , $y = 3^{x}$ aur $y = 5^{x}$ sketch karo; har ek mein tumhe yeh features dikhenge.

Har graph $(0, 1)$ se guzarta hai, kyunki har base ke liye $b^{0} = 1$ .
Har graph left se right poora upar chadhta hai: $b^{x}$ ek increasing function hai.
Bahut left mein graph negative $x$ -axis se chipka rehta hai lekin kabhi chhoota nahi, kyunki $b^{x}$ 0 ki taraf chhota hota hai lekin positive rehta hai.
Isliye $b^{x}$ har real input leta hai (domain $𝐑$ ) aur sirf positive outputs deta hai (range $(0, \infty)$ ).

Class XI mein tumne number $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ dekha tha, jo 2 aur 3 ke beech hai (lagbhag 2.71828). $e$ ko base lene par hamein natural exponential function $y = e^{x}$ milta hai, jo calculus ka sabse important exponential hai.

Kya exponential function ka inverse hota hai? Kyunki $b^{x}$ increasing hai, har positive number theek ek $x$ ke liye $b^{x}$ hai. Woh exponent hi uska logarithm hai.

Definition

Logarithm

Maan lo $b > 1$ aur $a > 0$ . Base $b$ par $a$ ka logarithm woh number $x$ hai jiske liye $b^{x} = a$ . Hum likhte hain $\log_{b} a = x$ .

Example $3^{4} = 81$ , isliye $\log_{3} 81 = 4$ .

Aur examples: $10^{- 2} = 0.01$ wahi kehta hai jo $\log_{10} 0.01 = - 2$ ; aur kyunki $64 = 4^{3} = 8^{2}$ , hamare paas $\log_{4} 64 = 3$ aur $\log_{8} 64 = 2$ hai. Base fix karne par $\log_{b}$ positive reals se saare reals tak ek function ban jaata hai: $\log_{b} x = y$ theek tab jab $b^{y} = x$ . Base 10 se common logarithms milte hain aur base $e$ se natural logarithms, jinhe aksar $\ln x$ likhte hain. Is chapter mein $\log x$ ka matlab hamesha natural logarithm $\log_{e} x$ hai.

$\log_{b} x$ ka graph line $y = x$ mein reflect kiya hua $b^{x}$ ka graph hai, jaisa Fig. 5.12 base 2 ke liye dikhata hai. Baaki features isi se nikalte hain.
$(0, 1)$ ko reflect karne par $(1, 0)$ milta hai: har base ke liye $\log_{b} 1 = 0$ .
Reflection ke baad bhi graph upar chadhta rehta hai, isliye $\log_{b}$ increasing hai.
Inputs aur outputs ke roles badal jaate hain: logarithm sirf positive numbers leta hai (domain $(0, \infty)$ ; $\log 0$ aur negatives ke logs undefined hain) aur koi bhi real number de sakta hai (range $𝐑$ ).
Jab $x$ , 0 ki taraf chhota hota hai, $\log x$ bina bound ke neeche girta hai, negative $y$ -axis se chipka hua lekin use chhoota nahi.

3/3Mirror points $y = \log_{2} x$ par hain, jo $(1, 0)$ se guzarta hai.

Fig. 5.12Coordinates swap karne se

2^{x}

ka graph

\log_{2} x

ka graph ban jaata hai.NCERT Figs 5.10–5.11 · p. 128

−2 se 5 tak square axes, dashed line y = x ke saath. Step 1 mein y = 2ˣ points (−1, 1/2), (0, 1), (1, 2) aur (2, 4) se guzarta hai. Step 2 in har point ko ek chhote dotted segment se, jo y = x ko right angle par kaatta hai, uske mirror point (1/2, −1), (1, 0), (2, 1) aur (4, 2) se jodta hai. Step 3 mirror points se guzarta y = log₂ x banata hai, jo x ke 0 ke paas aane par negative y-axis se chipka rehta hai.

Logarithms ke laws seedhe exponents ke laws se aate hain. Neeche ke derivation mein $p$ aur $q$ positive hain aur base $b > 1$ fixed hai.

DerivationLogarithms ke laws

Maan lo $\log_{b} p = β$ aur $\log_{b} q = γ$ , yaani $b^{β} = p$ aur $b^{γ} = q$ .

Product. $p q = b^{β} b^{γ} = b^{β + γ}$ , isliye $\log_{b} (p q) = β + γ$ :

\log_{b} (p q) = \log_{b} p + \log_{b} q

Quotient. $\frac{p}{q} = \frac{b^{β}}{b^{γ}} = b^{β - γ}$ , isliye

\log_{b} \frac{p}{q} = \log_{b} p - \log_{b} q

Power. $p \cdot p \dots p$ par product law $n$ baar lagane se har positive integer $n$ ke liye $\log_{b} p^{n} = n \log_{b} p$ milta hai. (Yeh har real $n$ ke liye sach hai, lekin woh hum yahan prove nahi karte.)

Change of base. Maan lo $a > 1$ ek aur base hai aur $\log_{a} p = α$ , yaani $a^{α} = p$ . Agar $\log_{b} a = δ$ , to $a = b^{δ}$ aur $p = (b^{δ})^{α} = b^{δ α}$ . $p = b^{β}$ se compare karne par $β = α δ$ milta hai, yaani

\log_{a} p = \frac{\log_{b} p}{\log_{b} a}

Example 25NCERT Example 25 · p. 129

e^(2 log x) kab x² ke barabar hai?

Kya har real $x$ ke liye $e^{2 \log x} = x^{2}$ hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Theorem 5eˣ aur log x ke derivatives

(1) $\frac{d}{d x} e^{x} = e^{x}$ . (2) $x > 0$ ke liye $\frac{d}{d x} \log x = \frac{1}{x}$ .

Part (1) hum prove nahi karte. Yeh kehta hai ki har point par $y = e^{x}$ ka slope uski height ke barabar hai, aur $e$ hi akela base hai jiski yeh property hai. Phir part (2) implicit differentiation se nikalta hai, yeh maan kar ki $\log x$ differentiable hai: agar $y = \log x$ to $x = e^{y}$ , isliye $1 = e^{y} \frac{d y}{d x}$ aur $\frac{d y}{d x} = \frac{1}{e^{y}} = \frac{1}{x}$ .

Example 26NCERT Example 26 · p. 129

Exponentials aur logarithms ke saath chain rule

Differentiate karo (i) $e^{x^{2}}$ (ii) $\log (1 + x^{2})$ (iii) $e^{\tan^{- 1} x}$ (iv) $\log (\sec x)$ , $- \frac{π}{2} < x < \frac{π}{2}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\log_{2} 32

nikaalo.

Answer dekho

Answer 5

2^{5} = 32

, isliye

\log_{2} 32 = 5

Practice · 2 questions

Practice 5.4: answers ke saath

NCERT Exercise 5.4 ki skills par naye questions.

1
Short answerMediumNCERT Exercise 5.4, Q1–Q5
$x$ ke respect mein differentiate karo.
1. $\frac{e^{x}}{1 + x^{2}}$
  Solution dikhao
  Answer $\frac{e^{x} (x - 1)^{2}}{(1 + x^{2})^{2}}$
  Quotient rule: $\frac{e^{x} (1 + x^{2}) - e^{x} \cdot 2 x}{(1 + x^{2})^{2}} = \frac{e^{x} (x^{2} - 2 x + 1)}{(1 + x^{2})^{2}}$ .
2. $e^{\cos^{- 1} x}$ , $- 1 < x < 1$
  Solution dikhao
  Answer $- \frac{e^{\cos^{- 1} x}}{\sqrt{1 - x^{2}}}$
  $e^{\cos^{- 1} x} \cdot (- \frac{1}{\sqrt{1 - x^{2}}})$ .
3. $e^{\sqrt{x}}$ , $x > 0$
  Solution dikhao
  Answer $\frac{e^{\sqrt{x}}}{2 \sqrt{x}}$
  $e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ .
4. $\sin^{- 1} (e^{- 2 x})$ , $x > 0$
  Solution dikhao
  Answer $- \frac{2 e^{- 2 x}}{\sqrt{1 - e^{- 4 x}}}$
  $\frac{1}{\sqrt{1 - (e^{- 2 x})^{2}}} \cdot (- 2 e^{- 2 x})$ . $x > 0$ ke liye $0 < e^{- 2 x} < 1$ , isliye formula lagta hai.
5. $\log (1 + \sin^{2} x)$
  Solution dikhao
  Answer $\frac{\sin 2 x}{1 + \sin^{2} x}$
  $\frac{1}{1 + \sin^{2} x} \cdot 2 \sin x \cos x = \frac{\sin 2 x}{1 + \sin^{2} x}$ .
2
Short answerMediumNCERT Exercise 5.4, Q6–Q10
$x$ ke respect mein differentiate karo.
1. $e^{x} + e^{2 x^{2}} + e^{3 x^{3}}$
  Solution dikhao
  Answer $e^{x} + 4 x e^{2 x^{2}} + 9 x^{2} e^{3 x^{3}}$
  Har term alag se: $e^{x}$ ; $e^{2 x^{2}} \cdot 4 x$ ; $e^{3 x^{3}} \cdot 9 x^{2}$ .
2. $\sqrt[3]{e^{x^{2}}}$
  Solution dikhao
  Answer $\frac{2 x}{3} e^{x^{2} / 3}$
  $\sqrt[3]{e^{x^{2}}} = e^{x^{2} / 3}$ , jiska derivative $e^{x^{2} / 3} \cdot \frac{2 x}{3}$ hai.
3. $\log (1 + \log x)$ , $x > \frac{1}{e}$
  Solution dikhao
  Answer $\frac{1}{x (1 + \log x)}$
  $x > \frac{1}{e}$ ke liye $1 + \log x > 0$ . Derivative $\frac{1}{1 + \log x} \cdot \frac{1}{x}$ hai.
4. $\frac{\log x}{x}$ , $x > 0$
  Solution dikhao
  Answer $\frac{1 - \log x}{x^{2}}$
  Quotient rule: $\frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^{2}}$ .
5. $\sin (x^{2} + \log x)$ , $x > 0$
  Solution dikhao
  Answer $(2 x + \frac{1}{x}) \cos (x^{2} + \log x)$
  Bahar $\sin t$ , andar $t = x^{2} + \log x$ , jahan $t^{'} = 2 x + \frac{1}{x}$ .

5.5 Logarithmic Differentiation

Tum $y = x^{x}$ ko kaise differentiate karoge? Power rule ko fixed exponent chahiye aur $a^{x}$ ke rule ko fixed base, lekin yahan dono badalte hain. Logarithm lene se mushkil exponent ek product ban jaata hai, jise product rule sambhaal leta hai. Yahi trick ek lambe product ya quotient ko sum mein badal deti hai. General case $y = [u (x)]^{v (x)}$ hai, jahan $u (x) > 0$ .

Derivationu to the power v ko differentiate karna

$y = u^{v}$ ka natural logarithm lo (dono sides positive hain):

\log y = v \log u

Dono sides ko $x$ ke respect mein differentiate karo, left par chain rule aur right par product rule lagakar:

\frac{1}{y} \frac{d y}{d x} = v^{'} \log u + v \cdot \frac{u^{'}}{u}

$y = u^{v}$ se multiply karo:

\frac{d y}{d x} = u^{v} [\frac{v}{u} u^{'} + v^{'} \log u]

Yeh logarithmic differentiation hai. Iski ek condition zaroori hai: $u (x)$ , aur isliye $y$ , positive hona chahiye, warna logarithms ka koi matlab nahi. Jo product ya quotient negative ho sakta hai, uske liye method ko $∣ y ∣$ par lagao: kyunki jahan bhi $y \neq 0$ , $\frac{d}{d x} \log ∣ y ∣ = \frac{y^{'}}{y}$ , tumhe wahi formula milta hai. $y = x^{x}$ ( $x > 0$ ) ke liye formula $\frac{d y}{d x} = x^{x} (1 + \log x)$ deta hai, aur Fig. 5.13 dikhata hai ki iska graph ke baare mein kya matlab hai.

Definition

Logarithmic differentiation

$y = f (x)$ ki dono sides ka pehle logarithm lekar, $\log y$ ko differentiate karke, aur phir $y$ se multiply karke $\frac{d y}{d x}$ nikaalna; variable powers aur lambe products aur quotients ke liye use hota hai.

3/3Slope wahan 0 hai jahan $\log x = - 1$ , yaani $x = \frac{1}{e} \approx 0.37$ , $y \approx 0.69$ par.

Fig. 5.13

y = x^{x}

aur uska derivative

x^{x} (1 + \log x)

.NCERT p. 130

Axes par x 0 se 2 tak aur y 0 se 4 tak hai. Step 1 y = x^x banata hai: yeh 0 ke theek right mein height 1 ke paas shuru hota hai, ek low point tak neeche jaata hai aur phir x = 2 par tezi se 4 tak chadhta hai. Step 2 (1, 1) par tangent jodta hai, jiska slope 1 hai. Step 3 x = 1/e, lagbhag 0.37, par sabse neeche ka point mark karta hai, jahan y lagbhag 0.69 hai, flat tangent ke saath, kyunki wahan 1 + log x = 0.

Example 27NCERT Example 27 · p. 131

Ek lamba quotient

$y = \frac{(x + 1)^{2} \sqrt{x - 2}}{(x + 4)^{3}}$ , $x > 2$ ko differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 28NCERT Example 28 · p. 131

7ˣ ka derivative, do tareekon se

$y = 7^{x}$ ko differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 29NCERT Example 29 · p. 132

Base bhi variable, exponent bhi variable

$y = x^{\sqrt{x}}$ , $x > 0$ ko differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 30NCERT Example 30 · p. 132

Implicit relation ke andar ek variable power

Agar $x^{y} + y^{2} = 5 x$ hai, jahan $x > 0$ aur $y > 0$ , to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

\frac{d}{d x} x^{x} =

(

x > 0

ke liye)

x \cdot x^{x - 1}

x^{x} \log x

x^{x}

x^{x} (1 + \log x)

Answer dekho

Answer (D) $x^{x} (1 + \log x)$

Logarithmic differentiation

x^{x} (1 + \log x)

deta hai. Answer

x \cdot x^{x - 1}

power rule aise use karta hai jaise exponent fixed ho, aur

x^{x} \log x

a^{x}

ka rule aise use karta hai jaise base fixed ho; sahi answer in dono ka sum hai.

Practice · 6 questions

Practice 5.5: answers ke saath

NCERT Exercise 5.5 ki skills par naye questions.

1
Short answerMediumNCERT Exercise 5.5, Q1–Q6
$x$ ke respect mein differentiate karo.
1. $x^{2} e^{x} \sin x$ , $0 < x < π$
  Solution dikhao
  Answer $2 x e^{x} \sin x + x^{2} e^{x} \sin x + x^{2} e^{x} \cos x$
  $\log y = 2 \log x + x + \log \sin x$ , isliye $\frac{y^{'}}{y} = \frac{2}{x} + 1 + \cot x$ .
2. $\sqrt[3]{\frac{(x + 1) (x + 2)}{(x + 3) (x + 4)}}$ , $x > - 1$
  Solution dikhao
  Answer $\frac{y}{3}$ times $\frac{1}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3} - \frac{1}{x + 4}$ , jahan $y$ diya gaya function hai
  $\log y = \frac{1}{3} [\log (x + 1) + \log (x + 2) - \log (x + 3) - \log (x + 4)]$ ; differentiate karo aur $y$ se multiply karo.
3. $(\tan x)^{x}$ , $0 < x < \frac{π}{2}$
  Solution dikhao
  Answer $(\tan x)^{x} \log \tan x + \frac{2 x (\tan x)^{x}}{\sin 2 x}$
  $\log y = x \log \tan x$ ; $\frac{y^{'}}{y} = \log \tan x + x \frac{\sec^{2} x}{\tan x}$ , aur $\frac{\sec^{2} x}{\tan x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2 x}$ .
4. $x^{2 x} + 3^{\cos x}$ , $x > 0$
  Solution dikhao
  Answer $2 x^{2 x} (1 + \log x) - 3^{\cos x} \sin x \log 3$
  $x^{2 x}$ ke liye: $\log u = 2 x \log x$ , $u^{'} = u (2 \log x + 2)$ . $3^{\cos x}$ ke liye: $3^{\cos x} \log 3 \cdot (- \sin x)$ .
5. $(2 x + 1)^{3} (x - 4)^{2} (x + 1)^{5}$
  Solution dikhao
  Answer $y$ times $\frac{6}{2 x + 1} + \frac{2}{x - 4} + \frac{5}{x + 1}$ , jahan $y$ diya gaya product hai
  $\log ∣ y ∣ = 3 \log ∣ 2 x + 1 ∣ + 2 \log ∣ x - 4 ∣ + 5 \log ∣ x + 1 ∣$ ; differentiate karo (jahan bhi $y \neq 0$ ) aur $y$ se multiply karo.
6. $x^{1 / x} + (x + 1)^{x}$ , $x > 0$
  Solution dikhao
  Answer $x^{1 / x} \frac{1 - \log x}{x^{2}} + (x + 1)^{x} \log (x + 1) + x (x + 1)^{x - 1}$
  $\log u = \frac{\log x}{x}$ se $\frac{u^{'}}{u} = \frac{1 - \log x}{x^{2}}$ milta hai; $\log v = x \log (x + 1)$ se $\frac{v^{'}}{v} = \log (x + 1) + \frac{x}{x + 1}$ milta hai.
2
Short answerHardNCERT Exercise 5.5, Q7–Q11
$x$ ke respect mein differentiate karo.
1. $x^{e^{x}} + (e^{x})^{x}$ , $x > 0$
  Solution dikhao
  Answer $x^{e^{x}} e^{x} (\log x + \frac{1}{x}) + 2 x e^{x^{2}}$
  $\log u = e^{x} \log x$ , isliye $\frac{u^{'}}{u} = e^{x} \log x + \frac{e^{x}}{x}$ . Aur $(e^{x})^{x} = e^{x^{2}}$ , jiska derivative $2 x e^{x^{2}}$ hai.
2. $(1 + x^{2})^{x} + \tan^{- 1} \sqrt{x}$ , $x > 0$
  Solution dikhao
  Answer $(1 + x^{2})^{x} \log (1 + x^{2}) + 2 x^{2} (1 + x^{2})^{x - 1} + \frac{1}{2 \sqrt{x} (1 + x)}$
  $\log u = x \log (1 + x^{2})$ se $\frac{u^{'}}{u} = \log (1 + x^{2}) + \frac{2 x^{2}}{1 + x^{2}}$ milta hai. Aur $\frac{d}{d x} \tan^{- 1} \sqrt{x} = \frac{1}{1 + x} \cdot \frac{1}{2 \sqrt{x}}$ .
3. $x^{\tan x} + (\cos x)^{x^{2}}$ , $0 < x < \frac{π}{2}$
  Solution dikhao
  Answer $x^{\tan x} \sec^{2} x \log x + x^{\tan x - 1} \tan x + 2 x (\cos x)^{x^{2}} \cdot \log \cos x - x^{2} (\cos x)^{x^{2}} \cdot \tan x$
  $\log u = \tan x \log x$ aur $\log v = x^{2} \log \cos x$ ; har ek ko product rule se differentiate karo.
4. $x^{x^{2}} + \frac{1 + x}{1 - x}$ , $0 < x < 1$
  Solution dikhao
  Answer $x^{x^{2}} x (1 + 2 \log x) + \frac{2}{(1 - x)^{2}}$
  $\log u = x^{2} \log x$ se $\frac{u^{'}}{u} = 2 x \log x + x$ milta hai. Quotient rule $\frac{(1 - x) + (1 + x)}{(1 - x)^{2}}$ deta hai.
5. $(x e^{x})^{x} + (\log x)^{1 / x}$ , $x > 1$
  Solution dikhao
  Answer $(x e^{x})^{x} (\log x + 1 + 2 x) + \frac{(\log x)^{1 / x}}{x^{2} \log x} - \frac{(\log x)^{1 / x} \log (\log x)}{x^{2}}$
  $\log u = x \log x + x^{2}$ , isliye $\frac{u^{'}}{u} = \log x + 1 + 2 x$ . $\log v = \frac{\log (\log x)}{x}$ , isliye $\frac{v^{'}}{v} = \frac{1}{x^{2} \log x} - \frac{\log (\log x)}{x^{2}}$ .
3
Short answerHardNCERT Exercise 5.5, Q12–Q15
$\frac{d y}{d x}$ nikaalo (saare variables positive hain).
1. $x^{y} \cdot y^{x} = 1$
  Solution dikhao
  Answer $- \frac{\frac{y}{x} + \log y}{\log x + \frac{x}{y}}$
  $y \log x + x \log y = 0$ . Differentiate karne par: $y^{'} \log x + \frac{y}{x} + \log y + \frac{x}{y} y^{'} = 0$ .
2. $y = x^{y}$
  Solution dikhao
  Answer $\frac{y^{2}}{x (1 - y \log x)}$
  $\log y = y \log x$ , isliye $\frac{y^{'}}{y} = y^{'} \log x + \frac{y}{x}$ , jisse $y^{'} (\frac{1}{y} - \log x) = \frac{y}{x}$ milta hai.
3. $(1 + x)^{y} = (1 + y)^{x}$
  Solution dikhao
  Answer $\frac{\log (1 + y) - \frac{y}{1 + x}}{\log (1 + x) - \frac{x}{1 + y}}$
  $y \log (1 + x) = x \log (1 + y)$ . Differentiate karne par: $y^{'} \log (1 + x) + \frac{y}{1 + x} = \log (1 + y) + \frac{x}{1 + y} y^{'}$ .
4. $x^{2} y = e^{x + y}$
  Solution dikhao
  Answer $\frac{y (x - 2)}{x (1 - y)}$
  $2 \log x + \log y = x + y$ . Differentiate karne par: $\frac{2}{x} + \frac{y^{'}}{y} = 1 + y^{'}$ , isliye $y^{'} (\frac{1}{y} - 1) = 1 - \frac{2}{x}$ .
Short answerMediumNCERT Exercise 5.5, Q16
Maan lo $f (x) = (1 + x) (1 + 2 x) (1 + 3 x) (1 + 4 x)$ . Logarithmic differentiation se $f^{'} (x)$ nikaalo, aur phir $f^{'} (1)$ nikaalo.
Solution dikhao
Jin $x$ ke liye har factor positive hai, wahan $\log f = \sum \log (1 + k x)$ , isliye $f^{'} (x)$ , $f (x)$ times $\frac{1}{1 + x} + \frac{2}{1 + 2 x} + \frac{3}{1 + 3 x} + \frac{4}{1 + 4 x}$ hai.
$x = 1$ par: $f (1) = 2 \cdot 3 \cdot 4 \cdot 5 = 120$ aur bracket $\frac{30 + 40 + 45 + 48}{60} = \frac{163}{60}$ hai. Isliye $f^{'} (1) = 120 \cdot \frac{163}{60} = 326$ .
Short answerMediumNCERT Exercise 5.5, Q17
$(x^{2} + 2 x - 1) (x^{3} - 4 x + 6)$ ko teen tareekon se differentiate karo: (i) product rule se, (ii) pehle expand karke, (iii) logarithmic differentiation se. Kya answers match karte hain?
Solution dikhao
AnswerTeeno $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ dete hain.
(i) $(2 x + 2) (x^{3} - 4 x + 6) + (x^{2} + 2 x - 1) (3 x^{2} - 4)$ . Har product expand karne par: $2 x^{4} + 2 x^{3} - 8 x^{2} + 4 x + 12$ aur $3 x^{4} + 6 x^{3} - 7 x^{2} - 8 x + 4$ , jinka sum $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ hai.
(ii) Product $x^{5} + 2 x^{4} - 5 x^{3} - 2 x^{2} + 16 x - 6$ hai, jiska derivative $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ hai.
(iii) Hint ko $y$ se multiply karne par theek (i) wala expression wapas milta hai. Isliye teeno match karte hain.
Short answerHardNCERT Exercise 5.5, Q18
Agar $u$ , $v$ aur $w$ , $x$ ke differentiable functions hain aur $w \neq 0$ , to do tareekon se dikhao ki $\frac{d}{d x} (\frac{u v}{w}) = \frac{u v}{w} (\frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w})$ (jahan $u, v \neq 0$ ): product aur quotient rules se, aur logarithmic differentiation se.
Solution dikhao
AnswerDono tareeke $\frac{u^{'} v w + u v^{'} w - u v w^{'}}{w^{2}}$ dete hain.
Rules: $\frac{(u v)^{'} w - u v w^{'}}{w^{2}} = \frac{u^{'} v w + u v^{'} w - u v w^{'}}{w^{2}}$ . $\frac{u v}{w}$ ko factor out karo: yeh $\frac{u v}{w} (\frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w})$ hai.
Logarithms: $y = \frac{u v}{w}$ ke saath $\log ∣ y ∣ = \log ∣ u ∣ + \log ∣ v ∣ - \log ∣ w ∣$ , isliye $\frac{y^{'}}{y} = \frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w}$ , aur $y$ se multiply karne par wahi result milta hai.

5.6 Parametric Forms mein Functions ke Derivatives

Ek batter cricket ball ko upar uchhaalta hai. Uski horizontal distance $x$ aur height $y$ , dono ko ek doosre ke terms ki jagah time $t$ ke terms mein describe karna aasaan hai. Time natural link hai: $x$ aur $y$ dono $t$ mein apne-apne formula ke roop mein aate hain.

Definition

Parameter

Agar ek teesre variable $t$ ke liye $x = f (t)$ aur $y = g (t)$ , to equations ka yeh pair $x$ aur $y$ ke beech relation ko parametric form mein deta hai, aur $t$ ko parameter kehte hain.

Example $x = 3 \cos t$ , $y = 2 \sin t$ parameter $t$ ke saath ek ellipse describe karta hai.

$\frac{d y}{d x}$ nikaalne ke liye $y$ ko $x$ ka function maan kar chain rule lagao, jahan $x$ khud $t$ ka function hai: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ . Divide karo, agar $\frac{d x}{d t} \neq 0$ :

Parametric form mein derivative

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{g^{'} (t)}{f^{'} (t)}, f^{'} (t) \neq 0

Shabdon mein: path ka slope vertical speed divided by horizontal speed hai. Fig. 5.14 ise ek ball ke liye dikhata hai jiske liye $x = 20 t$ aur $y = 20 t - 5 t^{2}$ (metres, seconds, air resistance ignore kiya, $g$ ko $10 m / s^{2}$ round kiya).

1/3 $\frac{d x}{d t} = 20$ , $\frac{d y}{d t} = 10$ : slope $\frac{1}{2}$ .

Fig. 5.14

x = 20 t

y = 20 t - 5 t^{2}

: slope matlab vertical speed ÷ horizontal speed.NCERT p. 135

Axes par horizontal distance 0 se 80 metres aur height 0 se 25 metres tak hai. Path ek symmetric arch hai jo (0, 0) se upar (40, 20) tak aur neeche (80, 0) tak jaata hai. Step 1 t = 1 s par ball ko (20, 15) par mark karta hai, 20 m/s ke horizontal speed arrow aur 10 m/s ke upward arrow ke saath, aur slope 10/20 = 1/2 wali tangent. Step 2 t = 2 s par ball ko top (40, 20) par mark karta hai, jahan vertical speed 0 hai aur tangent flat hai. Step 3 t = 3 s par ball ko (60, 15) par mark karta hai, 10 m/s ke downward arrow aur slope −1/2 wali tangent ke saath.

Example 31NCERT Example 31 · p. 135

Ek ellipse

Agar $x = 3 \cos t$ , $y = 2 \sin t$ hai to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 32NCERT Example 32 · p. 135

Cricket ball

$x = 20 t$ , $y = 20 t - 5 t^{2}$ wali ball ke liye $\frac{d y}{d x}$ nikaalo. Path kahan level hai, aur ball kis angle par girti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 33NCERT Example 33 · p. 136

Loop banata curve

Agar $x = 2 \cos t - \cos 2 t$ aur $y = 2 \sin t - \sin 2 t$ hai to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Dhyaan do ki upar ke answers sirf parameter mein likhe hain, pehle $x$ ya $y$ nikaalne ki zaroorat nahi padi. Parameter ko jaan-boojh kar bhi laaya ja sakta hai taaki implicit relation ko kaabu kiya ja sake, jaisa agla example dikhata hai.

Example 34NCERT Example 34 · p. 136

Parameter chunna

Agar $\sqrt{x} + \sqrt{y} = \sqrt{a}$ hai, jahan $a > 0$ aur $x, y > 0$ , to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Agar

x = t^{2}

aur

y = t^{3}

hai to

t = 2

par

\frac{d y}{d x}

nikaalo.

Answer dekho

Answer 3

t = 2

par

\frac{d y}{d x} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2} = 3

Practice · 3 questions

Practice 5.6: answers ke saath

NCERT Exercise 5.6 ki skills par naye questions.

1
Short answerMediumNCERT Exercise 5.6, Q1–Q5
Parameter ko eliminate kiye bina $\frac{d y}{d x}$ nikaalo.
1. $x = 3 t^{2}$ , $y = 2 t^{3}$
  Solution dikhao
  Answer $t$
  $t \neq 0$ ke liye $\frac{6 t^{2}}{6 t} = t$ .
2. $x = 2 \sin θ$ , $y = 5 \sin θ$
  Solution dikhao
  Answer $\frac{5}{2}$
  Jahan $\cos θ \neq 0$ , $\frac{5 \cos θ}{2 \cos θ} = \frac{5}{2}$ . (Points line $y = \frac{5}{2} x$ par hain.)
3. $x = \tan t$ , $y = \cos 2 t$
  Solution dikhao
  Answer $- 2 \sin 2 t \cos^{2} t$
  $\frac{- 2 \sin 2 t}{\sec^{2} t} = - 2 \sin 2 t \cos^{2} t$ .
4. $x = t + \frac{1}{t}$ , $y = t - \frac{1}{t}$
  Solution dikhao
  Answer $\frac{t^{2} + 1}{t^{2} - 1}$
  $t \neq \pm 1$ ke liye $\frac{1 + 1 / t^{2}}{1 - 1 / t^{2}} = \frac{t^{2} + 1}{t^{2} - 1}$ .
5. $x = 2 \cos θ + \cos 2 θ$ , $y = 2 \sin θ - \sin 2 θ$
  Solution dikhao
  Answer $- \tan \frac{θ}{2}$
  $\frac{d y}{d θ} = 2 (\cos θ - \cos 2 θ) = 4 \sin \frac{3 θ}{2} \sin \frac{θ}{2}$ aur $\frac{d x}{d θ} = - 2 (\sin θ + \sin 2 θ) = - 4 \sin \frac{3 θ}{2} \cos \frac{θ}{2}$ . Quotient $- \tan \frac{θ}{2}$ hai.
2
Short answerMediumNCERT Exercise 5.6, Q6–Q10
Parameter ko eliminate kiye bina $\frac{d y}{d x}$ nikaalo.
1. $x = 2 θ - \sin 2 θ$ , $y = 1 - \cos 2 θ$
  Solution dikhao
  Answer $\cot θ$
  $\frac{2 \sin 2 θ}{2 - 2 \cos 2 θ} = \frac{2 \sin θ \cos θ}{2 \sin^{2} θ} = \cot θ$ .
2. $x = \frac{1 - t^{2}}{1 + t^{2}}$ , $y = \frac{2 t}{1 + t^{2}}$
  Solution dikhao
  Answer $\frac{t^{2} - 1}{2 t}$
  $\frac{d x}{d t} = \frac{- 4 t}{(1 + t^{2})^{2}}$ , $\frac{d y}{d t} = \frac{2 (1 - t^{2})}{(1 + t^{2})^{2}}$ . Quotient $\frac{t^{2} - 1}{2 t}$ hai, jo $- \frac{x}{y}$ ke barabar hai: yeh points unit circle par hain.
3. $x = \log (\sec t + \tan t)$ , $y = \tan t$ , $0 < t < \frac{π}{2}$
  Solution dikhao
  Answer $\sec t$
  $\frac{d x}{d t} = \frac{\sec t \tan t + \sec^{2} t}{\sec t + \tan t} = \sec t$ aur $\frac{d y}{d t} = \sec^{2} t$ , isliye $\frac{d y}{d x} = \sec t$ .
4. $x = 2 \tan θ$ , $y = 3 \sec θ$
  Solution dikhao
  Answer $\frac{3}{2} \sin θ$
  $\frac{3 \sec θ \tan θ}{2 \sec^{2} θ} = \frac{3}{2} \cdot \frac{\tan θ}{\sec θ} = \frac{3}{2} \sin θ$ .
5. $x = e^{θ} (\sin θ + \cos θ)$ , $y = e^{θ} (\sin θ - \cos θ)$
  Solution dikhao
  Answer $\tan θ$
  $\frac{d x}{d θ} = e^{θ} (\sin θ + \cos θ + \cos θ - \sin θ) = 2 e^{θ} \cos θ$ aur $\frac{d y}{d θ} = 2 e^{θ} \sin θ$ .
Short answerMediumNCERT Exercise 5.6, Q11
Agar $x = e^{t} + e^{- t}$ aur $y = e^{t} - e^{- t}$ hai, to dikhao ki $\frac{d y}{d x} = \frac{x}{y}$ .
Solution dikhao
Answer $\frac{d x}{d t} = y$ aur $\frac{d y}{d t} = x$ , isliye $\frac{d y}{d x} = \frac{x}{y}$ ( $t \neq 0$ ke liye).
$\frac{d x}{d t} = e^{t} - e^{- t} = y$ aur $\frac{d y}{d t} = e^{t} + e^{- t} = x$ . Divide karne par, jahan bhi $y \neq 0$ , $\frac{d y}{d x} = \frac{x}{y}$ . (Points $x^{2} - y^{2} = 4$ satisfy karte hain, aur implicit differentiation bhi yahi slope deta hai.)

5.7 Second Order Derivative

$y = x^{4}$ ko differentiate karne par $4 x^{3}$ milta hai, jo ek naya function hai, aur use bhi differentiate karne se hamein kuch nahi rokta: result $12 x^{2}$ hai. Is operation ko aise dohraana itna common hai ki iska ek naam aur notation hona chahiye.

Definition

Second order derivative

Agar $y = f (x)$ aur $f^{'}$ differentiable hai, to $\frac{d}{d x} (\frac{d y}{d x})$ , $x$ ke respect mein $y$ ka second order derivative hai, jise $\frac{d^{2} y}{d x^{2}}$ , $f^{''} (x)$ , $D^{2} y$ , $y^{''}$ ya $y_{2}$ likhte hain.

Example $y = x^{4}$ : $\frac{d y}{d x} = 4 x^{3}$ aur $\frac{d^{2} y}{d x^{2}} = 12 x^{2}$ .

Isi notation mein first derivative ko aksar $y_{1}$ likhte hain, aur third aur usse upar ke derivatives bhi isi tarah define hote hain. Do tareeke $y^{''}$ ko concrete banate hain. Agar $s (t)$ position hai, to $\frac{d s}{d t}$ velocity hai aur $\frac{d^{2} s}{d t^{2}}$ acceleration, yaani velocity ke badalne ki rate. Graph par $y^{''}$ slope ke badalne ki rate hai, isliye iska sign batata hai ki curve kis taraf mudta hai (Fig. 5.15).

3/3 $y^{''} = 6 x$ : $x < 0$ ke liye negative (neeche mudta hai), $x > 0$ ke liye positive (upar mudta hai).

Fig. 5.15

y^{''}

slope ka slope hai.NCERT p. 137

Upar wala panel x −2.2 se 2.2 tak y = x³ − 3x dikhata hai, jo (−1, 2) par peak tak chadhta hai, origin se hote hue girta hai aur (1, −2) par valley tak pahunchta hai. Step 1 x = −1, 0 aur 1 par slopes 0, −3 aur 0 wali chhoti tangents banata hai. Step 2 neeche wala panel banata hai, slope graph y′ = 3x² − 3, ek U-shaped curve jo (−1, 0), (0, −3) aur (1, 0) se guzarta hai, matching tangents tak dotted links ke saath. Step 3 upar wale panel ke left half ko neeche mudta hua shade karta hai, jahan y″ = 6x negative hai, aur right half ko upar mudta hua, jahan y″ positive hai.

Example 35NCERT Example 35 · p. 138

Do baar differentiate karna

Agar $y = x^{4} - \sec x$ hai to $\frac{d^{2} y}{d x^{2}}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 36NCERT Example 36 · p. 138

Aisa function jise uska second derivative ulta kar deta hai

Agar $y = A \cos 3 x + B \sin 3 x$ ( $A$ , $B$ constants), to dikhao ki $\frac{d^{2} y}{d x^{2}} + 9 y = 0$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 37NCERT Example 37 · p. 138

Do exponentials

Agar $y = 2 e^{x} - e^{4 x}$ , to dikhao ki $\frac{d^{2} y}{d x^{2}} - 5 \frac{d y}{d x} + 4 y = 0$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 38NCERT Example 38 · p. 139

Inverse sine ka square

Agar $y = (\sin^{- 1} x)^{2}$ , $- 1 < x < 1$ , to dikhao ki $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Agar

y = x^{5}

hai to

x = 1

par

\frac{d^{2} y}{d x^{2}}

nikaalo.

Answer dekho

Answer 20

y^{'} = 5 x^{4}

aur

y^{''} = 20 x^{3}

, jo

x = 1

par 20 hai.

Practice · 4 questions

Practice 5.7: answers ke saath

NCERT Exercise 5.7 ki skills par naye questions.

1
Short answerEasyNCERT Exercise 5.7, Q1–Q5
Har function ka second order derivative nikaalo.
1. $4 x^{3} - 5 x + 7$
  Solution dikhao
  Answer $24 x$
  $y^{'} = 12 x^{2} - 5$ , $y^{''} = 24 x$ .
2. $x^{15}$
  Solution dikhao
  Answer $210 x^{13}$
  $y^{'} = 15 x^{14}$ , $y^{''} = 210 x^{13}$ .
3. $x^{2} \sin x$
  Solution dikhao
  Answer $2 \sin x + 4 x \cos x - x^{2} \sin x$
  $y^{'} = 2 x \sin x + x^{2} \cos x$ ; $y^{''} = 2 \sin x + 2 x \cos x + 2 x \cos x - x^{2} \sin x$ .
4. $\sqrt{x}$ , $x > 0$
  Solution dikhao
  Answer $- \frac{1}{4 x^{3 / 2}}$
  $y^{'} = \frac{1}{2} x^{- 1 / 2}$ , $y^{''} = - \frac{1}{4} x^{- 3 / 2}$ .
5. $x^{2} \log x$ , $x > 0$
  Solution dikhao
  Answer $2 \log x + 3$
  $y^{'} = 2 x \log x + x$ ; $y^{''} = 2 \log x + 2 + 1$ .
2
Short answerMediumNCERT Exercise 5.7, Q6–Q10
Har function ka second order derivative nikaalo.
1. $e^{2 x} \sin x$
  Solution dikhao
  Answer $e^{2 x} (3 \sin x + 4 \cos x)$
  $y^{'} = e^{2 x} (2 \sin x + \cos x)$ ; $y^{''} = e^{2 x} [2 (2 \sin x + \cos x) + (2 \cos x - \sin x)]$ .
2. $e^{- x} \cos 2 x$
  Solution dikhao
  Answer $e^{- x} (4 \sin 2 x - 3 \cos 2 x)$
  $y^{'} = e^{- x} (- \cos 2 x - 2 \sin 2 x)$ ; $y^{''} = e^{- x} [(\cos 2 x + 2 \sin 2 x) + (2 \sin 2 x - 4 \cos 2 x)]$ .
3. $\sin^{- 1} x$ , $- 1 < x < 1$
  Solution dikhao
  Answer $\frac{x}{(1 - x^{2})^{3 / 2}}$
  $y^{'} = (1 - x^{2})^{- 1 / 2}$ ; $y^{''} = - \frac{1}{2} (1 - x^{2})^{- 3 / 2} (- 2 x)$ .
4. $\log (1 + x^{2})$
  Solution dikhao
  Answer $\frac{2 (1 - x^{2})}{(1 + x^{2})^{2}}$
  $y^{'} = \frac{2 x}{1 + x^{2}}$ ; quotient rule: $\frac{2 (1 + x^{2}) - 2 x \cdot 2 x}{(1 + x^{2})^{2}}$ .
5. $\log (\sin x)$ , $0 < x < π$
  Solution dikhao
  Answer $- {cosec}^{2} x$
  $y^{'} = \frac{\cos x}{\sin x} = \cot x$ ; $y^{''} = - {cosec}^{2} x$ .
3
Short answerHardNCERT Exercise 5.7, Q11–Q14
Har identity prove karo.
1. Agar $y = 4 \sin 2 x - 3 \cos 2 x$ , to dikhao ki $y^{''} + 4 y = 0$ .
  Solution dikhao
  Answer $y^{''} = - 4 y$ .
  $y^{'} = 8 \cos 2 x + 6 \sin 2 x$ aur $y^{''} = - 16 \sin 2 x + 12 \cos 2 x = - 4 (4 \sin 2 x - 3 \cos 2 x)$ .
2. Agar $y = \sin^{- 1} x$ , to $\frac{d^{2} y}{d x^{2}}$ ko sirf $y$ ke terms mein nikaalo.
  Solution dikhao
  Answer $\sec^{2} y \tan y$
  $x = \sin y$ se $y^{'} = \frac{1}{\cos y} = \sec y$ milta hai (cosine $(- \frac{π}{2}, \frac{π}{2})$ par positive hai). Phir $y^{''} = \frac{d}{d x} \sec y = \sec y \tan y \cdot y^{'} = \sec^{2} y \tan y$ .
3. Agar $y = x \cos (\log x)$ , $x > 0$ , to dikhao ki $x^{2} y_{2} - x y_{1} + 2 y = 0$ .
  Solution dikhao
  AnswerSubstitute karne par 0 milta hai.
  $y_{1} = \cos (\log x) - \sin (\log x)$ aur $y_{2} = - \frac{\sin (\log x) + \cos (\log x)}{x}$ . $s = \sin (\log x)$ aur $c = \cos (\log x)$ likho; tab
  $x^{2} y_{2} - x y_{1} + 2 y = (- x s - x c) - (x c - x s) + 2 x c = 0$
4. Agar $y = (A + B x) e^{k x}$ , to dikhao ki $y^{''} - 2 k y^{'} + k^{2} y = 0$ .
  Solution dikhao
  AnswerSubstitute karne par 0 milta hai.
  $y^{'} = B e^{k x} + k y$ aur $y^{''} = k B e^{k x} + k y^{'} = 2 k B e^{k x} + k^{2} y$ . Isliye $y^{''} - 2 k y^{'} + k^{2} y = 2 k B e^{k x} + k^{2} y - 2 k B e^{k x} - 2 k^{2} y + k^{2} y = 0$ .
4
Short answerHardNCERT Exercise 5.7, Q15–Q17
Har identity prove karo.
1. Agar $y = 4 e^{- 3 x} + 7 e^{3 x}$ , to dikhao ki $\frac{d^{2} y}{d x^{2}} = 9 y$ .
  Solution dikhao
  Answer $y^{''} = 36 e^{- 3 x} + 63 e^{3 x} = 9 y$ .
  $y^{'} = - 12 e^{- 3 x} + 21 e^{3 x}$ aur $y^{''} = 36 e^{- 3 x} + 63 e^{3 x} = 9 (4 e^{- 3 x} + 7 e^{3 x})$ .
2. Agar $y = \log (\cos x)$ , $- \frac{π}{2} < x < \frac{π}{2}$ , to dikhao ki $y^{''} + (y^{'})^{2} + 1 = 0$ .
  Solution dikhao
  Answer $y^{''} = - \sec^{2} x = - (1 + (y^{'})^{2})$ .
  $y^{'} = - \tan x$ aur $y^{''} = - \sec^{2} x = - (1 + \tan^{2} x) = - (1 + (y^{'})^{2})$ .
3. Agar $y = e^{\tan^{- 1} x}$ , to dikhao ki $(1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0$ .
  Solution dikhao
  Answer $(1 + x^{2}) y_{1} = y$ ko differentiate karo.
  $y_{1} = \frac{y}{1 + x^{2}}$ , isliye $(1 + x^{2}) y_{1} = y$ . Differentiate karne par $2 x y_{1} + (1 + x^{2}) y_{2} = y_{1}$ , jise rearrange karne par result milta hai.

Miscellaneous examples

Yeh examples chapter ki techniques ko milate hain. Do aadatein madad karti hain: differentiate karne se pehle batao ki function kahan defined hai, aur jab koi identity available ho to pehle simplify karo, yeh check karte hue ki woh poore zaroori interval par sach hai.

Example 39NCERT Example 39 · p. 140

Roots aur change of base

Differentiate karo (i) $y = \sqrt{5 - 2 x} + \frac{1}{\sqrt{x^{2} + 1}}$ , $x < \frac{5}{2}$ (ii) $y = \log_{3} (\sin x)$ , $0 < x < π$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 40NCERT Example 40 · p. 141

Pehle simplify karo, lekin interval ka dhyaan rakho

Differentiate karo (i) $y = \cos^{- 1} (\cos x)$ , $0 < x < 2 π$ , $x \neq π$ (ii) $y = \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}$ , $- 1 < x < 1$ (iii) $y = \cos^{- 1} \frac{1 - 9^{x}}{1 + 9^{x}}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/2 $[0, π]$ ke bahar graph mudta hai: slopes $1$ aur $- 1$ ke beech badalte rehte hain.

Fig. 5.16

y = \cos^{- 1} (\cos x)

sirf

[0, π]

par

x

ke barabar hai.NCERT p. 141

Axes par x −π se 3π tak aur y 0 se π tak hai. Step 1 line y = x ko dashed banata hai aur y = cos⁻¹(cos x) ka graph sirf x 0 se π tak, jahan dono ek ho jaate hain. Step 2 poora graph banata hai: seedhe segments ka zigzag jo 0 se π tak slope 1 se chadhta hai aur π se 2π tak slope −1 se girta hai, aur yeh repeat hota hai, x = −π, π aur 3π par height π ke peaks ke saath. Line y = x, x = π par zigzag se alag ho jaati hai.

Example 41NCERT Example 41 · p. 143

Compound-interest function

Agar $f (x) = {(1 + \frac{1}{x})}^{x}$ , $x > 0$ hai to $f^{'} (x)$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 42NCERT Example 42 · p. 143

Variable powers wala ek parametric pair

Agar $x = (t^{2} + 1)^{3}$ aur $y = 3^{t^{2}}$ hai to $\frac{d y}{d x}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 43NCERT Example 43 · p. 144

Ek function ka doosre ke respect mein derivative

$- \frac{π}{2} < x < \frac{π}{2}$ ke liye $\tan x$ ko $\sin x$ ke respect mein differentiate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Kis interval par

\cos^{- 1} (\cos x) = x

hai?

[- \frac{π}{2}, \frac{π}{2}]

[0, π]

[0, 2 π]

har real

x

Answer dekho

Answer (B) $[0, π]$

\cos^{- 1}

[0, π]

mein values deta hai, isliye identity sirf wahin sach ho sakti hai.

[- \frac{π}{2}, \frac{π}{2}]

\sin^{- 1}

ki range hai, yahi aam confusion hai.

Practice · 13 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

NCERT ki Chapter 5 Miscellaneous Exercise ki skills par naye questions, saath mein teen quick self-checks.

1
Short answerMediumNCERT Miscellaneous Exercise 5, Q1–Q6
$x$ ke respect mein differentiate karo.
1. $(2 x^{3} - x + 4)^{7}$
  Solution dikhao
  Answer $7 (2 x^{3} - x + 4)^{6} (6 x^{2} - 1)$
  Chain rule: $u = 2 x^{3} - x + 4$ ke saath $7 u^{6} \cdot u^{'}$ .
2. $\sin^{4} x + \cos^{4} x$
  Solution dikhao
  Answer $- \sin 4 x$
  Derivative $4 \sin^{3} x \cos x - 4 \cos^{3} x \sin x$ hai. Factorise karo: $- 4 \sin x \cos x \cdot (\cos^{2} x - \sin^{2} x)$ , jo $- 2 \sin 2 x \cos 2 x = - \sin 4 x$ hai.
3. $(x^{2})^{\sin 3 x}$ , $x > 0$
  Solution dikhao
  Answer $3 \cos 3 x \log (x^{2}) \cdot (x^{2})^{\sin 3 x} + \frac{2 \sin 3 x}{x} \cdot (x^{2})^{\sin 3 x}$
  $\log y = \sin 3 x \cdot 2 \log x$ , isliye $\frac{y^{'}}{y} = 3 \cos 3 x \cdot 2 \log x + \frac{2 \sin 3 x}{x}$ , aur $2 \log x = \log (x^{2})$ .
4. $\tan^{- 1} (x \sqrt{x})$ , $x \geq 0$
  Solution dikhao
  Answer $\frac{3 \sqrt{x}}{2 (1 + x^{3})}$
  $\frac{1}{1 + x^{3}} \cdot \frac{d}{d x} x^{3 / 2} = \frac{1}{1 + x^{3}} \cdot \frac{3}{2} x^{1 / 2}$ .
5. $\frac{\sin^{- 1} (x / 3)}{\sqrt{x + 4}}$ , $- 3 < x < 3$
  Solution dikhao
  Answer $\frac{1}{\sqrt{9 - x^{2}} \sqrt{x + 4}} - \frac{\sin^{- 1} (x / 3)}{2 (x + 4)^{3 / 2}}$
  $\frac{d}{d x} \sin^{- 1} \frac{x}{3} = \frac{1}{\sqrt{1 - x^{2} / 9}} \cdot \frac{1}{3} = \frac{1}{\sqrt{9 - x^{2}}}$ . Phir $v = (x + 4)^{1 / 2}$ , $v^{'} = \frac{1}{2} (x + 4)^{- 1 / 2}$ ke saath quotient rule.
6. $\tan^{- 1} \frac{\cos x}{1 + \sin x}$ , $- \frac{π}{2} < x < \frac{π}{2}$
  Solution dikhao
  Answer $- \frac{1}{2}$
  $\frac{\cos x}{1 + \sin x} = \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2})^{2}} = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \tan (\frac{π}{4} - \frac{x}{2})$ , jahan $\frac{π}{4} - \frac{x}{2} \in (0, \frac{π}{2})$ . Isliye $y = \frac{π}{4} - \frac{x}{2}$ .
2
Short answerHardNCERT Miscellaneous Exercise 5, Q7–Q11
$x$ ke respect mein differentiate karo.
1. $(\sin x)^{\log x}$ , $0 < x < π$
  Solution dikhao
  Answer $\frac{\log (\sin x)}{x} \cdot (\sin x)^{\log x} + \log x \cot x \cdot (\sin x)^{\log x}$
  $\log y = \log x \cdot \log (\sin x)$ ; product rule: $\frac{y^{'}}{y} = \frac{\log \sin x}{x} + \log x \cdot \frac{\cos x}{\sin x}$ .
2. $e^{a \sin x + b \cos x}$ ( $a$ , $b$ constants)
  Solution dikhao
  Answer $(a \cos x - b \sin x) e^{a \sin x + b \cos x}$
  Andar ke derivative $a \cos x - b \sin x$ ke saath chain rule.
3. $(x^{2} + 1)^{x^{2} + 1}$
  Solution dikhao
  Answer $2 x (x^{2} + 1)^{x^{2} + 1} \cdot [\log (x^{2} + 1) + 1]$
  $\log y = (x^{2} + 1) \log (x^{2} + 1)$ , isliye $\frac{y^{'}}{y} = 2 x \log (x^{2} + 1) + (x^{2} + 1) \cdot \frac{2 x}{x^{2} + 1}$ .
4. $x^{x} - x^{e} + e^{x} - e^{e}$ , $x > 0$
  Solution dikhao
  Answer $x^{x} (1 + \log x) - e x^{e - 1} + e^{x}$
  Chaar alag rules: variable base aur exponent; ek constant power ( $e$ ek number hai); natural exponential; ek constant.
5. $x^{x^{2} + 1} + (x - 1)^{x}$ , $x > 1$
  Solution dikhao
  Answer $2 x \log x \cdot x^{x^{2} + 1} + (x^{2} + 1) x^{x^{2}} + (x - 1)^{x} \log (x - 1) + x (x - 1)^{x - 1}$
  $\log u = (x^{2} + 1) \log x$ aur $\log v = x \log (x - 1)$ ; har ek ko differentiate karo aur wapas multiply karo.
3
Short answerMediumNCERT Miscellaneous Exercise 5, Q12–Q14
Parametric aur implicit slopes.
1. Agar $x = 6 (t + \sin t)$ , $y = 4 (1 - \cos t)$ , $- π < t < π$ hai to $\frac{d y}{d x}$ nikaalo.
  Solution dikhao
  Answer $\frac{2}{3} \tan \frac{t}{2}$
  $\frac{4 \sin t}{6 (1 + \cos t)} = \frac{2}{3} \cdot \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos^{2} \frac{t}{2}}$ .
2. Agar $y = \tan^{- 1} x + \tan^{- 1} \frac{1}{x}$ , $x > 0$ hai to $\frac{d y}{d x}$ nikaalo.
  Solution dikhao
  Answer0
  Seedhe: dono derivatives cancel ho jaate hain. Ya dhyaan do ki $x > 0$ ke liye sum constant $\frac{π}{2}$ hai.
3. Agar $y \sqrt{x^{2} + 1} = \log (\sqrt{x^{2} + 1} - x)$ , to dikhao ki $(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ .
  Solution dikhao
  AnswerDono sides differentiate karo aur $\sqrt{x^{2} + 1}$ se multiply karo.
  Right side: $\frac{\frac{x}{\sqrt{x^{2} + 1}} - 1}{\sqrt{x^{2} + 1} - x} = - \frac{1}{\sqrt{x^{2} + 1}}$ . Left side: $y^{'} \sqrt{x^{2} + 1} + \frac{x y}{\sqrt{x^{2} + 1}}$ . Dono ko barabar rakho aur $\sqrt{x^{2} + 1}$ se multiply karo.
Short answerHardNCERT Miscellaneous Exercise 5, Q15
Ek circle ka centre $(a, b)$ aur radius $c$ hai: $x = a + c \cos t$ , $y = b + c \sin t$ . Dikhao ki $\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{∣ \frac{d^{2} y}{d x^{2}} ∣} = c$ , ek constant jo $a$ aur $b$ par depend nahi karta.
Solution dikhao
Answer $\frac{d y}{d x} = - \cot t$ aur $\frac{d^{2} y}{d x^{2}} = - \frac{1}{c \sin^{3} t}$ , isliye ratio $c$ hai.
$\frac{d y}{d x} = \frac{c \cos t}{- c \sin t} = - \cot t$ . Phir $\frac{d^{2} y}{d x^{2}} = \frac{{cosec}^{2} t}{- c \sin t} = - \frac{1}{c \sin^{3} t}$ .
Numerator: $(1 + \cot^{2} t)^{3 / 2} = ({cosec}^{2} t)^{3 / 2} = \frac{1}{∣ \sin t ∣^{3}}$ . $\frac{1}{c ∣ \sin t ∣^{3}}$ se divide karne par $c$ bachta hai. (Yeh ratio radius of curvature hai; circle ke liye yeh radius hi hai.)
Short answerHardNCERT Miscellaneous Exercise 5, Q16
Agar $\sin y = x \sin (a + y)$ , jahan $\sin a \neq 0$ , to prove karo ki $\frac{d y}{d x} = \frac{\sin^{2} (a + y)}{\sin a}$ .
Solution dikhao
Answer $\frac{d x}{d y} = \frac{\sin a}{\sin^{2} (a + y)}$ .
$x = \frac{\sin y}{\sin (a + y)}$ , isliye $\frac{d x}{d y} = \frac{\cos y \sin (a + y) - \sin y \cos (a + y)}{\sin^{2} (a + y)} = \frac{\sin ((a + y) - y)}{\sin^{2} (a + y)} = \frac{\sin a}{\sin^{2} (a + y)}$ .
Kyunki $\frac{d y}{d x} = 1 \div \frac{d x}{d y}$ , result nikal aata hai.
6
Short answerHardNCERT Miscellaneous Exercise 5, Q17–Q18
Alag tarah ki settings mein second derivatives.
1. Agar $x = e^{t} \cos t$ aur $y = e^{t} \sin t$ hai to $\frac{d^{2} y}{d x^{2}}$ nikaalo.
  Solution dikhao
  Answer $\frac{2}{e^{t} (\cos t - \sin t)^{3}}$
  $\frac{d y}{d x} = \frac{\sin t + \cos t}{\cos t - \sin t}$ . Quotient rule se iska $t$ -derivative hai
  $\frac{(\cos t - \sin t)^{2} + (\sin t + \cos t)^{2}}{(\cos t - \sin t)^{2}} = \frac{2}{(\cos t - \sin t)^{2}}$
  Ise $\frac{d x}{d t} = e^{t} (\cos t - \sin t)$ se divide karo.
2. Maan lo $f (x) = x ∣ x ∣$ . Dikhao ki $f^{'} (x)$ har real $x$ ke liye exist karta hai, lekin $f^{''} (0)$ exist nahi karta.
  Solution dikhao
  Answer $f^{'} (x) = 2 ∣ x ∣$ ; $f^{''}$ , $x < 0$ ke liye $- 2$ aur $x > 0$ ke liye $2$ hai.
  $x \geq 0$ ke liye $f (x) = x^{2}$ aur $x < 0$ ke liye $- x^{2}$ . 0 se door $f^{'} (x) = 2 x$ ya $- 2 x$ , yaani $2 ∣ x ∣$ . 0 par $\frac{f (h) - f (0)}{h} = ∣ h ∣ \to 0$ , isliye $f^{'} (0) = 0$ aur har jagah $f^{'} (x) = 2 ∣ x ∣$ . Lekin $2 ∣ x ∣$ mein 0 par corner hai (one-sided derivatives $- 2$ aur $2$ ), isliye $f^{''} (0)$ exist nahi karta.
Short answerMediumNCERT Miscellaneous Exercise 5, Q19
$\cos (x + B) = \cos x \cos B - \sin x \sin B$ se shuru karke, jahan $B$ ek constant hai, $x$ ke respect mein differentiate karo aur $\sin (x + B)$ ka addition formula nikaalo.
Solution dikhao
Answer $\sin (x + B) = \sin x \cos B + \cos x \sin B$ .
Differentiate karne par $- \sin (x + B) = - \sin x \cos B - \cos x \sin B$ . $- 1$ se multiply karo.
Short answerMediumNCERT Miscellaneous Exercise 5, Q20
Ek aisa function do jo har real number par continuous ho lekin theek teen points, $x = - 1$ , $0$ aur $2$ par differentiable na ho. Apna answer justify karo.
Solution dikhao
AnswerJaise $f (x) = ∣ x + 1 ∣ + ∣ x ∣ + ∣ x - 2 ∣$ .
Har term continuous hai, isliye $f$ continuous hai. $- 1, 0, 2$ se door, har point ke paas $f$ linear hai aur isliye differentiable.
$x = 0$ par baaki do terms differentiable hain, jabki $∣ x ∣$ ke one-sided derivatives $- 1$ aur $1$ hain. Isliye $f$ ke one-sided derivatives mein 2 ka fark hai aur $f^{'} (0)$ exist nahi karta. Yahi argument $- 1$ aur $2$ par bhi kaam karta hai.
Short answerHardNCERT Miscellaneous Exercise 5, Q21
Maan lo $y = ∣ \begin{array}{cc} x^{2} & \sin x \\ e^{x} & x \end{array} ∣$ . Expand karke $\frac{d y}{d x}$ nikaalo, aur check karo ki yeh $∣ \begin{array}{cc} 2 x & \cos x \\ e^{x} & x \end{array} ∣ + ∣ \begin{array}{cc} x^{2} & \sin x \\ e^{x} & 1 \end{array} ∣$ ke barabar hai (ek baar mein ek row differentiate karo).
Solution dikhao
AnswerDono tareekon se $3 x^{2} - e^{x} (\sin x + \cos x)$ .
$y = x^{3} - e^{x} \sin x$ , isliye $y^{'} = 3 x^{2} - e^{x} \sin x - e^{x} \cos x$ .
Pehla determinant $2 x^{2} - e^{x} \cos x$ hai aur doosra $x^{2} - e^{x} \sin x$ . Inka sum $3 x^{2} - e^{x} \cos x - e^{x} \sin x$ hai, wahi.
Short answerHardNCERT Miscellaneous Exercise 5, Q22
Agar $y = \sin (m \sin^{- 1} x)$ , $- 1 < x < 1$ , to dikhao ki $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ .
Solution dikhao
Answer $\sqrt{1 - x^{2}} y_{1} = m \cos (m \sin^{- 1} x)$ ko differentiate karne par result milta hai.
$y_{1} = \cos (m \sin^{- 1} x) \cdot \frac{m}{\sqrt{1 - x^{2}}}$ , isliye $\sqrt{1 - x^{2}} y_{1} = m \cos (m \sin^{- 1} x)$ .
Differentiate karo: $\sqrt{1 - x^{2}} y_{2} - \frac{x}{\sqrt{1 - x^{2}}} y_{1} = - m \sin (m \sin^{- 1} x) \cdot \frac{m}{\sqrt{1 - x^{2}}} = - \frac{m^{2} y}{\sqrt{1 - x^{2}}}$ . $\sqrt{1 - x^{2}}$ se multiply karo.
Quick answerEasy
Inmein se kaunsa function $x = 0$ par continuous hai lekin wahan differentiable nahi?
$x^{2}$ $[x]$ $∣ \sin x ∣$ $\frac{1}{x}$
Answer dekho
Answer (C) $∣ \sin x ∣$
$∣ \sin x ∣$ mein 0 par corner hai (one-sided derivatives $- 1$ aur $1$ ). $x^{2}$ differentiable hai, $[x]$ , 0 par jump karta hai, aur $\frac{1}{x}$ , 0 par defined nahi hai.
Quick answerEasy
Agar $x = 2 t^{2}$ aur $y = t^{4}$ hai, to $t = 3$ par $\frac{d y}{d x}$ nikaalo.
Answer dekho
Answer 9
$\frac{d y}{d x} = t^{2} = 9$ .
Quick answerEasy
True ya false: agar $f$ , $c$ par differentiable hai, to $\lim_{x \to c} f (x)$ exist karti hai.
SahiGalat
Answer dekho
Answer Sahi
Differentiability se continuity milti hai (Theorem 3), isliye limit exist karti hai aur $f (c)$ ke barabar hai.

Recap

Summary aur key points

Ek real function $f$ apne domain ke point $c$ par continuous hai agar $\lim_{x \to c} f (x) = f (c)$ : dono one-sided limits exist karti hain, barabar hain, aur $f (c)$ ke barabar hain (Section 5.2).
Continuous function ka apne domain mein koi point of discontinuity nahi hota; closed interval ke end par sirf one-sided limit use hoti hai (continuous functions).
Continuous functions ke sums, differences, products aur quotients (jahan denominator non-zero ho) continuous hote hain, aur composites bhi (Section 5.2.1).
Polynomials, rational functions, trigonometric functions, $∣ x ∣$ , exponentials aur logarithms apne domains par continuous hain; $[x]$ har integer par jump karta hai (Example 15).
$f$ , $c$ par differentiable hai jab $c$ par uske left-hand aur right-hand derivatives finite aur barabar hon. Differentiability se continuity milti hai, ulta nahi: $∣ x - 3 ∣$ unbroken hai lekin 3 par uska corner hai (Section 5.3).
Chain rule: agar $t = u (x)$ aur $f = v \circ u$ , to $\frac{d f}{d x} = \frac{d v}{d t} \cdot \frac{d t}{d x}$ , aur lambi chains mein aur rates multiply hote hain (Section 5.3.1).
Implicit relation ke liye dono sides ko $x$ ke respect mein differentiate karo, aur $y$ wale har term ke saath $\frac{d y}{d x}$ lagao (Section 5.3.2).
$(- 1, 1)$ par $\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}$ , aur $\frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}}$ (Section 5.3.3).
$\log_{b} a = x$ ka matlab $b^{x} = a$ hai; $\log$ products ko sums mein, quotients ko differences mein aur powers ko multiples mein badalta hai, aur $\log_{b} x$ , $y = x$ mein $b^{x}$ ki mirror image hai (Section 5.4).
$\frac{d}{d x} e^{x} = e^{x}$ , $\frac{d}{d x} \log x = \frac{1}{x}$ aur $\frac{d}{d x} a^{x} = a^{x} \log a$ (Theorem 5).
Logarithmic differentiation $[u (x)]^{v (x)}$ aur lambe products ko sambhaalta hai: $\frac{d y}{d x} = u^{v} [\frac{v}{u} u^{'} + v^{'} \log u]$ , jo $u > 0$ hone par valid hai (Section 5.5).
$x = f (t)$ , $y = g (t)$ ke liye: jahan bhi $f^{'} (t) \neq 0$ , $\frac{d y}{d x} = \frac{g^{'} (t)}{f^{'} (t)}$ (Section 5.6).
Second derivative $\frac{d^{2} y}{d x^{2}}$ , $\frac{d y}{d x}$ ka derivative hai: slope ke badalne ki rate (Section 5.7).

Flash cards · 10

Yaad rakhne wale derivatives

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Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 5.1, Q1–Q2	Practice 5.1 · Q1
NCERT Exercise 5.1, Q3	Practice 5.1 · Q2
NCERT Exercise 5.1, Q4	Practice 5.1 · Q3
NCERT Exercise 5.1, Q5	Practice 5.1 · Q4
NCERT Exercise 5.1, Q6–Q12	Practice 5.1 · Q5
NCERT Exercise 5.1, Q13–Q16	Practice 5.1 · Q6
NCERT Exercise 5.1, Q17–Q18	Practice 5.1 · Q7
NCERT Exercise 5.1, Q19	Practice 5.1 · Q8
NCERT Exercise 5.1, Q20–Q22	Practice 5.1 · Q9
NCERT Exercise 5.1, Q23–Q25	Practice 5.1 · Q10
NCERT Exercise 5.1, Q26–Q30	Practice 5.1 · Q11
NCERT Exercise 5.1, Q31–Q34	Practice 5.1 · Q12
NCERT Exercise 5.2, Q1–Q8	Practice 5.2 · Q1
NCERT Exercise 5.2, Q9	Practice 5.2 · Q2
NCERT Exercise 5.2, Q10	Practice 5.2 · Q3
NCERT Exercise 5.3, Q1–Q8	Practice 5.3 · Q1
NCERT Exercise 5.3, Q9–Q15	Practice 5.3 · Q2
NCERT Exercise 5.4, Q1–Q5	Practice 5.4 · Q1
NCERT Exercise 5.4, Q6–Q10	Practice 5.4 · Q2
NCERT Exercise 5.5, Q1–Q6	Practice 5.5 · Q1
NCERT Exercise 5.5, Q7–Q11	Practice 5.5 · Q2
NCERT Exercise 5.5, Q12–Q15	Practice 5.5 · Q3
NCERT Exercise 5.5, Q16	Practice 5.5 · Q4
NCERT Exercise 5.5, Q17	Practice 5.5 · Q5
NCERT Exercise 5.5, Q18	Practice 5.5 · Q6
NCERT Exercise 5.6, Q1–Q5	Practice 5.6 · Q1
NCERT Exercise 5.6, Q6–Q10	Practice 5.6 · Q2
NCERT Exercise 5.6, Q11	Practice 5.6 · Q3
NCERT Exercise 5.7, Q1–Q5	Practice 5.7 · Q1
NCERT Exercise 5.7, Q6–Q10	Practice 5.7 · Q2
NCERT Exercise 5.7, Q11–Q14	Practice 5.7 · Q3
NCERT Exercise 5.7, Q15–Q17	Practice 5.7 · Q4
NCERT Miscellaneous Exercise 5, Q1–Q6	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise 5, Q7–Q11	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise 5, Q12–Q14	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise 5, Q15	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise 5, Q16	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise 5, Q17–Q18	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise 5, Q19	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise 5, Q20	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercise 5, Q21	Miscellaneous exercises · Q9
NCERT Miscellaneous Exercise 5, Q22	Miscellaneous exercises · Q10

Quick check: MCQs

5.2 Continuity4 questions
5.2.1 Continuous functions ka algebra3 questions
5.3 Differentiability4 questions
5.3.3 Inverse trigonometric functions ke derivatives4 questions
5.4 Exponential aur Logarithmic Functions3 questions
5.5 Logarithmic Differentiation3 questions
5.6 Parametric Forms mein Functions ke Derivatives3 questions
5.7 Second Order Derivative3 questions
Miscellaneous examples3 questions

Glossary · 13

Key terms (English + Hindi)

chain ruleशृंखला नियमshrinkhala niyam: Composite ka derivative chain ke saath rates ka product hai: $\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}$ .Chapter mein pehli baar yahan aaya →
continuous (ek point par)संततcontinuous (at a point)santat: $f$ apne domain ke point $c$ par continuous hai jab $\lim_{x \to c} f (x)$ exist kare aur $f (c)$ ke barabar ho.Chapter mein pehli baar yahan aaya →
continuous functionसंतत फलनsantat phalan: Aisa function jo apne domain ke har point par continuous ho.

differentiableअवकलनीयavakalaneeya: $f$ , $c$ par differentiable hai jab $c$ par uske left-hand aur right-hand derivatives finite aur barabar hon.Chapter mein pehli baar yahan aaya →

exponential functionचरघातांकी फलनcharghaataanki phalan: Fixed base $b > 1$ ke liye $y = b^{x}$ ; base $e$ ke saath yeh natural exponential $e^{x}$ hai.

greatest integer functionमहत्तम पूर्णांक फलनmahattam poornaank phalan: $[x]$ , $x$ se chhota ya uske barabar sabse bada integer; jaise $[- 1.2] = - 2$ .Chapter mein pehli baar yahan aaya →

implicit functionअस्पष्ट फलनaspasht phalan: $x$ aur $y$ ke beech ke relation se fix hua $x$ ka function, jo $y = f (x)$ form mein nahi likha gaya.

logarithmलघुगणकlaghuganak: $\log_{b} a = x$ ka matlab $b^{x} = a$ hai. Is chapter mein $\log x$ natural logarithm hai, base $e$ .
logarithmic differentiationलघुगणकीय अवकलनlaghuganakiya avakalan: $y$ ki jagah $\log y$ ko differentiate karna, phir $y$ se multiply karna; $[u (x)]^{v (x)}$ aur lambe products ke liye use hota hai.

one-sided limitएकपक्षीय सीमाekpakshiya seema: Woh value jiski taraf $f (x)$ jaata hai jab $x \to c$ sirf left se ( $x < c$ ) ya sirf right se ( $x > c$ ) ho.Chapter mein pehli baar yahan aaya →

parameterप्राचलpraachal: Ek teesra variable $t$ jiske through $x = f (t)$ aur $y = g (t)$ dono diye jaate hain.
point of discontinuityअसांतत्य का बिंदुasaantatya ka bindu: Domain ka woh point jahan function continuous nahi hai.Chapter mein pehli baar yahan aaya →

second order derivativeद्वितीय कोटि का अवकलजdvitiya koti ka avakalaj: $\frac{d y}{d x}$ ka derivative, jise $\frac{d^{2} y}{d x^{2}}$ ya $y^{''}$ likhte hain.

5.1 Introduction

5.2 Continuity

5.2.1 Continuous functions ka algebra

5.3 Differentiability

5.3.1 Composite functions ke derivatives

5.3.2 Implicit functions ke derivatives

5.3.3 Inverse trigonometric functions ke derivatives

5.4 Exponential aur Logarithmic Functions

5.5 Logarithmic Differentiation

5.6 Parametric Forms mein Functions ke Derivatives

5.7 Second Order Derivative

Miscellaneous examples

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Quick check: MCQs

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