Limits and Derivatives: Class 11 Maths Chapter 12 Hinglish mein

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

Published 22 Oct 2026 · Updated 22 Oct 2026

95 min
1 simulation
52 practice questions
23 MCQs

Is chapter mein aap kya seekhenge

Chhote hote intervals par average rates se instantaneous rate of change ka estimate lagao, aur use tangent ke slope ki tarah samjho.
Tables, graphs aur formulas se left-hand aur right-hand limits nikalo, aur decide karo ki limit exist karti hai ya nahi.
Algebra of limits, substitution, common factors cancel karke aur (xⁿ − aⁿ)/(x − a) ki standard limit se limits evaluate karo.
Areas compare karke cos x < sin x/x < 1 prove karo, sandwich theorem se sin x/x aur (1 − cos x)/x ki limits nikalo, aur trigonometric limits evaluate karo.
Ek point par aur ek function ke roop mein derivative define karo, first principle se derivatives nikalo, aur unhe tangents ke slopes ki tarah samjho.
Sums, differences, products aur quotients, x ki powers, polynomials aur chhe trigonometric functions ko differentiate karo.

NCERT Mathematics Class 11, Chapter 12, pp. 217–256 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

12.1 Introduction

Car ka speedometer ek hi number dikhata hai: car is pal kitni tez chal rahi hai. Bukhar check karne wala doctor jaanna chahta hai ki temperature abhi kitni jaldi gir raha hai, raat bhar ka average girna nahi. Monsoon mein river gauge par khada engineer dekhta hai ki is ghante paani kitni tezi se badh raha hai. Har ek ek pal par rate of change hai, aur aise rates naapne ke liye bani mathematics ko calculus (कलन) kehte hain.

Calculus mein hum poochte hain ki jab kisi function ka input badalta hai to uska output kitni tezi se badalta hai, aur iska jawab do naye ideas se dete hain: limit aur derivative. Pehle hum ek pal ki speed ko informally samjhenge, phir tables, graphs aur algebra se limits padhenge, aur aakhir mein derivative define karke use polynomials aur trigonometric functions ke liye nikalenge. Class 12 isi base par continuity, kuch naye rules aur integration sikhati hai.

12.2 Derivatives ka Intuitive Idea

Ek metro train station se nikalti hai aur pehle 20 seconds tak barabar speed badhati hai. $t$ seconds baad starting point se uski distance $s$ (metres mein) $s = 0.5 t^{2}$ hai. Driver ka speedometer har pal ki speed dikhata hai. Kya hum sirf distance formula se $t = 10$ s par uski reading nikal sakte hain?

Mushkil yeh hai ki ek akele pal mein train koi distance cover nahi karti, isliye 'distance divided by time' se $\frac{0}{0}$ milta hai. Hum jo nikal sakte hain woh hai kisi time interval $[t_{1}, t_{2}]$ par average speed:

ek interval par average speed

\overset{ˉ}{v} = \frac{s (t_{2}) - s (t_{1})}{t_{2} - t_{1}}

(12.1)

Ab intervals ko $t = 10$ s ki taraf chhota karte jao, pehle left se (10 s par khatam hone wale intervals) aur phir right se (10 s se shuru hone wale intervals). Kyunki $s (10) = 50$ m hai, 9.9 s se 10 s wala interval 0.1 s mein $50 - 49.005 = 0.995$ m cover karta hai, yaani average 9.95 m/s. Table 12.1 mein aise aath averages hain; aakhri column distance ko interval ki length se divide karke milta hai.

Table 12.1

t = 10

s par khatam ya shuru hone wale intervals par train ki average speeds

Interval (s)	Distance (m)	Speed (m/s)
8–10	18	9
9–10	9.5	9.5
9.9–10	0.995	9.95
9.99–10	0.09995	9.995
10–10.01	0.10005	10.005
10–10.1	1.005	10.05
10–11	10.5	10.5
10–12	22	11

Aakhri column ko upar se neeche padho: jaise-jaise intervals left se 10 s ke paas aate hain, averages 9, 9.5, 9.95, 9.995 tak chadhte hain. Ab ise neeche se upar padho: jaise-jaise intervals right se paas aate hain, woh 11, 10.5, 10.05, 10.005 tak girte hain. Dono lists ek hi number ki taraf badhti hain, isliye $t = 10$ s par speedometer 10 m/s dikhayega, yaani 36 km/h. Koi bhi akela interval yeh value nahi deta. Yeh woh number hai jiske averages paas aate hain, yaani $t = 10$ s par train ki instantaneous speed (तात्क्षणिक चाल).

Isi idea ki ek picture bhi hai. $s = 0.5 t^{2}$ ke graph par point $A (10, 50)$ mark karo. $[10, 10 + h]$ par average speed, $A$ se graph ke us point $Q$ tak ki chord ka rise over run hai jo $t = 10 + h$ ke upar hai, isliye yeh chord $A Q$ ka slope hai. Jaise-jaise $h$ chhota hota hai, chord ghoom kar us line par tik jaati hai jo curve ko $A$ par bas chhooti hai: $A$ par tangent (Fig. 12.1).

4/4 $A$ par tangent ka slope 10 m/s hai, yahi instantaneous speed hai.

Fig. 12.1Time interval chhota hone par

A

se chords tangent ki taraf ghoomti hain.NCERT Fig. 12.1 · p. 219

Ek distance–time graph: curve s = 0.5t² origin se upar uthti hai aur A(10, 50) se guzarti hai. Step 1 mein A se t = 14 par Q tak dashed chord banti hai, aur ek panel mein slope 12 m/s dikhta hai. Step 2 mein Q t = 12 par aata hai; chord kam steep hai, slope 11 m/s. Step 3 mein Q t = 10.5 par aata hai; chord lagbhag curve se chipak jaati hai, slope 10.25 m/s. Step 4 mein chord ki jagah A par solid tangent aata hai, slope 10 m/s, jo time axis ko t = 5 par milta hai.

To $t = 10$ s par instantaneous speed us pal par distance–time graph ke tangent ke slope ke barabar hai. Section 12.5 ki bhasha mein, 10 m/s, $t = 10$ par $s = 0.5 t^{2}$ ka derivative hai. 'Woh number jiske averages paas aate hain' ko precise banane ke liye ab hum limits padhte hain.

Quick check

3 mein se 1

Train ke liye

s = 0.5 t^{2}

hai.

t = 4

s aur

t = 6

s ke beech uski average speed kya hai?

m/s

Answer dekho

Answer 5 m/s

s (6) = 18

m aur

s (4) = 8

m, isliye train 2 s mein 10 m cover karti hai: 5 m/s. 18 m ko 6 s se divide karne par 3 m/s milta hai, jo start se average speed hai, is interval ki nahi.

12.3 Limits

Speed nikalne mein humne ek pattern follow kiya jo hum baar-baar use karenge: average speed ke formula mein $t = 10$ rakhna allowed nahi tha, isliye humne dekha ki jab $t$ dheere-dheere 10 ke paas aata hai to kya hota hai. Is process ko limit lena kehte hain.

Ise $g (x) = \frac{x^{3} - 8}{x - 2}$ par try karo. $x = 2$ par formula $\frac{0}{0}$ deta hai, isliye $g (2)$ defined nahi hai. Lekin 2 ke paas $g$ evaluate karne se hamein koi nahi rokta:

Table 12.2

x = 2

ke paas

g (x) = \frac{x^{3} - 8}{x - 2}

ki values

$x$	$g (x)$
1.8	10.84
1.95	11.7025
1.995	11.970025
2.005	12.030025
2.05	12.3025
2.2	13.24

Dono taraf se values 12 ke aas-paas jama hoti hain, aur jab $x$ 2 ke kaafi paas ho to woh 12 ke jitna chahe utna paas rehti hain. Hum likhte hain $\lim_{x \to 2} g (x) = 12$ aur padhte hain ' $x$ ke 2 ki taraf jaane par $g (x)$ ki limit 12 hai'. Graph iska kaaran batata hai. $x \neq 2$ ke liye hum $x - 2$ cancel kar sakte hain, kyunki $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ , isliye $g$ parabola $y = x^{2} + 2 x + 4$ hai jismein se ek point nikaal diya gaya hai (Fig. 12.2).

Definition

Limit

Hum $\lim_{x \to a} f (x) = L$ tab likhte hain jab $x$ ko $a$ ke kaafi paas lekar ( $x \neq a$ ke saath) values $f (x)$ ko $L$ ke jitna chahein utna paas laaya ja sake. Number $L$ ko $a$ par $f$ ki limit kehte hain.

Example $\lim_{x \to 2} \frac{x^{3} - 8}{x - 2} = 12$ , halaanki function 2 par defined nahi hai.

4/4 $g (2) = 5$ rakho: limit ab bhi 12 hai.

Fig. 12.2Gayab ya galat jagah wala point limit nahi badalta.NCERT Fig. 12.2 · p. 220

g(x) = (x³ − 8)/(x − 2) ka graph parabola y = x² + 2x + 4 hai jismein (2, 12) par ek open circle hai; hole se x = 2 aur y = 12 tak dashed guides jaati hain. Step 1 curve dikhata hai aur batata hai ki g(2) undefined hai. Step 2 mein left branch par teen dots hole ki taraf badhte hain, aur x ke left se 2 ki taraf jaane par g(x) → 12. Step 3 yahi right se karta hai. Step 4 mein g(2) = 5 define hota hai: (2, 5) par ek bhara hua dot aata hai, hole se kaafi neeche, phir bhi limit 12 hi rehti hai.

Is definition ki do baatein dhyaan dene layak hain. Pehli, value $f (a)$ ka ismein koi role nahi hai. Function $a$ par undefined ho sakta hai, ya wahan kisi alag value ke saath defined ho sakta hai, jaise Fig. 12.2 ke aakhri step mein, aur limit wahi rehti hai. Doosri, $x$ kisi bhi taraf se $a$ ke paas aa sakta hai, aur definition ke liye dono taraf ka agree karna zaroori hai.

Left-hand aur right-hand limits

Pune ka ek courier counter weight ke hisaab se charge leta hai: 2 kg tak (2 kg bhi) ke parcel ke liye ₹60, aur 2 kg se zyada (5 kg tak) ke parcel ke liye ₹95. Maan lo $C (w)$ , $w$ kg ke parcel ka charge rupees mein hai.

C (w) = {\begin{cases} 60, & w \leq 2 \\ 95, & w > 2 \end{cases}

Weight ko neeche se 2 kg ki taraf le jao: 1.9 kg, 1.99 kg, 1.999 kg. Har baar charge ₹60 hai, isliye jab $w$ left se 2 ki taraf aata hai, $C (w)$ 60 ke paas aata hai. Yeh 2 par $C$ ki left-hand limit (बाएँ पक्ष की सीमा) hai, jise $\lim_{w \to 2^{-}} C (w) = 60$ likhte hain. Upar se (2.1 kg, 2.01 kg, 2.001 kg) charge hamesha ₹95 hai, isliye right-hand limit $\lim_{w \to 2^{+}} C (w) = 95$ hai (Fig. 12.3).

3/3One-sided limits alag hain, isliye $\lim_{w \to 2} C (w)$ exist nahi karti.

Fig. 12.3

w = 2

par dono one-sided limits alag hain.NCERT Fig. 12.3 · p. 221

Charge C (rupees mein) ko weight w (kilograms mein) ke against plot kiya gaya hai. Height 60 par ek solid horizontal segment w = 0 se w = 2 tak jaata hai, aur (2, 60) par bhare hue dot par khatam hota hai. Height 95 par ek dashed segment (2, 95) par open circle se shuru hokar w = 5 tak jaata hai. Step 1 neeche wale segment par w = 2 ki taraf arrow jodta hai: left-hand limit 60 hai. Step 2 upar wale segment par right se w = 2 ki taraf arrow jodta hai: right-hand limit 95 hai. Step 3 dono arrows aur 60 aur 95 ke beech ka gap dikhane wala double arrow dikhata hai: 2 par koi limit nahi hai.

Dono one-sided limits alag hain, isliye aisa koi ek number nahi hai jiske paas $w \to 2$ par $C (w)$ aata ho. 2 par $C$ ki limit exist nahi karti, jabki $C (2) = 60$ bilkul theek se defined hai.

Definition

Left- aur right-hand limits

Left-hand limit $\lim_{x \to a^{-}} f (x)$ woh number hai jiske paas $f (x)$ tab aata hai jab $x$ , $a$ se chhoti values se hokar $a$ ke paas aata hai. Right-hand limit $\lim_{x \to a^{+}} f (x)$ mein $a$ se badi values li jaati hain. Limit $\lim_{x \to a} f (x)$ theek tab exist karti hai jab dono one-sided limits exist karein aur barabar hon, aur tab yeh unki common value hoti hai.

ExampleCourier charge ke liye 2 par left-hand limit 60 hai aur right-hand limit 95, isliye 2 par koi limit nahi hai.

Tables aur graphs se limits

Values ki table ek bharosemand pehla tool hai. Table 12.3 chaar aur functions ko ek chune hue point $a$ ke 0.01 left aur 0.01 right par evaluate karti hai.

Table 12.3

f

ki values:

a

ke theek left (

a - 0.01

par) aur

a

ke theek right (

a + 0.01

par)

$f (x)$ aur $a$	Left	Right	Limit
$5 - 2 x$ , $3$	−0.98	−1.02	−1
$x^{2} - 3 x$ , $- 1$	4.0501	3.9501	4
$\tan x$ , $\frac{π}{4}$	0.9802	1.0202	1
$\frac{1}{(x - 1)^{2}}$ , $1$	$10^{4}$	$10^{4}$	koi nahi

Pehli teen rows seedhi-saadi hain. Values dono taraf se ek number ke paas aati hain, aur woh number bas $f (a)$ hai: $5 - 2 (3) = - 1$ , $(- 1)^{2} - 3 (- 1) = 4$ aur $\tan \frac{π}{4} = 1$ . Polynomials aur trigonometric functions ke liye, jin points par formula ka matlab banta hai, wahan limit hi value hai. Section 12.3.2 polynomials ke liye yeh prove karta hai.

Aakhri row alag hai. 1 ke paas jaate hi values phat padti hain: $x = 0.999$ par $10^{6}$ aur $x = 0.9999$ par $10^{8}$ , tum jo bhi bound socho usse bhi aage. Kisi real number ke paas nahi aaya ja raha, isliye limit exist nahi karti. Is behaviour ko kabhi-kabhi $\lim_{x \to 1} \frac{1}{(x - 1)^{2}} = + \infty$ likhte hain, lekin $+ \infty$ sirf yeh batata hai ki values bina kisi bound ke badh rahi hain; yeh koi number nahi hai. Aisi limits is course se bahar hain.

InteractiveLimit aur slope lab

Do modes. Approach: chhe functions mein se ek chuno (hole wala parabola, courier charge, π/4 ke paas tan x, 1 ke paas 1/(x − 1)², 0 ke paas sin x / x aur 0 ke paas (1 − cos x)/x); ek slider set karta hai ki x point se kitna door hai, graph point ke theek left aur right ke do points mark karta hai, aur ek table har taraf 0.1, 0.01, 0.001 aur 0.0001 doori par f(x) dikhati hai, saath mein faisla ki limit exist karti hai ya nahi. Chord to tangent: f(x) = 0.5x², x³/3 − x, 6/x ya sin x chuno, point a aur step h (positive ya negative) set karo; graph chord PQ dikhata hai, panel difference quotient dikhata hai, aur ek button tangent aur derivative ki value dikhata hai.

Fig. 12.4Approach mode mein ek function chuno aur

x

ko kisi bhi taraf se point ki taraf slide karo, saath mein table bharti jaayegi. Chord to tangent mode mein

h

ko chhota karo aur dekho ki chord ka slope kaise derivative par tik jaata hai.

Tables aur graphs ne hamein yeh sikhaya hai:

$a$ par limit sirf $a$ ke paas ki values par depend karti hai, khud $f (a)$ par kabhi nahi.
Limit tabhi exist karti hai jab left-hand aur right-hand limits exist karein aur barabar hon.
$f (a)$ undefined ho sakta hai jabki limit exist kare (Fig. 12.2), defined ho sakta hai jabki limit exist na kare (Fig. 12.3), ya defined aur limit se alag ho sakta hai.
Bina bound ke badhne wali values ki koi limit nahi hoti.

Quick check

3 mein se 1

Maan lo

x < 2

ke liye

f (x) = x + 1

aur

x \geq 2

ke liye

f (x) = 7 - x

\lim_{x \to 2} f (x)

kya hai?

354, yaani 3 aur 5 ka averageYeh exist nahi karti.

Answer dekho

Answer (D) Yeh exist nahi karti.

Left se

f (x) \to 3

aur right se

f (x) \to 5

. One-sided limits alag hain, isliye koi limit nahi hai. Value

f (2) = 5

isse decide nahi karti, aur dono one-sided limits ka average lena allowed nahi hai.

12.3.1 Limits ka algebra

Tables slow hoti hain, aur achhi baat yeh hai ki hamein unki kam hi zaroorat padti hai. Agar do functions ki $a$ par limits hain, to unke sum, difference, product aur quotient ki bhi hongi, aur limits seedhe tareeke se judti hain. Bas ek shart: hum 0 limit se divide nahi kar sakte. Yahan hum in rules ko bina proof maan lete hain, kyunki unhe prove karne ke liye limit ki exact definition chahiye, jo aage ke courses mein aati hai.

Theorem 12.1Limits ka algebra

Maan lo $\lim_{x \to a} f (x) = l$ aur $\lim_{x \to a} g (x) = m$ dono exist karti hain. Tab $x \to a$ par: (i) $f (x) + g (x) \to l + m$ ; (ii) $f (x) - g (x) \to l - m$ ; (iii) $f (x) g (x) \to l m$ ; (iv) $\frac{f (x)}{g (x)} \to \frac{l}{m}$ , bashart $m \neq 0$ .

Agar $g$ constant function $c$ hai, to rule (iii) kehta hai $\lim_{x \to a} c f (x) = c l$ : constants limits se bahar aa jaate hain. Jaise, (iii) se $\lim_{x \to - 1} x^{2} = (- 1) (- 1) = 1$ milta hai, aur phir (ii) se $\lim_{x \to - 1} (x^{2} - 3 x) = 1 - 3 (- 1) = 4$ , jaisa table ne bataya tha. Rule (iv) mein ek asli shart hai: $\lim_{x \to 0} \frac{\sin x}{x}$ ko $\frac{\lim \sin x}{\lim x}$ mein nahi toda ja sakta, kyunki neeche ki limit 0 hai. Aisi limits ke liye Section 12.4 chahiye.

12.3.2 Polynomials aur rational functions ki limits

Do limits to saaf hain: constant function $c$ , $c$ par hi rehta hai, isliye uski limit $c$ hai, aur function $x$ , $a$ ke paas aata hai jaise $x$ aata hai. Har polynomial inhi do se multiply aur add karke banta hai. Kyunki $x^{2} = x \cdot x$ , rule (iii) se $\lim_{x \to a} x^{2} = a^{2}$ milta hai; yeh step dohraane se ( $n$ par induction) har positive integer $n$ ke liye $\lim_{x \to a} x^{n} = a^{n}$ milta hai.

DerivationPolynomial ki limit

Maan lo $p (x) = c_{0} + c_{1} x + \dots + c_{n} x^{n}$ aur har term ko ek function maano. Sum rule se hum term by term limit le sakte hain, constant rule $c_{k} x^{k}$ ki limit ko $c_{k} \lim_{x \to a} x^{k}$ bana deta hai, aur yeh $c_{k} a^{k}$ hai. Saare terms jodne par $c_{0} + c_{1} a + \dots + c_{n} a^{n}$ milta hai, jo $p (a)$ hai.

polynomial ki limit

\lim_{x \to a} p (x) = p (a)

(12.3)

Rational function do polynomials ka quotient $f (x) = \frac{g (x)}{h (x)}$ hota hai, jo har jagah defined hai jahan $h (x) \neq 0$ . Agar $h (a) \neq 0$ , to quotient rule aur (12.3) se milta hai

rational function, $h(a) \ne 0$

\lim_{x \to a} \frac{g (x)}{h (x)} = \frac{g (a)}{h (a)}

(12.4)

Mushkil tab shuru hoti hai jab $h (a) = 0$ . Agar $g (a) \neq 0$ , to upar wala part $g (a)$ ke paas rehta hai jabki neeche wala 0 tak simat jaata hai, isliye quotient bina bound ke badhta hai aur koi limit nahi hoti. Agar $g (a) = 0$ bhi ho, to hamare paas form $\frac{0}{0}$ hai, jo apne aap kuch nahi batata. Tab $x - a$ dono polynomials ko divide karta hai (factor theorem), aur hum $x - a$ ke saare factors bahar nikaal sakte hain:

\begin{aligned} g (x) & = (x - a)^{k} g_{1} (x) \\ h (x) & = (x - a)^{l} h_{1} (x) \end{aligned}

Yahan $g_{1} (a) \neq 0$ aur $h_{1} (a) \neq 0$ . $x \neq a$ ke liye cancel karne par teen results bachte hain (Fig. 12.5):

$k > l$ : factor $(x - a)^{k - l}$ upar reh jaata hai, isliye $f (x) \to 0$ .
$k = l$ : sab cancel ho jaata hai, isliye $f (x) \to \frac{g_{1} (a)}{h_{1} (a)}$ .
$k < l$ : ek factor $(x - a)^{l - k}$ neeche reh jaata hai, values bina bound ke badhti hain, aur koi limit nahi hoti.

2/3Factors cancel ho jaate hain: limit $\frac{g_{1} (a)}{h_{1} (a)}$ hai.

Fig. 12.5

x - a

ke factors ginne se

\frac{0}{0}

wali limit decide hoti hai.

Axes ka ek set jismein x = 1 par dashed vertical guide hai. Step 1 (k = 2, l = 1): (x − 1)²/(x − 1) ka graph line y = x − 1 hai jismein (1, 0) par hole hai, isliye limit 0 hai. Step 2 (k = l = 1): (x² − 1)/(x − 1) ka graph line y = x + 1 hai jismein (1, 2) par hole hai, isliye limit 2 hai. Step 3 (k = 1, l = 2): (x − 1)/(x − 1)² ka graph hyperbola y = 1/(x − 1) hai, jiski do branches x = 1 ke bagal mein minus aur plus infinity ki taraf bhaagti hain, isliye koi limit nahi hai.

Cancel karna sahi hai kyunki limit sirf $x \neq a$ wali values use karti hai, aur wahan $x - a$ zero nahi hai.

Example 12.1NCERT Example 1 · p. 230

Polynomials: substitute karo

Evaluate karo (i) $\lim_{x \to 2} (x^{4} - 3 x^{2} + 5)$ , (ii) $\lim_{x \to - 2} (x + 1) (x^{2} - 4 x)$ aur (iii) $\lim_{x \to 2} (1 + x + x^{2} + \dots + x^{5})$ .

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

Example 12.2NCERT Example 2 · p. 230

Rational functions

Har limit evaluate karo, ya dikhao ki woh exist nahi karti.

$\lim_{x \to 3} \frac{x^{2} + 1}{x + 2}$
$\lim_{x \to - 3} \frac{x^{2} + x - 6}{x^{2} - 9}$
$\lim_{x \to 1} \frac{x^{2} - 1}{x^{3} - 3 x^{2} + 3 x - 1}$
$\lim_{x \to 1} \frac{x^{3} - x^{2} - x + 1}{x^{2} + 2 x - 3}$
$\lim_{x \to 2} (\frac{1}{x - 2} - \frac{4}{x^{2} - 4})$

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

Theorem 12.2Ek standard limit

Har positive integer $n$ aur har real number $a$ ke liye, $\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$ .

$x^{n} - a^{n}$ ko $x - a$ se long division karne par koi remainder nahi bachta: $x^{n} - a^{n} = (x - a) S (x)$ , jahan $S (x)$ , $n$ terms $x^{n - 1}$ , $x^{n - 2} a$ , …, $x a^{n - 2}$ , $a^{n - 1}$ ka sum hai. ( $(x - a) S (x)$ ko multiply karke check karo: $x^{n} - a^{n}$ ke alawa sab cancel ho jaata hai.)

$x \neq a$ ke liye hum $x - a$ cancel karte hain. Jo bachta hai woh $n$ terms wala ek polynomial hai, aur $x = a$ par har term $a^{n - 1}$ ke barabar hai. (12.3) se limit $a^{n - 1}$ ki $n$ copies hai, yaani $n a^{n - 1}$ .

difference of powers

\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}

(12.5)

Yahi formula tab bhi sahi hai jab $n$ koi bhi rational number ho aur $a > 0$ ; ise hum bina proof use karte hain. Jaise, $n = \frac{1}{2}$ ke saath yeh kehta hai ki $x \to a$ par $\frac{\sqrt{x} - \sqrt{a}}{x - a} \to \frac{1}{2 \sqrt{a}}$ .

Example 12.3NCERT Example 3 · p. 233

Difference-of-powers limit ka use

Evaluate karo (i) $\lim_{x \to 2} \frac{x^{5} - 32}{x^{3} - 8}$ aur (ii) $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$ .

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

12.4 Trigonometric Functions ki Limits

Quotient $\frac{\sin x}{x}$ , $x = 0$ par $\frac{0}{0}$ deta hai, lekin cancel karne ke liye koi polynomial factor nahi hai. Hamein inequalities aur limits ke baare mein do facts chahiye, aur phir ek picture.

Theorem 12.3Limits inequalities ko banaye rakhti hain

Maan lo $a$ ke paas ke saare $x$ ke liye ( $a$ par chhod kar, shayad) $f (x) \leq g (x)$ hai, aur $\lim_{x \to a} f (x)$ aur $\lim_{x \to a} g (x)$ dono exist karti hain. Tab $\lim_{x \to a} f (x) \leq \lim_{x \to a} g (x)$ .

Conclusion mein $\leq$ hi aata hai, tab bhi jab $f (x) < g (x)$ strictly ho: har $x \neq 0$ ke liye $x^{2} < 2 x^{2}$ , phir bhi $x \to 0$ par dono 0 ki taraf jaate hain.

Theorem 12.4Sandwich theorem

Maan lo jab bhi $x$ , $a$ ke paas ho (khud $a$ par chhod kar, shayad) to $f (x) \leq g (x) \leq h (x)$ , aur $\lim_{x \to a} f (x) = l = \lim_{x \to a} h (x)$ . Tab $\lim_{x \to a} g (x)$ exist karti hai aur $l$ ke barabar hai.

Floor $f$ aur ceiling $h$ dono $l$ ke paas aate hain, aur $g$ unke beech phansa hua hai, isliye uske paas aur kahin jaane ka raasta nahi. Yeh sandwich theorem woh limits suljhata hai jo tables nahi kar paati. Fig. 12.6 mein $g (x) = x^{2} \sin \frac{1}{x}$ 0 ke paas anginat baar upar-neeche hota hai. Kyunki $- 1 \leq \sin \frac{1}{x} \leq 1$ , hamare paas $- x^{2} \leq g (x) \leq x^{2}$ hai, aur dono bounds 0 ki taraf jaate hain. Isliye $\lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0$ .

2/3 $g (x) = x^{2} \sin \frac{1}{x}$ hamesha inke beech rehta hai.

Fig. 12.6Dono bounds origin par

g

ko 0 tak dabaa dete hain.NCERT Fig. 12.9 · p. 234

−1 se 1 tak axes. Step 1 do dashed parabolas banata hai, upar khulne wala y = x² aur neeche khulne wala y = −x²; yeh sirf origin par milte hain. Step 2 solid curve g(x) = x² sin(1/x) jodta hai, jo 0 ke paas tez aur tez oscillate karti hai lekin hamesha dono parabolas ke beech rehti hai. Step 3 origin ko gher kar arrows jodta hai: dono bounds 0 ki taraf jaate hain, isliye g(x) ko bhi 0 ki taraf jaana padta hai.

Ab sabse zaroori inequality. $0 < ∣ x ∣ < \frac{π}{2}$ ke liye, jahan $x$ radians mein naapa gaya hai,

$0 < |x| < \frac{\pi}{2}$ ke liye

\cos x < \frac{\sin x}{x} < 1

(12.6)

DerivationAreas compare karke (12.6) ka proof

Pehle $0 < x < \frac{π}{2}$ lo. Origin $O$ par centre wale unit circle par $A (1, 0)$ aur $P (\cos x, \sin x)$ mark karo, taaki $∠ A O P = x$ ho. $A$ par circle ka tangent $A$ se guzarne wali vertical line hai; yeh ray $O P$ ko $T (1, \tan x)$ par milti hai (Fig. 12.7).

Teen regions ek hi base $O A = 1$ par khade hain, har ek agle ke andar: height $\sin x$ wala triangle $O A P$ , angle $x$ wala sector $O A P$ (area $\frac{x}{2}$ , kyunki radius 1 hai), aur height $\tan x$ wala triangle $O A T$ . Isliye unke areas, $\frac{1}{2} \sin x$ , $\frac{x}{2}$ aur $\frac{1}{2} \tan x$ , isi order mein badhte hain. Inhe double karke,

\sin x < x < \tan x

Poori inequality ko positive number $\sin x$ se divide karo:

1 < \frac{x}{\sin x} < \frac{1}{\cos x}

Ab har (positive) quantity ko ulta karo, jisse signs palat jaate hain:

\cos x < \frac{\sin x}{x} < 1

$- \frac{π}{2} < x < 0$ ke liye kuch naya nahi chahiye: $\cos x$ aur $\frac{\sin x}{x}$ , $x$ aur $- x$ par same values lete hain.

4/4Triangle $O A T$ ka area $\frac{1}{2} \tan x$ hai.

Fig. 12.7Teen nested regions se

\sin x < x < \tan x

milta hai.NCERT Fig. 12.10 · p. 235

Centre O wale unit circle ka ek quarter, jismein horizontal radius par A = (1, 0) hai. Radius OP, OA ke saath angle x banata hai, aur A par dashed tangent badhaye gaye OP ko T par milta hai. Step 1 set-up dikhata hai, jismein P = (cos x, sin x) aur AT = tan x hai. Step 2 triangle OAP ko shade karta hai, area ½ · 1 · sin x. Step 3 sector OAP ko shade karta hai, area ½ x. Step 4 bade triangle OAT ko shade karta hai, area ½ tan x, aur sin x < x < tan x likhta hai.

Theorem 12.5Do standard trigonometric limits

$x$ radians mein ho to, (i) $\lim_{x \to 0} \frac{\sin x}{x} = 1$ aur (ii) $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ .

(i) Theorem 12.4 mein floor $\cos x$ , ceiling 1 aur $g (x) = \frac{\sin x}{x}$ lo; (12.6) 0 ke paas $g$ ko inke beech rakhta hai. $x \to 0$ par $\cos x \to 1$ , aur ceiling pehle se hi 1 hai, isliye $g (x) \to 1$ .

(ii) Upar aur neeche dono ko $1 + \cos x$ se multiply karo, jo 0 ke paas 2 ke kareeb hai, aur $1 - \cos^{2} x = \sin^{2} x$ use karo. Quotient $\frac{\sin x}{x}$ aur $\frac{\sin x}{1 + \cos x}$ ka product ban jaata hai. $x \to 0$ par pehla factor 1 ki taraf jaata hai aur doosra $\frac{0}{2} = 0$ ki taraf, isliye product 0 ki taraf jaata hai.

\lim_{x \to 0} \frac{\sin x}{x} = 1

(12.7)

\lim_{x \to 0} \frac{1 - \cos x}{x} = 0

(12.8)

(12.7) ko dekhne ka ek badhiya tareeka hai. Radius $r$ wale circle par, centre par angle $θ$ , length $r θ$ ka arc aur length $2 r \sin \frac{θ}{2}$ ki chord kaatta hai. Inka ratio $\frac{\sin (θ / 2)}{θ / 2}$ hai, jo (12.7) ke hisaab se 1 ki taraf jaata hai: chhote angles ke liye chord aur arc lagbhag same length ke hote hain. Ise Fig. 12.8 mein try karo.

SimulationAngle chhota hone par arc aur chord

Centre O wala ek circle, jis par do draggable points A aur B hain, jo radii se O se jude hain; shuru mein yeh 120 degrees door hain. Panel angle AOB degrees mein, arc AB ki length aur chord AB ki length centimetres mein, aur radius dikhata hai, jise ek handle se badla ja sakta hai. Points ko paas drag karne par arc aur chord ki lengths paas aati jaati hain, jab tak woh ek decimal place tak barabar na ho jaayein.

Fig. 12.8

A

aur

B

ko ek doosre ki taraf drag karo aur arc

A B

ko chord

A B

se compare karo. Inka ratio

\frac{\sin (θ / 2)}{θ / 2}

hai, jo 1 ki taraf jaata hai.

Try karo

5 cm ke circle par 120° par arc lagbhag 10.5 cm aur chord lagbhag 8.7 cm hai. Angle ko 30° tak chhota karo. Ab dono readouts mein kitna fark hai?
Radius ko jitna ho sake bada karo aur angle ko phir se chhota karo. Kya chord-to-arc ratio radius par depend karta hai?

Example 12.4NCERT Example 4 · p. 236

Standard pattern banana

Evaluate karo (i) $\lim_{x \to 0} \frac{\sin 5 x}{\sin 3 x}$ , (ii) $\lim_{x \to 0} \frac{\tan 3 x}{x}$ aur (iii) $\lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}}$ .

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

Practice · 22 questions

Practice 12.1: answers ke saath

1
Short answerEasyNCERT Exercise 12.1, Q1–Q3
Har limit evaluate karo.
1. $\lim_{x \to 4} (2 x - 5)$
  Solution dikhao
  Answer3
  Yeh polynomial hai, isliye substitute karo: $2 (4) - 5 = 3$ .
2. $\lim_{x \to π / 2} (\frac{2 x}{π} + 1)$
  Solution dikhao
  Answer2
  Substitute karo: $\frac{2}{π} \cdot \frac{π}{2} + 1 = 2$ .
3. $\lim_{r \to 3} \frac{4}{3} π r^{3}$ (ball ka volume jab uska radius 3 ki taraf jaata hai)
  Solution dikhao
  Answer $36 π$
  Substitute karo: $\frac{4}{3} π \cdot 27 = 36 π$ .
2
Short answerEasyNCERT Exercise 12.1, Q4–Q5
Har limit evaluate karo.
1. $\lim_{x \to 3} \frac{5 x - 1}{x + 1}$
  Solution dikhao
  Answer $\frac{7}{2}$
  $x = 3$ par denominator $4 \neq 0$ hai, isliye (12.4) se limit $\frac{14}{4} = \frac{7}{2}$ hai.
2. $\lim_{x \to - 1} \frac{x^{6} - x^{3} + 1}{x - 2}$
  Solution dikhao
  Answer−1
  Denominator $- 3 \neq 0$ hai. Numerator $1 + 1 + 1 = 3$ hai, isliye limit $\frac{3}{- 3} = - 1$ hai.
Short answerMediumNCERT Exercise 12.1, Q6
$\lim_{x \to 0} \frac{(x + 2)^{4} - 16}{x}$ evaluate karo.
Solution dikhao
Answer32
$y = x + 2$ rakho; tab $x \to 0$ par $y \to 2$ , aur $x = y - 2$ :
$\frac{(x + 2)^{4} - 16}{x} = \frac{y^{4} - 2^{4}}{y - 2}$
(12.5) se limit $4 \cdot 2^{3} = 32$ hai. ( $(x + 2)^{4}$ ko expand karke $x$ cancel karne par bhi yahi jawab milta hai.)
4
Short answerMediumNCERT Exercise 12.1, Q7–Q8
Har limit evaluate karo.
1. $\lim_{x \to 3} \frac{2 x^{2} - 5 x - 3}{x^{2} - 9}$
  Solution dikhao
  Answer $\frac{7}{6}$
  $2 x^{2} - 5 x - 3 = (2 x + 1) (x - 3)$ aur $x^{2} - 9 = (x - 3) (x + 3)$ . $x - 3$ cancel karne par $\frac{2 x + 1}{x + 3}$ bachta hai, jo $\frac{7}{6}$ ki taraf jaata hai.
2. $\lim_{x \to 2} \frac{x^{4} - 16}{3 x^{2} - 4 x - 4}$
  Solution dikhao
  Answer4
  $x^{4} - 16 = (x - 2) (x + 2) (x^{2} + 4)$ aur $3 x^{2} - 4 x - 4 = (3 x + 2) (x - 2)$ . $x - 2$ cancel karne par $\frac{(x + 2) (x^{2} + 4)}{3 x + 2}$ bachta hai, jo $\frac{4 \cdot 8}{8} = 4$ ki taraf jaata hai.
5
Short answerMediumNCERT Exercise 12.1, Q9, Q11
Har limit evaluate karo; letters constants hain.
1. $\lim_{x \to 0} \frac{a x^{2} + b x + c}{p x + q}$ , jahan $q \neq 0$
  Solution dikhao
  Answer $\frac{c}{q}$
  Denominator $q \neq 0$ ki taraf jaata hai, isliye substitute karo: limit $\frac{c}{q}$ hai.
2. $\lim_{x \to 2} \frac{x^{2} + a x}{x^{2} + b x}$ , jahan $b \neq - 2$
  Solution dikhao
  Answer $\frac{2 + a}{2 + b}$
  $x = 2$ par denominator $4 + 2 b = 2 (2 + b) \neq 0$ hai. Substitute karne par $\frac{4 + 2 a}{4 + 2 b} = \frac{2 + a}{2 + b}$ milta hai.
Short answerMediumNCERT Exercise 12.1, Q10
$\lim_{x \to 16} \frac{x^{1 / 4} - 2}{x^{1 / 2} - 4}$ evaluate karo.
Solution dikhao
Answer $\frac{1}{4}$
Upar aur neeche dono ko $x - 16$ se divide karo aur $a = 16$ ke saath (12.5) use karo:
$\begin{aligned} \frac{x^{1 / 4} - 16^{1 / 4}}{x^{1 / 2} - 16^{1 / 2}} & \to \frac{\frac{1}{4} \cdot 16^{- 3 / 4}}{\frac{1}{2} \cdot 16^{- 1 / 2}} \\ = \frac{1 / 32}{1 / 8} = \frac{1}{4} \end{aligned}$
Check: $t = x^{1 / 4}$ ke saath quotient $\frac{t - 2}{t^{2} - 4} = \frac{1}{t + 2}$ hai, jo $t \to 2$ par $\frac{1}{4}$ ki taraf jaata hai.
Short answerMediumNCERT Exercise 12.1, Q12
$\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ evaluate karo.
Solution dikhao
Answer $- \frac{1}{9}$
Numerator $\frac{3 - x}{3 x} = - \frac{x - 3}{3 x}$ hai. $x - 3$ se divide karne par ( $x \neq 3$ ke liye allowed) $- \frac{1}{3 x}$ bachta hai, jo $- \frac{1}{9}$ ki taraf jaata hai.
8
Short answerMediumNCERT Exercise 12.1, Q13–Q14
Har limit evaluate karo.
1. $\lim_{x \to 0} \frac{\sin π x}{2 x}$
  Solution dikhao
  Answer $\frac{π}{2}$
  $\frac{\sin π x}{2 x} = \frac{π}{2} \cdot \frac{\sin π x}{π x} \to \frac{π}{2}$ .
2. $\lim_{x \to 0} \frac{\sin a x}{\tan b x}$ , jahan $a, b \neq 0$
  Solution dikhao
  Answer $\frac{a}{b}$
  Ise $\frac{a}{b} \cdot \frac{\sin a x}{a x} \cdot \frac{b x}{\tan b x}$ likho. Dono fractions 1 ki taraf jaate hain (Example 12.4), isliye limit $\frac{a}{b}$ hai.
9
Short answerMediumNCERT Exercise 12.1, Q15–Q16
Har limit evaluate karo.
1. $\lim_{x \to π / 2} \frac{\cos x}{\frac{π}{2} - x}$
  Solution dikhao
  Answer1
  $y = \frac{π}{2} - x$ rakho, to $y \to 0$ aur $\cos x = \sin y$ . Quotient $\frac{\sin y}{y} \to 1$ hai.
2. $\lim_{x \to 0} \frac{1 + \cos x}{π + x}$
  Solution dikhao
  Answer $\frac{2}{π}$
  Yahan kuch bhi 0 nahi hai: denominator $π$ ki taraf jaata hai. Substitute karo: $\frac{1 + 1}{π} = \frac{2}{π}$ . Har trigonometric limit ko (12.7) ki zaroorat nahi hoti.
10
Short answerHardNCERT Exercise 12.1, Q17–Q18
Har limit evaluate karo.
1. $\lim_{x \to 0} \frac{1 - \cos 6 x}{1 - \cos 2 x}$
  Solution dikhao
  Answer9
  $1 - \cos 6 x = 2 \sin^{2} 3 x$ aur $1 - \cos 2 x = 2 \sin^{2} x$ , isliye quotient ${(\frac{\sin 3 x}{\sin x})}^{2}$ hai. Example 12.4 ki tarah $\frac{\sin 3 x}{\sin x} \to 3$ , isliye limit 9 hai.
2. $\lim_{x \to 0} \frac{2 x + x \cos 2 x}{3 \sin x}$
  Solution dikhao
  Answer1
  $x$ bahar nikalo: quotient $\frac{2 + \cos 2 x}{3} \cdot \frac{x}{\sin x}$ hai. Yeh $\frac{3}{3} \cdot 1 = 1$ ki taraf jaata hai.
11
Short answerMediumNCERT Exercise 12.1, Q19–Q21
Har limit evaluate karo.
1. $\lim_{x \to π} x \cos x$
  Solution dikhao
  Answer $- π$
  Yahan $\frac{0}{0}$ nahi hai: substitute karke $π \cos π = - π$ milta hai.
2. $\lim_{x \to 0} \frac{\sin 2 x + 4 x}{x + \sin 3 x}$
  Solution dikhao
  Answer $\frac{3}{2}$
  Upar aur neeche dono ko $x$ se divide karo: $\frac{\frac{\sin 2 x}{x} + 4}{1 + \frac{\sin 3 x}{x}}$ . Kyunki $\frac{\sin 2 x}{x} \to 2$ aur $\frac{\sin 3 x}{x} \to 3$ , limit $\frac{2 + 4}{1 + 3} = \frac{3}{2}$ hai.
3. $\lim_{x \to 0} \frac{\sec x - 1}{x}$
  Solution dikhao
  Answer0
  $\sec x - 1 = \frac{1 - \cos x}{\cos x}$ , isliye (12.8) se quotient $\frac{1 - \cos x}{x} \cdot \frac{1}{\cos x} \to 0 \cdot 1 = 0$ .
Short answerHardNCERT Exercise 12.1, Q22
$\lim_{x \to π / 4} \frac{1 - \tan x}{x - \frac{π}{4}}$ evaluate karo.
Solution dikhao
Answer−2
$y = x - \frac{π}{4}$ rakho, to $y \to 0$ . Tab $\tan x = \frac{1 + \tan y}{1 - \tan y}$ aur
$1 - \tan x = \frac{- 2 \tan y}{1 - \tan y}$
Quotient $\frac{- 2}{1 - \tan y} \cdot \frac{\tan y}{y} \to \frac{- 2}{1} \cdot 1 = - 2$ ban jaata hai.
13
Short answerMediumNCERT Exercise 12.1, Q23–Q24
Tukdon (pieces) mein defined yeh functions: inki limits nikalo.
1. $x < 1$ ke liye $f (x) = 4 - x^{2}$ aur $x \geq 1$ ke liye $f (x) = 2 x + 1$ . $\lim_{x \to 0} f (x)$ aur $\lim_{x \to 1} f (x)$ nikalo.
  Solution dikhao
  Answer4 aur 3
  0 ke paas sirf pehla piece use hota hai, isliye limit $4 - 0 = 4$ hai.
  1 par: left se $4 - 1 = 3$ , right se $2 + 1 = 3$ . Dono barabar hain, isliye $\lim_{x \to 1} f (x) = 3$ .
2. $x \leq 2$ ke liye $g (x) = x^{2} + 1$ aur $x > 2$ ke liye $g (x) = 3 x - 2$ . Kya $\lim_{x \to 2} g (x)$ exist karti hai?
  Solution dikhao
  AnswerNahi
  Left se $g (x) \to 5$ ; right se $g (x) \to 4$ . Dono alag hain, isliye limit exist nahi karti.
14
Short answerMediumNCERT Exercise 12.1, Q25–Q27
Har limit nikalo, ya dikhao ki woh exist nahi karti.
1. $\lim_{x \to 3} \frac{∣ x - 3 ∣}{x - 3}$
  Solution dikhao
  Answerexist nahi karti
  $x > 3$ ke liye quotient $1$ hai; $x < 3$ ke liye $- 1$ . One-sided limits $- 1$ aur $1$ hain, isliye koi limit nahi hai.
2. $\lim_{x \to - 4} (∣ x ∣ + 2 x)$
  Solution dikhao
  Answer−4
  $- 4$ ke paas $∣ x ∣ = - x$ , isliye function $x$ hai, jo $- 4$ ki taraf jaata hai. (Ya substitute karo: $4 - 8 = - 4$ .)
3. $\lim_{x \to 0} \frac{∣ x ∣ + x^{2}}{x}$
  Solution dikhao
  Answerexist nahi karti
  $x > 0$ ke liye yeh $1 + x \to 1$ hai; $x < 0$ ke liye $- 1 + x \to - 1$ . One-sided limits alag hain, isliye koi limit nahi hai.
Short answerMediumNCERT Exercise 12.1, Q28
Maan lo $x < 2$ ke liye $f (x) = a x + b$ , $f (2) = 6$ , aur $x > 2$ ke liye $f (x) = a x^{2} - b$ . $a$ aur $b$ aise chuno ki 2 par $f$ ki limit $f (2)$ ke barabar ho.
Solution dikhao
Answer $a = 2$ , $b = 2$
Left-hand limit $2 a + b$ ; right-hand limit $4 a - b$ . Dono 6 ke barabar honi chahiye.
$2 a + b = 6$ aur $4 a - b = 6$ ko jodne par $6 a = 12$ milta hai, isliye $a = 2$ aur $b = 2$ .
16
Short answerMediumNCERT Exercise 12.1, Q29
Maan lo $f (x) = (x - 1) (x - 2) (x - 3) (x - 4)$ .
1. $\lim_{x \to 3} f (x)$ nikalo.
  Solution dikhao
  Answer0
  Substitute karo: ek factor $3 - 3 = 0$ hai, isliye limit 0 hai.
2. $\lim_{x \to 5} f (x)$ nikalo.
  Solution dikhao
  Answer24
  $4 \cdot 3 \cdot 2 \cdot 1 = 24$ .
3. $\lim_{x \to 3} \frac{f (x)}{x - 3}$ nikalo.
  Solution dikhao
  Answer−2
  $x - 3$ cancel karo: quotient $(x - 1) (x - 2) (x - 4) \to 2 \cdot 1 \cdot (- 1) = - 2$ hai.
Short answerMediumNCERT Exercise 12.1, Q30
Maan lo $x < 1$ ke liye $f (x) = x + 2$ , $f (1) = 0$ , aur $x > 1$ ke liye $f (x) = 4 - x$ . Kin points $a$ par $f$ ki limit hai?
Solution dikhao
Answerhar real a
$a \neq 1$ ke liye $a$ ke paas $f$ ek hi polynomial se diya gaya hai, isliye limit exist karti hai.
1 par left-hand limit $1 + 2 = 3$ hai aur right-hand limit $4 - 1 = 3$ . Dono barabar hain, isliye wahan bhi limit exist karti hai aur 3 ke barabar hai, jabki $f (1) = 0$ hai.
Limit har real $a$ ke liye exist karti hai.
18
Short answerMediumNCERT Exercise 12.1, Q31
Ek function $\lim_{x \to 2} \frac{f (x) - 5}{x - 2} = 3$ satisfy karta hai.
1. $\lim_{x \to 2} f (x)$ nikalo.
  Solution dikhao
  Answer5
  $f (x) - 5 = \frac{f (x) - 5}{x - 2} \cdot (x - 2) \to 3 \cdot 0 = 0$ , isliye $f (x) \to 5$ .
2. $\lim_{x \to 2} \frac{f (x) - 5}{x^{2} - 4}$ nikalo.
  Solution dikhao
  Answer $\frac{3}{4}$
  $\frac{f (x) - 5}{x^{2} - 4} = \frac{f (x) - 5}{x - 2} \cdot \frac{1}{x + 2} \to 3 \cdot \frac{1}{4} = \frac{3}{4}$ .
Short answerHardNCERT Exercise 12.1, Q32
Maan lo $x < 0$ ke liye $f (x) = m x + n$ , $0 \leq x \leq 1$ ke liye $f (x) = n x^{2} + 3$ , aur $x > 1$ ke liye $f (x) = m x + 1$ . Agar $\lim_{x \to 0} f (x)$ aur $\lim_{x \to 1} f (x)$ dono exist karti hain to $m$ aur $n$ nikalo.
Solution dikhao
Answer $m = 5$ , $n = 3$
0 par: left-hand limit $n$ hai aur right-hand limit $3$ , isliye $n = 3$ .
1 par: left-hand limit $n + 3 = 6$ hai aur right-hand limit $m + 1$ , isliye $m = 5$ .
Quick answerMedium
(Entrance-style) $\lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}}$ barabar hai
0 $\frac{1}{2}$ 1 $\frac{3}{2}$
Answer dekho
Answer (B) $\frac{1}{2}$
$\tan x - \sin x = \tan x (1 - \cos x)$ , isliye quotient $\frac{\tan x}{x} \cdot \frac{1 - \cos x}{x^{2}} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ hai. Ise $\frac{\tan x}{x^{3}} - \frac{\sin x}{x^{3}}$ mein todna fail hota hai kyunki har piece bina bound ke badhta hai.
Quick answerMedium
(Assertion–reason) A: $\lim_{x \to 0} \frac{∣ x ∣}{x}$ exist nahi karti. R: 0 par $\frac{∣ x ∣}{x}$ ki left-hand aur right-hand limits $- 1$ aur $1$ hain.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
$x < 0$ ke liye quotient $- 1$ hai aur $x > 0$ ke liye $1$ , isliye R true hai. Alag one-sided limits hi woh kaaran hain jisse limit exist nahi karti, isliye R, A ko explain karta hai.
Quick answerHard
(Entrance-style) $\lim_{x \to 1} \frac{x + x^{2} + x^{3} + \dots + x^{10} - 10}{x - 1}$ evaluate karo.
Answer dekho
Answer 55
Quotient ko $\sum_{k = 1}^{10} \frac{x^{k} - 1}{x - 1}$ mein todo. (12.5) se har term $k$ ki taraf jaata hai, isliye limit $1 + 2 + \dots + 10 = 55$ hai.

12.5 Derivatives

Section 12.2 mein train ki ek pal ki speed average speeds ki limit ke roop mein nikli. Sawaal yeh quantity abhi kitni tezi se badal rahi hai? har jagah aata hai: tablet lene ke ek ghante baad patient ka bukhar kitni jaldi utarta hai, bowler ke haath se chhootte waqt cricket ball kitni tez chal rahi hai, ek aur unit banane par factory ki cost kitni badhti hai. Har ek ka jawab isi tarah ki limit se milta hai.

Definition

Ek point par derivative

Maan lo $f$ ek real function hai aur $a$ uske domain ka ek point. Jab limit $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ exist karti hai, to hum use $a$ par $f$ ka derivative kehte hain aur $f^{'} (a)$ likhte hain.

Example $s = 0.5 t^{2}$ ke liye $t = 10$ par limit 10 hai, yaani train ki speed.

$a$ par derivative

\begin{aligned} f^{'} (a) = \\ \lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \end{aligned}

(12.9)

Limit ke andar wala fraction $a$ aur $a + h$ ke beech $f$ ka average rate of change hai, jise difference quotient kehte hain. Positive $h$ , $a$ ke right ki taraf dekhta hai aur negative $h$ left ki taraf, isliye definition ke liye dono one-sided limits ka barabar hona zaroori hai. $f^{'} (a)$ ko hum ' $f$ dash $a$ ' ya ' $f$ prime of $a$ ' padhte hain.

Example 12.5NCERT Example 5 · p. 240

Ek seedhi line

$x = 4$ par $f (x) = 5 - 2 x$ ka derivative nikalo.

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

Example 12.6NCERT Example 6 · p. 240

Do points par ek quadratic

$f (x) = 3 x^{2} - 4 x + 1$ ke liye $f^{'} (2)$ aur $f^{'} (- 1)$ nikalo, aur dikhao ki $5 f^{'} (2) + 4 f^{'} (- 1) = 0$ .

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

Example 12.7NCERT Example 7 · p. 241

Zero par cosine

$x = 0$ par $\cos x$ ka derivative nikalo.

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

Example 12.8NCERT Example 8 · p. 241

Ek constant function

$x = 5$ aur $x = - 2$ par $f (x) = - 7$ ka derivative nikalo.

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

Tangent ke slope ki tarah derivative

Chapter 9 mein do points se guzarne wali line ka slope rise over run tha. $y = f (x)$ ke graph par $P (a, f (a))$ ko $Q (a + h, f (a + h))$ se jodne wali chord ke liye run $h$ hai aur rise $f (a + h) - f (a)$ : yeh Fig. 12.9 ke right triangle ki legs $P M$ aur $M Q$ hain. Isliye difference quotient theek chord $P Q$ ka slope hai.

Ab $h$ ko chhota karo. $Q$ curve par fisal kar $P$ tak aata hai, aur chord $P$ ke around ghoomti hai jab tak woh $P$ par tangent ke saath na aa jaaye. Slopes bhi saath chalte hain, isliye derivative tangent ka slope hai:

f^{'} (a) = \tan ψ

jahan

$ψ$: $P$ par tangent positive $x$ -axis ke saath jo angle banata hai

4/4 $P$ par tangent: slope $- 1.5 = \tan ψ$ .

Fig. 12.9Jaise

Q

fisal kar

P

tak aata hai, chord

P Q

P

par tangent ban jaati hai.NCERT Fig. 12.11 · p. 241

Lagbhag x = 1 se 6.5 tak girti hui curve y = 6/x, jismein P(2, 3) mark hai. Steps 1 se 3 mein ek dashed chord P ko x = 5, 3.5 aur 2.5 par Q se jodti hai, aur uske neeche right triangle PMQ bana hai: horizontal side PM ki length h hai aur vertical side MQ neeche jaati hai. Panel chord slopes −0.6, −0.86 aur −1.2 deta hai. Step 4 mein chord ki jagah P par solid tangent aata hai, jo x-axis ko x = 4 par kaatta hai; ek arc positive x-axis ke saath uska angle ψ mark karta hai, aur panel slope −1.5 = tan ψ deta hai.

Girti hui curve ka derivative negative hota hai, aur tab $ψ$ obtuse hota hai. Yeh Section 12.2 ki do pictures ko jodta hai: rate of change aur tangent ka slope ek hi number hain.

Derivative function

$a$ ko fixed rakhne ki koi majboori nahi hai. Har us point par $f^{'}$ nikalne se jahan limit exist karti hai, ek naya function banta hai.

Definition

Derivative (first principle se)

Real function $f$ ke liye derivative woh function $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ hai, jo har us $x$ par defined hai jahan yeh limit exist karti hai. Seedhe is limit se kaam karne ko first principle se derivative nikalna kehte hain.

first principle

\begin{aligned} f^{'} (x) = \\ \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \end{aligned}

(12.10)

Kai notations use hote hain. Agar $y = f (x)$ , to uska derivative $f^{'} (x)$ , $\frac{d}{d x} f (x)$ , $\frac{d y}{d x}$ ya $D (f (x))$ likha ja sakta hai, jise ' $x$ ke respect mein $f (x)$ ka, ya $y$ ka, derivative' padhte hain. $x = a$ par iski value $f^{'} (a)$ , ${\frac{d f}{d x} ∣}_{x = a}$ ya ${(\frac{d y}{d x})}_{x = a}$ likhi jaati hai. Derivative nikalne ko differentiation (अवकलन) kehte hain.

Fig. 12.10 $f (x) = \frac{x^{3}}{3} - x$ ke liye $f^{'}$ ko point by point banata hai. Jahan $f$ chadhta hai, $f^{'}$ positive hai. Do level jagahon, $x = \pm 1$ , par $f^{'}$ 0 hai, aur jahan $f$ girta hai, $f^{'}$ negative hai. Section 12.5.2 ke baad tum check kar sakte ho ki $f^{'} (x) = x^{2} - 1$ .

4/4Slopes parabola $y = x^{2} - 1$ banate hain.

Fig. 12.10

f

ka har tangent slope

f^{'}

ke graph par ek point ban jaata hai.

Do panels ek hi x-axis share karte hain. Upar wala panel y = f(x) = x³/3 − x dikhata hai, jo x = −1 par peak tak chadhta hai, origin se hokar x = 1 par trough tak girta hai, phir chadhta hai. Neeche wale panel mein y = f′(x) aayega. Step 1 x = −1 par level tangent banata hai aur uske neeche height 0 par ek diamond plot karta hai. Step 2 x = 0 par slope −1 wala tangent banata hai aur −1 par diamond plot karta hai. Step 3 x = 2 par slope 3 wala steep tangent banata hai aur 3 par diamond plot karta hai. Step 4 x = −2 aur x = 1 par diamonds jodta hai aur paanchon ko dashed parabola y = x² − 1 se jodta hai.

Example 12.9NCERT Example 9 · p. 242

Ek linear function

$f (x) = 4 x - 9$ ka derivative nikalo.

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

Example 12.10NCERT Example 10 · p. 243

First principle se ek quadratic

First principle se $f (x) = 2 x^{2} - x$ ka derivative nikalo.

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

Example 12.11NCERT Example 11 · p. 243

Constants ka derivative 0 hota hai

Dikhao ki constant function $f (x) = c$ ka derivative 0 hai, aur isse $g (x) = \sin^{2} x + \cos^{2} x$ ko differentiate karo.

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

Example 12.12NCERT Example 12 · p. 243

Ek reciprocal

$f (x) = \frac{6}{x}$ , $x \neq 0$ , ka derivative nikalo, aur Fig. 12.9 mein tangent slope check karo.

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

Quick check

3 mein se 1

Definition use karke

f (x) = x^{2} - 3 x

ke liye

f^{'} (4)

nikalo.

Answer dekho

Answer 5

f (4) = 4

aur

f (4 + h) = 4 + 5 h + h^{2}

, isliye difference quotient

5 + h \to 5

hai. Value

f (4) = 4

graph ki height hai, uska slope nahi.

12.5.1 Functions ke derivative ka algebra

Har derivative difference quotients ki limit hai, isliye Theorem 12.1 ko derivatives par bhi lagaya ja sakta hai. Neeche $u$ aur $v$ , $x$ ke functions hain jinke ek common domain par derivatives $u^{'}$ aur $v^{'}$ hain.

Theorem 12.6Differentiation ke rules

(i) $(u + v)^{'} = u^{'} + v^{'}$ ; (ii) $(u - v)^{'} = u^{'} - v^{'}$ ; (iii) $(u v)^{'} = u^{'} v + u v^{'}$ , product rule (ise Leibniz rule bhi kehte hain); (iv) jahan bhi $v \neq 0$ ho, ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ , quotient rule.

(i) aur (ii) limits ke sum aur difference rules se seedhe nikal aate hain. (iii) ke liye $u (x)$ , $v (x)$ ko $u$ , $v$ likho aur $u (x + h)$ , $v (x + h)$ ko $u_{h}$ , $v_{h}$ . $u v$ ke difference quotient ke numerator mein $u_{h} v$ jodne aur ghatane par milta hai

\begin{aligned} \frac{u_{h} v_{h} - u v}{h} \\ = u_{h} \cdot \frac{v_{h} - v}{h} \\ + v \cdot \frac{u_{h} - u}{h} \end{aligned}

$h \to 0$ par dono fractions $v^{'}$ aur $u^{'}$ ki taraf jaate hain. Saath hi $u_{h} \to u$ , kyunki $u_{h} - u = h \cdot \frac{u_{h} - u}{h} \to 0 \cdot u^{'} = 0$ . Isliye limit $u v^{'} + u^{'} v$ hai.

Rule (iv) bhi isi tarah, ek suitable term jod kar aur ghata kar, prove kiya ja sakta hai.

product rule

(u v)^{'} = u^{'} v + u v^{'}

(12.11)

quotient rule

{(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}

(12.12)

2/3Naya area do strips plus ek corner hai.

Fig. 12.11Jab

u

aur

v

thoda badhte hain, to area

u v

mein do strips aur ek chhota sa corner judta hai.

Width u aur height v wale rectangle ka area uv hai. Step 1 sirf yeh rectangle dikhata hai. Step 2 mein u, Δu se aur v, Δv se badhta hai: right mein area v Δu ki ek strip aati hai, upar area u Δv ki ek dashed strip, aur area Δu Δv ka ek chhota corner. Step 3 chhote change ke liye patli strips dikhata hai; h se divide karke h → 0 karne par strips v u′ + u v′ deti hain, jabki corner wala term Δu · v′ 0 ki taraf jaata hai.

Shabdon mein: pehle ka derivative guna doosra, plus pehla guna doosre ka derivative. Quotient rule mein order yaad rakho, $u^{'} v$ pehle, aur denominator ka square karo. Ek aam galti hai $(u v)^{'} = u^{'} v^{'}$ ; Fig. 12.11 dikhata hai ki yeh sahi kyun nahi ho sakta, kyunki area mein do strips judti hain, sirf ek corner nahi.

Kuch pehle nateeje. Function $x$ ka derivative 1 hai, kyunki $\frac{(x + h) - x}{h} = 1$ . Constant $c$ ke liye product rule $(c u)^{'} = 0 \cdot u + c u^{'} = c u^{'}$ deta hai: constants derivatives se bahar aa jaate hain. $x^{2} = x \cdot x$ par product rule lagane se $1 \cdot x + x \cdot 1 = 2 x$ milta hai, aur $x^{3} = x^{2} \cdot x$ par lagane se $2 x \cdot x + x^{2} \cdot 1 = 3 x^{2}$ . Pattern aage bhi chalta hai.

Theorem 12.7Power rule

Har positive integer $n$ ke liye, $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ .

power rule

\frac{d}{d x} (x^{n}) = n x^{n - 1}

(12.13)

Power rule har real power $n$ ke liye sahi hai; ise hum bina proof use karte hain. $n = - 1$ ke saath yeh $\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x^{2}}$ deta hai, jo Example 12.12 se match karta hai, aur $n = \frac{1}{2}$ ke saath $\frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}}$ deta hai.

12.5.2 Polynomials aur trigonometric functions ka derivative

Theorem 12.8Polynomial ka derivative

Agar $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ , to $f^{'} (x) = n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \dots + 2 a_{2} x + a_{1}$ .

Example 12.13NCERT Example 13 · p. 246

Har term alag se

$4 x^{25} - 3 x^{10} + 7 x - 2$ ko differentiate karo.

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

Example 12.14NCERT Example 14 · p. 247

Do points par ek lamba polynomial

$f (x) = x + x^{2} + x^{3} + \dots + x^{20}$ ke liye $f^{'} (1)$ aur $f^{'} (- 1)$ nikalo.

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

Example 12.15NCERT Example 15 · p. 247

Quotient rule

$f (x) = \frac{2 x - 1}{x + 3}$ , $x \neq - 3$ , ko differentiate karo.

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

Example 12.16NCERT Example 16 · p. 247

sin x ka derivative

First principle se dikhao ki $\frac{d}{d x} (\sin x) = \cos x$ .

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

\frac{d}{d x} (\sin x) = \cos x

(12.14)

Yahi method, $\cos (x + h) = \cos x \cos h - \sin x \sin h$ se shuru karke, saathi result prove karta hai (Practice 12.2, Q9):

\frac{d}{d x} (\cos x) = - \sin x

(12.15)

4/4Plot kiye gaye slopes $y = \cos x$ banate hain.

Fig. 12.12Sine curve ke slopes cosine curve banate hain.NCERT Example 16 · p. 247

0 se 2π tak curve y = sin x. Step 1 origin par slope 1 wala tangent banata hai aur x = 0 ke upar height 1 par ek diamond plot karta hai. Step 2 crest x = π/2 par level tangent banata hai aur height 0 par diamond plot karta hai. Step 3 x = π par slope −1 wala tangent banata hai aur −1 par diamond plot karta hai. Step 4 3π/2 (height 0) aur 2π (height 1) par diamonds jodta hai aur saare diamonds ko ek dashed curve, y = cos x, se jodta hai.

Example 12.17NCERT Example 17 · p. 247

tan x ka derivative

First principle se dikhao ki jahan bhi $\cos x \neq 0$ , wahan $\frac{d}{d x} (\tan x) = \sec^{2} x$ .

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

\frac{d}{d x} (\tan x) = \sec^{2} x

(12.16)

Example 12.18NCERT Example 18 · p. 248

Sine ke saath product rule

$f (x) = x^{2} \sin x$ ko differentiate karo, aur $f^{'} (π)$ nikalo.

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

Table 12.4 is chapter ke standard derivatives ek jagah rakhti hai. Aakhri teen quotient rule se aate hain, Example 12.21 aur Practice 12.2 mein.

Table 12.4Standard derivatives

$f (x)$	$f^{'} (x)$	Kahan mila
$x^{n}$	$n x^{n - 1}$	Theorem 12.7
$\sin x$	$\cos x$	Example 12.16
$\cos x$	$- \sin x$	Practice 12.2
$\tan x$	$\sec^{2} x$	Example 12.17
$\cot x$	$- {cosec}^{2} x$	Example 12.21
$\sec x$	$\sec x \tan x$	Practice 12.2
$cosec x$	$- cosec x \cot x$	Practice 12.2

Quick check

4 mein se 1

\frac{d}{d x} (x^{- 3})

hai

- 3 x^{- 2}

- 3 x^{- 4}

3 x^{- 4}

- \frac{1}{3} x^{- 4}

Answer dekho

Answer (B) $- 3 x^{- 4}$

n = - 3

ke saath power rule:

- 3 x^{- 3 - 1} = - 3 x^{- 4}

. Power ko 1 kam karne par

- 3

- 4

ban jaata hai,

- 2

nahi.

Practice · 14 questions

Practice 12.2: answers ke saath

1
Short answerEasyNCERT Exercise 12.2, Q1–Q3
Har derivative nikalne ke liye definition (12.9) use karo.
1. $x = - 3$ par $f (x) = x^{2} + 5 x$
  Solution dikhao
  Answer−1
  $f (- 3 + h) - f (- 3) = (9 - 6 h + h^{2} - 15 + 5 h) - (- 6) = h^{2} - h$ . $h$ se divide karne par $h - 1 \to - 1$ milta hai.
2. $x = - 4$ par $f (x) = x$
  Solution dikhao
  Answer1
  Har $h \neq 0$ ke liye $\frac{(- 4 + h) - (- 4)}{h} = 1$ , isliye $f^{'} (- 4) = 1$ .
3. $x = 40$ par $f (x) = 25 x$
  Solution dikhao
  Answer25
  $\frac{25 (40 + h) - 25 \cdot 40}{h} = 25$ , isliye $f^{'} (40) = 25$ . Dhyaan do ki point 40 ka ismein koi role nahi hai.
2
Short answerMediumNCERT Exercise 12.2, Q4
Har function ko differentiate karne ke liye first principle (12.10) use karo.
1. $4 - x^{3}$
  Solution dikhao
  Answer $- 3 x^{2}$
  $(4 - (x + h)^{3}) - (4 - x^{3}) = - (3 x^{2} h + 3 x h^{2} + h^{3})$ . $h$ se divide karo aur $h \to 0$ hone do: $- 3 x^{2}$ .
2. $(x + 3) (x - 1)$
  Solution dikhao
  Answer $2 x + 2$
  Function $x^{2} + 2 x - 3$ hai. Difference $2 x h + h^{2} + 2 h$ hai; $h$ se divide karo aur $h \to 0$ hone do: $2 x + 2$ .
3. $\frac{1}{x^{3}}$
  Solution dikhao
  Answer $- \frac{3}{x^{4}}$
  $\frac{1}{(x + h)^{3}} - \frac{1}{x^{3}} = \frac{x^{3} - (x + h)^{3}}{x^{3} (x + h)^{3}} = \frac{- h (3 x^{2} + 3 x h + h^{2})}{x^{3} (x + h)^{3}}$ . $h$ se divide karo aur $h \to 0$ hone do: $\frac{- 3 x^{2}}{x^{6}} = - \frac{3}{x^{4}}$ .
4. $\frac{2 x + 1}{x + 1}$
  Solution dikhao
  Answer $\frac{1}{(x + 1)^{2}}$
  Common denominator $(x + h + 1) (x + 1)$ par numerator $(2 x + 2 h + 1) (x + 1) - (2 x + 1) (x + h + 1) = h$ hai. Isliye quotient $\frac{1}{(x + h + 1) (x + 1)} \to \frac{1}{(x + 1)^{2}}$ hai.
Short answerMediumNCERT Exercise 12.2, Q5
Maan lo $f (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{60}}{60}$ . Dikhao ki $f^{'} (1) = 60 f^{'} (0)$ , aur $f^{'} (- 1)$ nikalo.
Solution dikhao
Answer $f^{'} (- 1) = 0$
$f^{'} (x) = 1 + x + x^{2} + \dots + x^{59}$ , 60 terms ka sum.
$f^{'} (1) = 60$ aur $f^{'} (0) = 1$ , isliye $f^{'} (1) = 60 f^{'} (0)$ .
$f^{'} (- 1) = 1 - 1 + 1 - \dots - 1 = 0$ , kyunki 60 terms pairs mein cancel ho jaate hain.
Short answerMediumNCERT Exercise 12.2, Q6
Ek fixed real number $a$ ke liye $g (x) = x^{5} + a x^{4} + a^{2} x^{3} + a^{3} x^{2} + a^{4} x + a^{5}$ ko differentiate karo, aur $g^{'} (a)$ nikalo.
Solution dikhao
Answer $g^{'} (a) = 15 a^{4}$
$\begin{aligned} g^{'} (x) = 5 x^{4} & + 4 a x^{3} + 3 a^{2} x^{2} \\ + 2 a^{3} x + a^{4} \end{aligned}$
$x = a$ par har term $a^{4}$ ka multiple hai: $g^{'} (a) = (5 + 4 + 3 + 2 + 1) a^{4} = 15 a^{4}$ .
5
Short answerMediumNCERT Exercise 12.2, Q7
Constants $p$ aur $q$ ke liye har function ko differentiate karo.
1. $(x + p) (2 x - q)$
  Solution dikhao
  Answer $4 x + 2 p - q$
  Product rule: $1 \cdot (2 x - q) + (x + p) \cdot 2 = 4 x + 2 p - q$ .
2. $(p x^{3} - q)^{2}$
  Solution dikhao
  Answer $6 p^{2} x^{5} - 6 p q x^{2}$
  Expand karo: $p^{2} x^{6} - 2 p q x^{3} + q^{2}$ . Derivative $6 p^{2} x^{5} - 6 p q x^{2}$ hai.
3. $\frac{x + p}{x + q}$
  Solution dikhao
  Answer $\frac{q - p}{(x + q)^{2}}$
  Quotient rule: $\frac{1 \cdot (x + q) - (x + p) \cdot 1}{(x + q)^{2}} = \frac{q - p}{(x + q)^{2}}$ .
Short answerMediumNCERT Exercise 12.2, Q8
Constant $a$ ke liye $\frac{x^{4} - a^{4}}{x - a}$ ko do tareekon se differentiate karo: pehle simplify karke, aur quotient rule se.
Solution dikhao
Answer $3 x^{2} + 2 a x + a^{2}$
Simplify karne par function $x^{3} + a x^{2} + a^{2} x + a^{3}$ hai ( $x \neq a$ ke liye), jiska derivative $3 x^{2} + 2 a x + a^{2}$ hai.
Quotient rule se derivative $\frac{4 x^{3} (x - a) - (x^{4} - a^{4})}{(x - a)^{2}} = \frac{3 x^{4} - 4 a x^{3} + a^{4}}{(x - a)^{2}}$ hai.
Dono match karte hain: $(3 x^{2} + 2 a x + a^{2}) (x - a)^{2}$ ko multiply karne par $3 x^{4} - 4 a x^{3} + a^{4}$ milta hai.
7
Short answerEasyNCERT Exercise 12.2, Q9 (i)–(iii)
Har function ko differentiate karo.
1. $3 x + \frac{2}{5}$
  Solution dikhao
  Answer3
  Constant $\frac{2}{5}$ ka derivative 0 hai, isliye jawab 3 hai.
2. $(2 x^{2} - x + 4) (x + 3)$
  Solution dikhao
  Answer $6 x^{2} + 10 x + 1$
  Product rule: $(4 x - 1) (x + 3) + (2 x^{2} - x + 4) = 6 x^{2} + 10 x + 1$ .
3. $x^{- 2} (4 + 5 x)$
  Solution dikhao
  Answer $- 8 x^{- 3} - 5 x^{- 2}$
  Function $4 x^{- 2} + 5 x^{- 1}$ hai, isliye derivative $- 8 x^{- 3} - 5 x^{- 2}$ hai.
8
Short answerMediumNCERT Exercise 12.2, Q9 (iv)–(vi)
Har function ko differentiate karo.
1. $x^{4} (2 - 3 x^{- 6})$
  Solution dikhao
  Answer $8 x^{3} + 6 x^{- 3}$
  Function $2 x^{4} - 3 x^{- 2}$ hai, isliye derivative $8 x^{3} + 6 x^{- 3}$ hai.
2. $x^{- 3} (5 - 2 x^{- 4})$
  Solution dikhao
  Answer $- 15 x^{- 4} + 14 x^{- 8}$
  Function $5 x^{- 3} - 2 x^{- 7}$ hai, isliye derivative $- 15 x^{- 4} + 14 x^{- 8}$ hai.
3. $\frac{3}{x - 2} - \frac{x^{2}}{2 x + 1}$
  Solution dikhao
  Answersolution dekho
  Pehla term: $- \frac{3}{(x - 2)^{2}}$ . Doosra term, quotient rule se: $\frac{2 x (2 x + 1) - 2 x^{2}}{(2 x + 1)^{2}} = \frac{2 x^{2} + 2 x}{(2 x + 1)^{2}}$ .
  Derivative $- \frac{3}{(x - 2)^{2}} - \frac{2 x^{2} + 2 x}{(2 x + 1)^{2}}$ hai.
ProofMediumNCERT Exercise 12.2, Q10
First principle se (12.15) prove karo, phir $x = \frac{π}{6}$ par $y = \cos x$ ka slope nikalo.
Solution dikhao
Answer $- \frac{1}{2}$
$\cos (x + h) - \cos x = \cos x (\cos h - 1) - \sin x \sin h$ . $h$ se divide karne par:
$\begin{aligned} - \cos x \cdot \frac{1 - \cos h}{h} \\ - \sin x \cdot \frac{\sin h}{h} \end{aligned}$
$h \to 0$ par yeh $- \cos x \cdot 0 - \sin x \cdot 1 = - \sin x$ ki taraf jaata hai.
$x = \frac{π}{6}$ par slope $- \sin \frac{π}{6} = - \frac{1}{2}$ hai.
10
Short answerMediumNCERT Exercise 12.2, Q11 (i)–(iv)
Har function ko differentiate karne ke liye quotient ya product rule use karo.
1. $\sec x$
  Solution dikhao
  Answer $\sec x \tan x$
  $\frac{0 \cdot \cos x - 1 \cdot (- \sin x)}{\cos^{2} x} = \frac{\sin x}{\cos^{2} x} = \sec x \tan x$ .
2. $cosec x$
  Solution dikhao
  Answer $- cosec x \cot x$
  $\frac{0 - \cos x}{\sin^{2} x} = - cosec x \cot x$ .
3. $\sin x \tan x$
  Solution dikhao
  Answer $\sin x (1 + \sec^{2} x)$
  Product rule: $\cos x \tan x + \sin x \sec^{2} x = \sin x + \sin x \sec^{2} x = \sin x (1 + \sec^{2} x)$ .
11
Short answerEasyNCERT Exercise 12.2, Q11 (v)–(vii)
Har function ko differentiate karo.
1. $4 \sec x - 3 \cos x$
  Solution dikhao
  Answer $4 \sec x \tan x + 3 \sin x$
  $4 \sec x \tan x + 3 \sin x$ .
2. $2 \cot x - 3 cosec x$
  Solution dikhao
  Answersolution dekho
  $- 2 {cosec}^{2} x + 3 cosec x \cot x$ .
3. $3 \cos x - 4 \sin x + 9$
  Solution dikhao
  Answer $- 3 \sin x - 4 \cos x$
  $- 3 \sin x - 4 \cos x$ ; constant 9 se 0 aata hai.
4. $5 \tan x + 2 \sec x$
  Solution dikhao
  Answer $5 \sec^{2} x + 2 \sec x \tan x$
  $5 \sec^{2} x + 2 \sec x \tan x$ .
Quick answerMedium
(Entrance-style) Agar $y = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$ , to $x = 0$ par $\frac{d y}{d x}$ hai
01415
Answer dekho
Answer (B) 1
Multiply karke kholne par $y = 1 + x + x^{2} + \dots + x^{15}$ , isliye $y^{'} = 1 + 2 x + \dots + 15 x^{14}$ aur $y^{'} (0) = 1$ . Jawab 15, $y^{'}$ ka sabse bada coefficient hai, 0 par uski value nahi.
Quick answerEasy
(Assertion–reason) A: Har $x$ ke liye $\sin^{2} x + \cos^{2} x$ ka derivative 0 hai. R: Constant function ka derivative 0 hota hai.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko explain karta hai
$\sin^{2} x + \cos^{2} x = 1$ , ek constant, isliye A seedhe R se nikalta hai. (Term by term differentiate karne par bhi $2 \sin x \cos x - 2 \cos x \sin x = 0$ milta hai.)
Quick answerMedium
(Entrance-style) $f (x) = x^{3} - 6 x^{2} + 9 x$ ke liye equation $f^{'} (x) = 0$ ke do roots hain. Unka sum kya hai?
Answer dekho
Answer 4
$f^{'} (x) = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ , isliye roots 1 aur 3 hain, sum 4. Yahi woh jagah hain jahan $f$ ke graph ke tangents level hain.

Miscellaneous Examples

Yeh examples first principle aur rules dono ko milate hain. Jab question kahe 'first principle se', to (12.10) se kaam karo; warna jo rule sabse jaldi kaam kare woh use karo.

Example 12.19NCERT Example 19 · p. 249

First principle se do derivatives

First principle se differentiate karo (i) $f (x) = \frac{3 x - 1}{x + 2}$ , $x \neq - 2$ , aur (ii) $g (x) = \sqrt{x}$ , $x > 0$ .

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

Fig. 12.13 dikhata hai ki formula $g^{'} (x) = \frac{1}{2 \sqrt{x}}$ ka graph ke liye kya matlab hai: jaise-jaise point of contact origin ki taraf fisalta hai, $y = \sqrt{x}$ ke tangents aur steep hote jaate hain, aur khud 0 par difference quotient $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$ bina bound ke badhta hai, isliye wahan koi derivative nahi hai.

3/4 $x = \frac{1}{4}$ par slope 1 hai.

Fig. 12.130 ke paas slope

\frac{1}{2 \sqrt{x}}

bina bound ke badhta hai.

y = √x ka graph origin se upar uthta hai, pehle tezi se aur phir dheere. Step 1 x = 4 par tangent banata hai, slope 1/4. Step 2 x = 1 par tangent banata hai, slope 1/2. Step 3 x = 1/4 par tangent banata hai, slope 1. Step 4 x = 0.04 par steep tangent banata hai, slope 2.5: x ke 0 ke paas aane par slope 1/(2√x) bina bound ke badhta hai.

Example 12.20NCERT Example 20 · p. 250

First principle se trigonometric functions

First principle se differentiate karo (i) $\sin 2 x$ aur (ii) $x \cos x$ .

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

Example 12.21NCERT Example 21 · p. 251

Rules ka use

Differentiate karo (i) $\cos 2 x$ , $\cos 2 x = \cos^{2} x - \sin^{2} x$ use karke, aur (ii) $\cot x$ .

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

Example 12.22NCERT Example 22 · p. 252

Trigonometric parts wale quotients

Jahan bhi defined hon, differentiate karo (i) $\frac{x^{3} + \sin x}{\cos x}$ aur (ii) $\frac{1 + \sin x}{x + \cos x}$ .

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

Quick check

2 mein se 1

x \neq 0

ke liye

\frac{d}{d x} (\frac{\sin x}{x})

hai

\frac{\cos x}{1}

\frac{\sin x - x \cos x}{x^{2}}

\frac{x \cos x - \sin x}{x^{2}}

\frac{x \cos x + \sin x}{x^{2}}

Answer dekho

Answer (C) $\frac{x \cos x - \sin x}{x^{2}}$

Quotient rule:

\frac{(\cos x) (x) - (\sin x) (1)}{x^{2}}

. Option B mein numerator ke terms galat order mein hain, jisse sign palat jaata hai, aur option A dono derivatives ko divide karta hai.

Practice · 16 questions

Practice questions answers ke saath (NCERT “Chapter 12 par miscellaneous exercises”)

1
Short answerMediumNCERT Miscellaneous Exercise, Q1
Har function ko differentiate karne ke liye definition (12.10) se kaam karo.
1. $3 - 2 x$
  Solution dikhao
  Answer−2
  $\frac{(3 - 2 x - 2 h) - (3 - 2 x)}{h} = - 2$ .
2. $(2 x)^{- 1}$
  Solution dikhao
  Answer $- \frac{1}{2 x^{2}}$
  $\frac{1}{2 (x + h)} - \frac{1}{2 x} = \frac{- h}{2 x (x + h)}$ . $h$ se divide karo aur $h \to 0$ hone do: $- \frac{1}{2 x^{2}}$ .
3. $\cos (x + 2)$
  Solution dikhao
  Answer $- \sin (x + 2)$
  $u = x + 2$ ke saath difference $\cos (u + h) - \cos u$ hai, isliye Practice 12.2 Q9 ki tarah limit $- \sin u = - \sin (x + 2)$ hai.
4. $\sin (x - \frac{π}{6})$
  Solution dikhao
  Answer $\cos (x - \frac{π}{6})$
  $u = x - \frac{π}{6}$ ke saath Example 12.16 ke steps $\cos u = \cos (x - \frac{π}{6})$ dete hain.
2
Short answerMediumNCERT Miscellaneous Exercise, Q2–Q4
Differentiate karo; $x$ ke alawa saare letters non-zero constants hain.
1. $\frac{2 x + a}{3}$
  Solution dikhao
  Answer $\frac{2}{3}$
  $\frac{2}{3} x + \frac{a}{3}$ ka derivative $\frac{2}{3}$ hai.
2. $(a x + b) (\frac{c}{x} + d)$
  Solution dikhao
  Answer $a d - \frac{b c}{x^{2}}$
  Expand karo: $a c + a d x + \frac{b c}{x} + b d$ . Derivative $a d - \frac{b c}{x^{2}}$ hai.
3. $(p x - q)^{2} (r x + s)$
  Solution dikhao
  Answersolution dekho
  Product rule, $\frac{d}{d x} (p x - q)^{2} = 2 p (p x - q)$ ke saath: $2 p (p x - q) (r x + s) + r (p x - q)^{2}$ .
  $p x - q$ bahar nikalne par: $(p x - q) (3 p r x + 2 p s - q r)$ .
3
Short answerMediumNCERT Miscellaneous Exercise, Q5–Q7
Differentiate karo; $x$ ke alawa letters constants hain.
1. $\frac{p x + q}{r x + s}$
  Solution dikhao
  Answer $\frac{p s - q r}{(r x + s)^{2}}$
  $\frac{p (r x + s) - r (p x + q)}{(r x + s)^{2}} = \frac{p s - q r}{(r x + s)^{2}}$ .
2. $\frac{x - \frac{1}{x}}{x + \frac{1}{x}}$
  Solution dikhao
  Answer $\frac{4 x}{(x^{2} + 1)^{2}}$
  Upar aur neeche dono ko $x$ se multiply karo: function $\frac{x^{2} - 1}{x^{2} + 1}$ hai.
  Quotient rule: $\frac{2 x (x^{2} + 1) - 2 x (x^{2} - 1)}{(x^{2} + 1)^{2}} = \frac{4 x}{(x^{2} + 1)^{2}}$ .
3. $\frac{1}{p x^{2} + q}$
  Solution dikhao
  Answersolution dekho
  $\frac{0 - 2 p x}{(p x^{2} + q)^{2}} = - \frac{2 p x}{(p x^{2} + q)^{2}}$ .
4
Short answerMediumNCERT Miscellaneous Exercise, Q8–Q9
Differentiate karo; $c$ , $d$ , $p$ , $q$ , $r$ constants hain.
1. $\frac{x + c}{x^{2} + d}$
  Solution dikhao
  Answersolution dekho
  $\frac{(x^{2} + d) - (x + c) (2 x)}{(x^{2} + d)^{2}} = \frac{d - x^{2} - 2 c x}{(x^{2} + d)^{2}}$ .
2. $\frac{x^{2} + p x + q}{x + r}$
  Solution dikhao
  Answersolution dekho
  $\frac{(2 x + p) (x + r) - (x^{2} + p x + q)}{(x + r)^{2}} = \frac{x^{2} + 2 r x + p r - q}{(x + r)^{2}}$ .
5
Short answerMediumNCERT Miscellaneous Exercise, Q10–Q12
Differentiate karo; $p$ aur $q$ constants hain aur $n$ ek positive integer hai.
1. $\frac{p}{x^{3}} + \frac{q}{x} + \sin x$
  Solution dikhao
  Answersolution dekho
  $- \frac{3 p}{x^{4}} - \frac{q}{x^{2}} + \cos x$ .
2. $3 \sqrt{x} + 5$
  Solution dikhao
  Answer $\frac{3}{2 \sqrt{x}}$
  $3 \cdot \frac{1}{2} x^{- 1 / 2} = \frac{3}{2 \sqrt{x}}$ .
3. $(2 x + 1)^{n}$
  Solution dikhao
  Answer $2 n (2 x + 1)^{n - 1}$
  Claim: derivative $2 n (2 x + 1)^{n - 1}$ hai. $n = 1$ ke liye yeh 2 hai. Agar yeh $n - 1$ ke liye sahi hai, to $(2 x + 1)^{n - 1} (2 x + 1)$ par product rule $2 (n - 1) (2 x + 1)^{n - 1} + 2 (2 x + 1)^{n - 1} = 2 n (2 x + 1)^{n - 1}$ deta hai.
Short answerHardNCERT Miscellaneous Exercise, Q13
$(x + 1)^{4} (x - 1)^{3}$ ko differentiate karo aur jawab ko factorise karo.
Solution dikhao
Answer $(x + 1)^{3} (x - 1)^{2} (7 x - 1)$
Product rule: $4 (x + 1)^{3} (x - 1)^{3} + 3 (x + 1)^{4} (x - 1)^{2}$ .
$(x + 1)^{3} (x - 1)^{2}$ bahar nikalo: bracket $4 (x - 1) + 3 (x + 1) = 7 x - 1$ hai.
$(x + 1)^{3} (x - 1)^{2} (7 x - 1)$
7
Short answerMediumNCERT Miscellaneous Exercise, Q14–Q16
Har function ko differentiate karo.
1. $\cos (x + a)$ , $a$ ek constant
  Solution dikhao
  Answer $- \sin (x + a)$
  $\cos x \cos a - \sin x \sin a$ ka derivative $- \sin x \cos a - \cos x \sin a = - \sin (x + a)$ hai.
2. $\sec x \tan x$
  Solution dikhao
  Answersolution dekho
  Product rule: $\sec x \tan x \cdot \tan x + \sec x \cdot \sec^{2} x = \sec x \tan^{2} x + \sec^{3} x$ .
3. $\frac{\sin x}{1 + \cos x}$
  Solution dikhao
  Answer $\frac{1}{1 + \cos x}$
  $\frac{\cos x (1 + \cos x) + \sin^{2} x}{(1 + \cos x)^{2}} = \frac{1 + \cos x}{(1 + \cos x)^{2}} = \frac{1}{1 + \cos x}$ .
8
Short answerMediumNCERT Miscellaneous Exercise, Q17–Q18
Har function ko differentiate karo.
1. $\frac{1 + \tan x}{1 - \tan x}$
  Solution dikhao
  Answersolution dekho
  $\frac{\sec^{2} x (1 - \tan x) + (1 + \tan x) \sec^{2} x}{(1 - \tan x)^{2}} = \frac{2 \sec^{2} x}{(1 - \tan x)^{2}}$ .
2. $\frac{1 - \cos x}{1 + \cos x}$
  Solution dikhao
  Answersolution dekho
  $\frac{\sin x (1 + \cos x) + (1 - \cos x) \sin x}{(1 + \cos x)^{2}} = \frac{2 \sin x}{(1 + \cos x)^{2}}$ .
ProofHardNCERT Miscellaneous Exercise, Q19
Dikhao ki har positive integer $n$ ke liye $\frac{d}{d x} (\cos^{n} x) = - n \cos^{n - 1} x \sin x$ .
Solution dikhao
Answer $- n \cos^{n - 1} x \sin x$
$n = 1$ ke liye yeh (12.15) hai. Maan lo yeh $n - 1$ ke liye sahi hai. $\cos^{n} x = \cos^{n - 1} x \cdot \cos x$ likho aur product rule use karo:
$\begin{aligned} - (n - 1) \cos^{n - 2} x \sin x \cdot \cos x \\ + \cos^{n - 1} x \cdot (- \sin x) \end{aligned}$
Yeh $- n \cos^{n - 1} x \sin x$ hai, isliye rule $n$ ke liye bhi sahi hai.
10
Short answerHardNCERT Miscellaneous Exercise, Q20–Q21
Differentiate karo; $a$ , $p$ , $q$ , $r$ , $s$ constants hain.
1. $\frac{p + q \cos x}{r + s \sin x}$
  Solution dikhao
  Answersolution dekho
  Numerator $- q \sin x (r + s \sin x) - (p + q \cos x) s \cos x$ hai.
  $\sin^{2} x + \cos^{2} x = 1$ se yeh $- (q r \sin x + p s \cos x + q s)$ ban jaata hai; $(r + s \sin x)^{2}$ se divide karo.
2. $\frac{\cos (x + a)}{\sin x}$
  Solution dikhao
  Answer $- \frac{\cos a}{\sin^{2} x}$
  Numerator $- \sin (x + a) \sin x - \cos (x + a) \cos x = - \cos ((x + a) - x) = - \cos a$ hai.
  Derivative $- \frac{\cos a}{\sin^{2} x}$ hai.
11
Short answerMediumNCERT Miscellaneous Exercise, Q22–Q24
Har function ko differentiate karo; $a$ aur $b$ constants hain.
1. $x^{3} (2 \cos x + 3 \sin x)$
  Solution dikhao
  Answersolution dekho
  $3 x^{2} (2 \cos x + 3 \sin x) + x^{3} (3 \cos x - 2 \sin x)$ .
2. $(x^{2} - 2) \sin x$
  Solution dikhao
  Answersolution dekho
  $2 x \sin x + (x^{2} - 2) \cos x$ .
3. $(x^{2} + \cos x) (a + b \sin x)$
  Solution dikhao
  Answersolution dekho
  $(2 x - \sin x) (a + b \sin x) + b \cos x (x^{2} + \cos x)$ .
12
Short answerMediumNCERT Miscellaneous Exercise, Q25, Q29
Har product ko differentiate karo.
1. $(x + \sin x) (x + \cot x)$
  Solution dikhao
  Answersolution dekho
  $(1 + \cos x) (x + \cot x) + (x + \sin x) (1 - {cosec}^{2} x)$ .
2. $(x + \tan x) (1 - \sec x)$
  Solution dikhao
  Answersolution dekho
  $(1 + \sec^{2} x) (1 - \sec x) - (x + \tan x) \sec x \tan x$ .
13
Short answerHardNCERT Miscellaneous Exercise, Q26–Q28, Q30
Har quotient ko differentiate karo; $n$ ek positive integer hai.
1. $\frac{2 x - 3 \cos x}{5 x + 4 \sin x}$
  Solution dikhao
  Answersolution dekho
  Quotient rule ka numerator $(2 + 3 \sin x) (5 x + 4 \sin x) - (2 x - 3 \cos x) (5 + 4 \cos x)$ hai.
  Expand karke aur $\sin^{2} x + \cos^{2} x = 1$ use karke yeh $15 x \sin x - 8 x \cos x + 8 \sin x + 15 \cos x + 12$ hai; $(5 x + 4 \sin x)^{2}$ se divide karo.
2. $\frac{x^{3} \sin \frac{π}{3}}{\cos x}$
  Solution dikhao
  Answersolution dekho
  $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ ek constant hai. Derivative $\frac{\sqrt{3}}{2} \cdot \frac{3 x^{2} \cos x + x^{3} \sin x}{\cos^{2} x}$ hai.
3. $\frac{x}{1 + \cot x}$
  Solution dikhao
  Answersolution dekho
  $\frac{(1 + \cot x) - x (- {cosec}^{2} x)}{(1 + \cot x)^{2}} = \frac{1 + \cot x + x {cosec}^{2} x}{(1 + \cot x)^{2}}$ .
4. $\frac{x}{\cos^{n} x}$
  Solution dikhao
  Answersolution dekho
  Q9 se, quotient rule $\frac{\cos^{n} x + n x \cos^{n - 1} x \sin x}{\cos^{2 n} x}$ deta hai.
  $\cos^{n - 1} x$ cancel karo: $\frac{\cos x + n x \sin x}{\cos^{n + 1} x}$ .
Quick answerMedium
(Entrance-style) Agar $y = (\sin x + \cos x)^{2}$ , to $x = \frac{π}{4}$ par $\frac{d y}{d x}$ hai
012 $\sqrt{2}$
Answer dekho
Answer (A) 0
$y = 1 + \sin 2 x$ , isliye $y^{'} = 2 \cos 2 x$ , aur $\cos \frac{π}{2} = 0$ . Jawab 2, $\frac{π}{4}$ par khud $y$ ki value hai.
Quick answerMedium
(Assertion–reason) A: $x = 0$ par $x \sin x$ ka derivative 0 hai. R: Product ka derivative derivatives ka product hota hai.
A aur R dono true hain, aur R, A ko explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai, lekin R false haiA false hai, lekin R true hai
Answer dekho
Answer (C) A true hai, lekin R false hai
Product rule se derivative $\sin x + x \cos x$ hai, jo $x = 0$ par 0 hai, isliye A true hai. R false hai: $(u v)^{'} = u^{'} v + u v^{'}$ , $u^{'} v^{'}$ nahi.
Quick answerMedium
(Entrance-style) $f (x) = (x + 1) (x + 2) (x + 3)$ ke liye $f^{'} (0)$ nikalo.
Answer dekho
Answer 11
$f^{'} (x) = (x + 2) (x + 3) + (x + 1) (x + 3) + (x + 1) (x + 2)$ , isliye $f^{'} (0) = 6 + 3 + 2 = 11$ . (Ya expand karo: $f (x) = x^{3} + 6 x^{2} + 11 x + 6$ .)

Recap

Summary aur key points

Instantaneous speed woh number hai jiske paas chhote hote intervals par average speeds aati hain; yeh distance–time graph ke tangent ka slope hai (Section 12.2).
$\lim_{x \to a} f (x) = L$ ka matlab hai ki $x$ ko $a$ ke kaafi paas lene par ( $x \neq a$ ), $f (x)$ , $L$ ke jitna chahein utna paas aa jaata hai; value $f (a)$ ka ismein koi role nahi (Section 12.3).
Limit theek tab exist karti hai jab left-hand aur right-hand limits exist karein aur barabar hon (Left-hand aur right-hand limits).
Sums, differences, products aur quotients (non-zero neeche wali limit ke saath) ki limits, limits ke sums, differences, products aur quotients hoti hain (Theorem 12.1).
Polynomial ke liye $\lim_{x \to a} p (x) = p (a)$ ; $\frac{0}{0}$ form wale rational function ke liye $x - a$ ke factors cancel karo (Section 12.3.2).
$\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$ ((12.5)).
$0 < ∣ x ∣ < \frac{π}{2}$ ke liye $\cos x < \frac{\sin x}{x} < 1$ ; sandwich theorem ke saath isse $\lim_{x \to 0} \frac{\sin x}{x} = 1$ aur $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ milte hain (Section 12.4).
$f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ , $(a, f (a))$ par tangent ka slope hai ((12.9)).
Derivative function $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ hai, jise $\frac{d y}{d x}$ bhi likhte hain ((12.10)).
$(u \pm v)^{'} = u^{'} \pm v^{'}$ , $(u v)^{'} = u^{'} v + u v^{'}$ aur ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ (Section 12.5.1).
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$ ; polynomials ko term by term differentiate kiya jaata hai ((12.13)).
$\frac{d}{d x} (\sin x) = \cos x$ , $\frac{d}{d x} (\cos x) = - \sin x$ aur $\frac{d}{d x} (\tan x) = \sec^{2} x$ (Section 12.5.2).

History seNCERT Historical Note · p. 255

Calculus kisne banaya?

Calculus ke do inventors hain jinhone alag-alag kaam kiya. Isaac Newton (1642–1727) ne apne rates of change ko 'fluxions' kaha; Gottfried Wilhelm Leibniz (1646–1716) ne symbols $d x$ aur $d y$ banaye jo hum aaj bhi likhte hain. Unki main discoveries 1660s aur 1680s ke beech ki hain.

Ek sadi se zyada tak iski neev kamzor rahi, kyunki koi theek se nahi bata sakta tha ki 'infinitely small' change kya hota hai. Joseph-Louis Lagrange ne hamein 'derived function' shabd aur notation $f^{'} (x)$ diya. Augustin-Louis Cauchy ne subject ko limits par dobara khada kiya aur derivative ko difference quotient ki limit ke roop mein define kiya, jaise humne kiya. Baad mein Karl Weierstrass ne limit ki definition ko poori tarah exact banaya. 1901 mein engineer aur teacher John Perry ne Glasgow ke ek address mein kaha ki calculus school mathematics ka hissa hona chahiye, aur yeh ek kaaran hai ki tum ise Class 11 mein padhte ho.

Newton se bahut pehle, Kerala school ke mathematicians ne, Saṅgamagrāma ke Mādhava (lagbhag 1340–1425) se shuru karke, sine aur cosine ke liye infinite series dhoondh li thi. Class 12 unke kaam par wapas aati hai.

Timeline

Fluxions se limits tak

1665–1666
Newton apna method of fluxions develop karte hain
1684
Leibniz $d x$ aur $d y$ ke saath apna differential calculus publish karte hain
1797
Lagrange ke derived functions aur notation $f^{'} (x)$
1823
Cauchy derivative ko ek limit ke roop mein define karte hain
1860s
Weierstrass limit ki exact definition dete hain
1901
Perry ka Glasgow address schools mein calculus ki wakalat karta hai

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 12.1, Q1–Q3	Practice 12.1 · Q1
NCERT Exercise 12.1, Q4–Q5	Practice 12.1 · Q2
NCERT Exercise 12.1, Q6	Practice 12.1 · Q3
NCERT Exercise 12.1, Q7–Q8	Practice 12.1 · Q4
NCERT Exercise 12.1, Q9, Q11	Practice 12.1 · Q5
NCERT Exercise 12.1, Q10	Practice 12.1 · Q6
NCERT Exercise 12.1, Q12	Practice 12.1 · Q7
NCERT Exercise 12.1, Q13–Q14	Practice 12.1 · Q8
NCERT Exercise 12.1, Q15–Q16	Practice 12.1 · Q9
NCERT Exercise 12.1, Q17–Q18	Practice 12.1 · Q10
NCERT Exercise 12.1, Q19–Q21	Practice 12.1 · Q11
NCERT Exercise 12.1, Q22	Practice 12.1 · Q12
NCERT Exercise 12.1, Q23–Q24	Practice 12.1 · Q13
NCERT Exercise 12.1, Q25–Q27	Practice 12.1 · Q14
NCERT Exercise 12.1, Q28	Practice 12.1 · Q15
NCERT Exercise 12.1, Q29	Practice 12.1 · Q16
NCERT Exercise 12.1, Q30	Practice 12.1 · Q17
NCERT Exercise 12.1, Q31	Practice 12.1 · Q18
NCERT Exercise 12.1, Q32	Practice 12.1 · Q19
NCERT Exercise 12.2, Q1–Q3	Practice 12.2 · Q1
NCERT Exercise 12.2, Q4	Practice 12.2 · Q2
NCERT Exercise 12.2, Q5	Practice 12.2 · Q3
NCERT Exercise 12.2, Q6	Practice 12.2 · Q4
NCERT Exercise 12.2, Q7	Practice 12.2 · Q5
NCERT Exercise 12.2, Q8	Practice 12.2 · Q6
NCERT Exercise 12.2, Q9 (i)–(iii)	Practice 12.2 · Q7
NCERT Exercise 12.2, Q9 (iv)–(vi)	Practice 12.2 · Q8
NCERT Exercise 12.2, Q10	Practice 12.2 · Q9
NCERT Exercise 12.2, Q11 (i)–(iv)	Practice 12.2 · Q10
NCERT Exercise 12.2, Q11 (v)–(vii)	Practice 12.2 · Q11
NCERT Miscellaneous Exercise, Q1	Chapter 12 par miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2–Q4	Chapter 12 par miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q5–Q7	Chapter 12 par miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q8–Q9	Chapter 12 par miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q10–Q12	Chapter 12 par miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q13	Chapter 12 par miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q14–Q16	Chapter 12 par miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q17–Q18	Chapter 12 par miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q19	Chapter 12 par miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q20–Q21	Chapter 12 par miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q22–Q24	Chapter 12 par miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q25, Q29	Chapter 12 par miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q26–Q28, Q30	Chapter 12 par miscellaneous exercises · Q13

Quick check: MCQs

12.2 Derivatives ka Intuitive Idea3 questions
Tables aur graphs se limits3 questions
12.3.2 Polynomials aur rational functions ki limits4 questions
12.4 Trigonometric Functions ki Limits4 questions
Derivative function3 questions
12.5.2 Polynomials aur trigonometric functions ka derivative4 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English + Hindi)

derivativeअवकलजavakalaj: $f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ , $a$ par $f$ ka instantaneous rate of change aur wahan tangent ka slope.
difference quotientअंतर भागफलantar bhagphal: $\frac{f (a + h) - f (a)}{h}$ , $a$ aur $a + h$ ke beech $f$ ka average rate of change; ek chord ka slope.Chapter mein pehli baar yahan aaya →

first principleप्रथम सिद्धांतpratham siddhant: Derivative ko seedhe uski definition se, yaani difference quotients ki limit se, nikalna.

instantaneous rate of changeतात्क्षणिक परिवर्तन दरtatkshanik parivartan dar: Ek pal par koi quantity jis rate se badal rahi hai: woh number jiske paas chhote hote intervals par average rates aate hain.Chapter mein pehli baar yahan aaya →

left-hand limitबाएँ पक्ष की सीमाbaayein paksh ki seema: Woh number jiske paas $f (x)$ tab aata hai jab $x \to a$ , $a$ se chhoti values se hokar; ise $\lim_{x \to a^{-}} f (x)$ likhte hain.Chapter mein pehli baar yahan aaya →
limitसीमाseema: $\lim_{x \to a} f (x) = L$ : jab $x$ , $a$ ke kaafi paas ho ( $x \neq a$ ), to values $f (x)$ , $L$ ke jitna chahein utna paas aa jaati hain.

product ruleगुणनफल नियमgunanphal niyam: $(u v)^{'} = u^{'} v + u v^{'}$ , ise Leibniz rule bhi kehte hain.Chapter mein pehli baar yahan aaya →

quotient ruleभागफल नियमbhagphal niyam: Jahan bhi $v \neq 0$ ho, ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ .Chapter mein pehli baar yahan aaya →

rational functionपरिमेय फलनparimey phalan: Do polynomials ka quotient $\frac{g (x)}{h (x)}$ , jo wahan defined hai jahan $h (x) \neq 0$ .Chapter mein pehli baar yahan aaya →
right-hand limitदाएँ पक्ष की सीमाdaayein paksh ki seema: Woh number jiske paas $f (x)$ tab aata hai jab $x \to a$ , $a$ se badi values se hokar; ise $\lim_{x \to a^{+}} f (x)$ likhte hain.Chapter mein pehli baar yahan aaya →

sandwich theoremसैंडविच प्रमेयsandwich pramey: Agar $a$ ke paas $f \leq g \leq h$ ho aur $f$ aur $h$ dono $l$ ki taraf jaayein, to $g$ bhi $l$ ki taraf jaata hai.Chapter mein pehli baar yahan aaya →

tangentस्पर्श रेखाsparsh rekha: Woh line jo kisi curve ko ek point par bas chhooti hai; us point se guzarne wali chords ki limiting position.Chapter mein pehli baar yahan aaya →

12.1 Introduction

12.2 Derivatives ka Intuitive Idea

12.3 Limits

Left-hand aur right-hand limits

Tables aur graphs se limits

12.3.1 Limits ka algebra

12.3.2 Polynomials aur rational functions ki limits

12.4 Trigonometric Functions ki Limits

12.5 Derivatives

Tangent ke slope ki tarah derivative

Derivative function

12.5.1 Functions ke derivative ka algebra

12.5.2 Polynomials aur trigonometric functions ka derivative

Miscellaneous Examples

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