Continuity and Differentiability: Class 12 Maths · Chapter 5

Mathematics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

145 min
45 practice questions
30 MCQs

What you will learn in Continuity and Differentiability

Decide whether a function is continuous at a point by comparing its one-sided limits with its value, and find every point of discontinuity of a piecewise function.
Use the algebra of continuous functions and composition to show that polynomial, rational, trigonometric and modulus functions are continuous on their domains.
Explain why differentiability implies continuity, and use one-sided derivatives to show that a corner such as |x − a| is not differentiable.
Differentiate composite, implicit and inverse trigonometric functions with the chain rule.
Describe exponential and logarithmic functions and their laws, and differentiate eˣ, log x, aˣ and their composites.
Use logarithmic differentiation for variable powers and long products, and find dy/dx for curves given in parametric form.
Find second order derivatives and verify relations between a function and its derivatives.

Original companion text aligned to NCERT Mathematics Class 12, Chapter 5, pp. 104–146. Not affiliated with or endorsed by NCERT.

5.1 Introduction

Picture two graphs from an ordinary day. The temperature in Jaipur climbs smoothly through the morning, never by a sudden leap. A car-park charge sits at ₹40 for a while, then leaps to ₹60 the instant another hour is completed.

This chapter turns that everyday difference into mathematics. A function whose graph has no breaks at a point is continuous (संतत) there. A function whose graph also has a single, well-defined slope at a point is differentiable (अवकलनीय) there. We will see exactly how the two ideas are linked.

Then we widen your Class XI toolkit for derivatives: composite functions, relations never solved for $y$ , inverse trigonometric functions, exponentials and logarithms, curves given through a third variable, and derivatives of derivatives.

5.2 Continuity

Slide a pencil along a graph from left to right and watch what happens as you pass a particular point $x = c$ . If the pencil never leaves the paper near $c$ , the graph is unbroken there. Fig. 5.1 shows the ways a graph can break at $c$ . Each break is a different failure, and our definition must catch every one of them.

3/5The left side and the right side head for different heights.

Fig. 5.1Four ways to break a graph at

x = c

, and the unbroken case.NCERT Figs 5.1–5.2 · p. 104

Axes with a dashed vertical guide at x = c. Step 1 shows an unbroken rising curve that passes through a filled dot at x = c. Step 2 shows the same curve with an open circle at x = c and no dot anywhere above c, so f(c) is not defined. Step 3 cuts the curve at c: the left piece ends at a filled dot and the right piece starts higher up at an open circle, so the two sides approach different heights. Step 4 shows the curve again with an open circle on it at c, while the filled dot giving f(c) sits well above the curve. Step 5 shows a curve that shoots up on the right of c and down on the left of c, hugging the dashed guide, with no finite height to approach.

Recall from Class XI that $\lim_{x \to c^{-}} f (x)$ , the left-hand limit, is the value $f (x)$ approaches as $x$ comes towards $c$ through numbers smaller than $c$ . The right-hand limit $\lim_{x \to c^{+}} f (x)$ uses numbers larger than $c$ . When both exist and are equal, their common value is the limit $\lim_{x \to c} f (x)$ . These one-sided limits are the tools that separate the cases in the figure.

Definition

Continuity at a point

Take a real function $f$ and a number $c$ at which $f$ is defined. We say $f$ is continuous at $c$ when $\lim_{x \to c} f (x) = f (c)$ .

Example $f (x) = x^{2}$ is continuous at $c = 3$ : as $x \to 3$ , $x^{2} \to 9 = f (3)$ .

Continuity at c, in full

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

Written out, continuity at $c$ asks three questions, and all three answers must be yes:

Is $f (c)$ defined? (A hole fails here.)
Do the left-hand and right-hand limits exist and agree? (A jump or a blow-up fails here.)
Is that common value equal to $f (c)$ ? (A misplaced point fails here.)

When $f$ fails this test at a point $c$ of its domain, we call $c$ a point of discontinuity and say that $f$ is discontinuous there. The words of its domain matter: a number at which $f$ has no value at all is never called a point of discontinuity in this chapter.

Example 1NCERT Example 1 · p. 106

A straight line at one point

Is $f (x) = 4 - 3 x$ continuous at $x = 2$ ?

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 106

A cubic at a negative point

Check whether $p (x) = x^{3} - 2 x$ is continuous at $x = - 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 106

A modulus at its kink

Discuss the continuity of $f (x) = ∣ x + 3 ∣$ at $x = - 3$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 106

A point lifted off the curve

Let $f (x) = x^{2} - 1$ for $x \neq 1$ and $f (1) = 2$ . Show that $f$ is not continuous at $x = 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Use the probe below to run the three checks yourself. It walks $x$ towards the point from both sides, ten times closer at each press, and lists the heights it finds.

InteractiveContinuity probe

A graph from x = −1 to 5 with a dashed guide at c = 2 and six function buttons: joined pieces, a jump, a hole, a misplaced point, a blow-up and a corner. Two probe dots sit at 2 − h and 2 + h on the graph. Each press of Move closer divides h by 10, from 1 down to 0.001, and adds a row to a table of f(2 − h) and f(2 + h); a line shows f(2) or says it is undefined. After the last press three checks are ticked or crossed (f(2) defined, both sides settle on one number, that number equals f(2)) and a verdict says whether the function is continuous at 2. Reset starts again.

Fig. 5.2Choose a function, press Move closer to bring

x

towards

c = 2

from both sides, then read the verdict.

Example 5NCERT Example 5 · p. 107

A constant function

Show that the constant function $f (x) = - 4$ is continuous at every real number.

Try it yourself first. Then reveal the solution one step at a time.

Example 6NCERT Example 6 · p. 107

The identity function

Show that $f (x) = x$ , the identity function, has no point of discontinuity.

Try it yourself first. Then reveal the solution one step at a time.

So far we have tested one point at a time. Usually we want to know about every point of the domain at once.

Definition

Continuous function

A real function that is continuous at each point of its domain.

Example $f (x) = x^{2}$ is a continuous function: it is continuous at every real number.

Two edge cases are worth a moment. If the domain of $f$ is just one point, $f$ passes the test automatically, since no nearby values exist to disagree with $f (c)$ . If $f$ lives on a closed interval $[a, b]$ , each end has neighbours on one side only, so we use one-sided limits there: $\lim_{x \to a^{+}} f (x) = f (a)$ at the left end and $\lim_{x \to b^{-}} f (x) = f (b)$ at the right end. Asking for the left-hand limit at $a$ would mean asking for values that $f$ does not have.

Example 7NCERT Example 7 · p. 108

Modulus everywhere

Is $f (x) = ∣ x + 3 ∣$ a continuous function?

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 108

A polynomial of degree four

Discuss the continuity of $f (x) = 2 x^{4} - 5 x + 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 108

A reciprocal on its domain

Discuss the continuity of $f (x) = \frac{1}{x - 2}$ , $x \neq 2$ .

Try it yourself first. Then reveal the solution one step at a time.

What does happen near $x = 2$ ? Try values ever closer to 2 from each side, as in Table 5.1. The table shows two runaway trends.

Table 5.1Values of

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

x

closes in on 2

$x$ (right of 2)	$f (x)$	$x$ (left of 2)	$f (x)$
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

On the right of 2 the values grow past every bound: name a number, a million say, and $f (x)$ beats it once $x$ is within a millionth of 2. The shorthand is $\lim_{x \to 2^{+}} f (x) = + \infty$ . On the left the values sink past every bound in the same way, written $\lim_{x \to 2^{-}} f (x) = - \infty$ . Be careful with the symbol: $+ \infty$ and $- \infty$ are not real numbers. The equations say that the one-sided limits do not exist as real numbers, and they describe how they fail.

2/2As $x \to 2^{-}$ the dots fall without bound: $f (x) \to - \infty$ .

Fig. 5.3The graph of

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

near

x = 2

.NCERT Fig 5.3 · p. 109

Axes with a dashed vertical guide at x = 2. In the first step the right branch is highlighted, with dots at x = 3, 2.5, 2.25 and 2.2 whose heights 1, 2, 4 and 5 climb towards the top of the frame, labelled plus infinity. In the second step the left branch is highlighted, with dots at x = 1, 1.5, 1.75 and 1.8 whose heights −1, −2, −4 and −5 fall towards the bottom, labelled minus infinity. The two branches never touch the guide.

Example 10NCERT Example 10 · p. 110

Two pieces that miss

Discuss the continuity of $f (x) = {\begin{cases} 2 x - 1, & x \leq 3 \\ x + 4, & x > 3 \end{cases}$

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 110

A three-part rule

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

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 111

A gap in the domain is not a break

Discuss the continuity of $f (x) = {\begin{cases} 3 - x, & x < 0 \\ x^{2} + 1, & x > 0 \end{cases}$

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 111

Pieces that meet

Show that $f (x) = {\begin{cases} x^{2} + 2, & x \leq 1 \\ 4 x - 1, & x > 1 \end{cases}$ is a continuous function.

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 112

Every polynomial is continuous

Prove that every polynomial $p (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ (with real $a_{i}$ ) is a continuous function.

Try it yourself first. Then reveal the solution one step at a time.

The greatest integer function $[x]$ gives the largest integer that is less than or equal to $x$ . For example $[2.7] = 2$ , $[5] = 5$ and $[- 1.2] = - 2$ (not $- 1$ , because $- 1$ is larger than $- 1.2$ ). Between two neighbouring integers $[x]$ stays constant, and at each integer it steps up by 1. Charges that rise in steps behave the same way, as Fig. 5.4 shows.

2/3At $x = 2$ the left side approaches ₹40 but the fee is ₹60: a jump.

Fig. 5.4The fee

F (x) = 20 + 20 [x]

rupees for

x

hours,

0 \leq x < 4

.NCERT Fig 5.8 · p. 112

Axes show hours parked from 0 to 4 and the fee in rupees from 0 to 100. The graph is four flat steps at 20, 40, 60 and 80 rupees; each step starts with a filled dot on the left and ends with an open circle on the right. The second step highlights the whole hour x = 2: an arrow shows the left step approaching ₹40, while the fee at exactly 2 hours, ₹60, is marked on the step above. The third step highlights x = 2.5, in the middle of a flat step, where the graph is level on both sides and nothing breaks.

Example 15NCERT Example 15 · p. 112

Where does the parking fee jump?

A car park charges ₹20 on entry and another ₹20 at the end of every completed hour, up to 4 hours. So the fee for $x$ hours is $F (x) = 20 + 20 [x]$ , $0 \leq x < 4$ . Find the points where $F$ is discontinuous.

Try it yourself first. Then reveal the solution one step at a time.

5.2.1 Algebra of continuous functions

Testing every function from scratch would be slow. Since continuity is a statement about limits, the limit laws from Class XI carry straight over: combine continuous functions with $+$ , $-$ , $\times$ or $\div$ and the result is continuous too.

Theorem 1Algebra of continuous functions

If two real functions $f$ and $g$ are both continuous at $c$ , so are their sum $f + g$ , their difference $f - g$ and their product $f \cdot g$ . Their quotient $\frac{f}{g}$ is continuous at $c$ too, as long as $g (c) \neq 0$ .

We prove the product part; the others follow the same pattern with the matching limit law. Both $f$ and $g$ are defined at $c$ , so $f \cdot g$ is too.

\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)

The first step is the product law for limits, and the second uses the continuity of $f$ and $g$ at $c$ . For the quotient, the condition $g (c) \neq 0$ is exactly what the quotient law for limits needs.

Taking $f$ to be a constant $λ$ in the product part: if $g$ is continuous, so is $λ g$ . With $λ = - 1$ , continuity of $g$ gives continuity of $- g$ .
Taking $f = λ$ in the quotient part: $\frac{λ}{g}$ is continuous wherever $g (x) \neq 0$ . In particular $\frac{1}{g}$ is.

Example 16NCERT Example 16 · p. 114

Rational functions

Show that $r (x) = \frac{x^{2} + 1}{x - 3}$ is continuous, and explain why the same holds for every rational function.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 114

The cosine function

Prove that $\cos x$ is continuous. Use two facts that are clear from the graphs of sine and cosine near 0: $\lim_{h \to 0} \sin h = 0$ and $\lim_{h \to 0} \cos h = 1$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 115

The secant function

Show that $\sec x$ is continuous wherever it is defined.

Try it yourself first. Then reveal the solution one step at a time.

Composition is the other way of building functions: feed the output of $g$ into $f$ to get $(f \circ g) (x) = f (g (x))$ , which needs every output of $g$ to be an allowed input of $f$ . Fig. 5.5 shows why continuity survives this too.

3/3 $f$ is continuous at $g (c)$ , so $f (g (x))$ stays close to $f (g (c))$ .

Fig. 5.5Small changes stay small through each continuous stage.NCERT p. 115

Three horizontal number lines are stacked, labelled x, u = g(x) and y = f(u). In step 1 a short highlighted interval surrounds the point c on the x line. In step 2 curved arrows labelled g carry it to a short highlighted interval around g(c) on the middle line. In step 3 arrows labelled f carry that to a short highlighted interval around f(g(c)) on the bottom line, so a small change at the start causes only a small change at the end.

Theorem 2Continuity of a composite

Suppose $f \circ g$ is defined at $c$ . Continuity of $g$ at $c$ , together with continuity of $f$ at the point $g (c)$ , makes $f \circ g$ continuous at $c$ .

We use this theorem without proof; the figure carries the idea.

Example 19NCERT Example 19 · p. 115

Sine of a polynomial

Show that $h (x) = \sin (2 x^{3} - x)$ is a continuous function.

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 115

A modulus inside a modulus

Show that $f (x) = ∣ ∣ x ∣ - 2 ∣$ is a continuous function.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 12 questions

Practice 5.1: questions with answers

Fresh questions on the skills of NCERT Exercise 5.1.

1
Short answerEasyNCERT Exercise 5.1, Q1–Q2
Use the definition of continuity.
1. Show that $f (x) = 7 - 2 x$ is continuous at $x = - 1$ , $x = 0$ and $x = 4$ .
  Show solution
  AnswerContinuous at all three points.
  $f$ is a polynomial, so $\lim_{x \to c} f (x) = 7 - 2 c = f (c)$ for every $c$ . In particular $f (- 1) = 9$ , $f (0) = 7$ and $f (4) = - 1$ equal the limits there.
2. Examine the continuity of $g (x) = 3 x^{2} + x - 4$ at $x = - 2$ .
  Show solution
  AnswerContinuous at $x = - 2$ .
  $g (- 2) = 12 - 2 - 4 = 6$ , and $\lim_{x \to - 2} (3 x^{2} + x - 4) = 6$ . The limit equals the value.
2
Short answerEasyNCERT Exercise 5.1, Q3
Examine each function for continuity.
1. $h (x) = 4 x + 9$
  Show solution
  AnswerContinuous everywhere.
  A polynomial is continuous at every real number.
2. $h (x) = \frac{1}{x + 3}$ , $x \neq - 3$
  Show solution
  AnswerContinuous on its domain.
  For $c \neq - 3$ , $\lim_{x \to c} \frac{1}{x + 3} = \frac{1}{c + 3} = h (c)$ . The point $- 3$ is not in the domain, so $h$ is a continuous function.
3. $h (x) = \frac{x^{2} - 9}{x - 3}$ , $x \neq 3$
  Show solution
  AnswerContinuous on its domain.
  For $x \neq 3$ , $h (x) = x + 3$ . At every $c \neq 3$ , $\lim_{x \to c} h (x) = c + 3 = h (c)$ .
4. $h (x) = ∣ x + 1 ∣$
  Show solution
  AnswerContinuous everywhere.
  For $c < - 1$ , $h = - (x + 1)$ near $c$ ; for $c > - 1$ , $h = x + 1$ near $c$ ; both are continuous. At $- 1$ both one-sided limits are 0, which equals $h (- 1)$ .
Short answerEasyNCERT Exercise 5.1, Q4
Prove that $f (x) = x^{m}$ is continuous at $x = 2$ for every positive integer $m$ .
Show solution
Answer $\lim_{x \to 2} x^{m} = 2^{m} = f (2)$ .
$x^{m}$ is a product of $m$ copies of $x$ , and $\lim_{x \to 2} x = 2$ . By the product law, $\lim_{x \to 2} x^{m} = 2^{m}$ , which is $f (2)$ . So $f$ is continuous at 2.
Short answerMediumNCERT Exercise 5.1, Q5
Let $f (x) = x + 1$ for $x \leq 2$ and $f (x) = 4$ for $x > 2$ . Is $f$ continuous at $x = 0$ ? At $x = 2$ ? At $x = 3$ ?
Show solution
AnswerYes at 0, no at 2, yes at 3.
At 0 and at 3, $f$ agrees with a polynomial or a constant near the point, so it is continuous there.
At 2: $f (2) = 3$ , $\lim_{x \to 2^{-}} (x + 1) = 3$ but $\lim_{x \to 2^{+}} 4 = 4$ . The sides disagree, so $f$ is discontinuous at 2.
5
Short answerMediumNCERT Exercise 5.1, Q6–Q12
Find all points of discontinuity of each function.
1. $f (x) = 3 x - 1$ for $x \leq 1$ ; $f (x) = 3 x + 2$ for $x > 1$
  Show solution
  Answer $x = 1$ only.
  At 1: left-hand limit $2$ , right-hand limit $5$ . They differ, so 1 is a point of discontinuity; elsewhere $f$ is continuous.
2. $f (x) = ∣ x ∣ + 2$ for $x \leq - 1$ ; $x^{2} + 2$ for $- 1 < x < 2$ ; $3 x + 1$ for $x \geq 2$
  Show solution
  Answer $x = 2$ only.
  At $- 1$ : $f (- 1) = 3$ , left-hand limit $3$ , right-hand limit $1 + 2 = 3$ , so continuous. At 2: left-hand limit $4 + 2 = 6$ , right-hand limit $f (2) = 7$ , so discontinuous.
3. $f (x) = \frac{∣ x - 2 ∣}{x - 2}$ for $x \neq 2$ ; $f (2) = 1$
  Show solution
  Answer $x = 2$ only.
  For $x > 2$ , $f (x) = 1$ ; for $x < 2$ , $f (x) = - 1$ . At 2 the one-sided limits are $- 1$ and $1$ , so $f$ is discontinuous there. Elsewhere $f$ is constant near each point.
4. $f (x) = \frac{∣ x ∣}{x}$ for $x > 0$ ; $f (x) = 1$ for $x \leq 0$
  Show solution
  AnswerNone.
  For $x > 0$ , $∣ x ∣ / x = x / x = 1$ . So $f (x) = 1$ for every real $x$ , a constant function, continuous everywhere.
5. $f (x) = 2 x + 1$ for $x \geq 2$ ; $f (x) = x^{2} + 1$ for $x < 2$
  Show solution
  AnswerNone.
  At 2: $f (2) = 5$ , left-hand limit $4 + 1 = 5$ , right-hand limit 5. All agree, so $f$ is continuous everywhere.
6. $f (x) = x^{3} + 1$ for $x \leq 1$ ; $f (x) = 3 x - 1$ for $x > 1$
  Show solution
  AnswerNone.
  At 1: $f (1) = 2$ , left-hand limit 2, right-hand limit $3 - 1 = 2$ . Continuous everywhere.
7. $f (x) = x^{5} + 1$ for $x \leq - 1$ ; $f (x) = x^{2}$ for $x > - 1$
  Show solution
  Answer $x = - 1$ only.
  At $- 1$ : $f (- 1) = - 1 + 1 = 0$ , but the right-hand limit is $(- 1)^{2} = 1$ . Discontinuous at $- 1$ only.
6
Short answerMediumNCERT Exercise 5.1, Q13–Q16
Discuss the continuity of each function.
1. $f (x) = 4 - x$ for $x \leq 0$ ; $f (x) = x + 4$ for $x > 0$ . Is $f$ a continuous function?
  Show solution
  AnswerYes.
  At 0: $f (0) = 4$ and both one-sided limits are 4. Elsewhere $f$ is a polynomial near each point. So $f$ is continuous.
2. A courier charges ₹10 for parcels up to 2 kg, ₹15 for more than 2 kg up to 5 kg, and ₹25 for more than 5 kg up to 8 kg. Discuss the continuity of the charge $C (w)$ , $0 \leq w \leq 8$ .
  Show solution
  AnswerDiscontinuous at $w = 2$ and $w = 5$ only.
  $C$ is constant on each interval. At 2: $C (2) = 10$ , left-hand limit 10, right-hand limit 15, so discontinuous. At 5: $C (5) = 15$ , right-hand limit 25, so discontinuous. At the end points 0 and 8 the one-sided limits match the values.
3. $f (x) = - 3 x$ for $x < 0$ ; $0$ for $0 \leq x \leq 2$ ; $2 x - 3$ for $x > 2$
  Show solution
  AnswerDiscontinuous at $x = 2$ only.
  At 0: left-hand limit 0, $f (0) = 0$ , right-hand limit 0, so continuous. At 2: $f (2) = 0$ , but the right-hand limit is $4 - 3 = 1$ , so discontinuous.
4. $f (x) = 1$ for $x \leq 0$ ; $\cos x$ for $0 < x \leq π$ ; $- 1$ for $x > π$
  Show solution
  AnswerContinuous everywhere.
  At 0: $f (0) = 1$ and $\lim_{x \to 0^{+}} \cos x = 1$ . At $π$ : $f (π) = \cos π = - 1$ and the right-hand limit is $- 1$ . Each piece is continuous, so $f$ is continuous everywhere.
7
Short answerMediumNCERT Exercise 5.1, Q17–Q18
Joining pieces with unknown constants.
1. Find the relation between $a$ and $b$ for which $f (x) = a x - 2$ ( $x \leq 2$ ), $f (x) = b x^{2} + 1$ ( $x > 2$ ) is continuous at $x = 2$ .
  Show solution
  Answer $2 a - 4 b = 3$ .
  $f (2) = 2 a - 2$ and the right-hand limit is $4 b + 1$ . Continuity needs $2 a - 2 = 4 b + 1$ , that is $2 a - 4 b = 3$ .
2. For which $λ$ is $f (x) = λ (x^{2} + 3)$ ( $x \leq 1$ ), $f (x) = 2 x + 6$ ( $x > 1$ ) continuous at $x = 1$ ? Is $f$ continuous at $x = 0$ for every $λ$ ?
  Show solution
  Answer $λ = 2$ ; yes, $f$ is continuous at 0 for every $λ$ .
  At 1: $f (1) = 4 λ$ and the right-hand limit is 8, so $λ = 2$ . Near 0, $f (x) = λ (x^{2} + 3)$ for all $x$ close to 0 on both sides, a polynomial, so $f$ is continuous at 0 whatever $λ$ is.
Short answerHardNCERT Exercise 5.1, Q19
Show that $h (x) = [x] + [- x]$ is discontinuous at every integer.
Show solution
Answer $h = 0$ at integers and $h = - 1$ elsewhere, so $\lim_{x \to n} h (x) = - 1 \neq 0 = h (n)$ .
If $x$ is not an integer, write $[x] = n$ with $n < x < n + 1$ . Then $- n - 1 < - x < - n$ , so $[- x] = - n - 1$ and $h (x) = - 1$ .
If $x = n$ is an integer, $h (n) = n + (- n) = 0$ .
Near any integer $n$ (but not at it), $h$ equals $- 1$ , so $\lim_{x \to n} h (x) = - 1 \neq h (n)$ . Every integer is a point of discontinuity.
9
Short answerMediumNCERT Exercise 5.1, Q20–Q22
Trigonometric combinations.
1. Is $f (x) = 2 \cos x + x^{3}$ continuous at $x = \frac{π}{2}$ ?
  Show solution
  AnswerYes.
  It is a sum of continuous functions, so it is continuous everywhere, including at $\frac{π}{2}$ , where its value is $\frac{π^{3}}{8}$ .
2. Discuss the continuity of $3 \sin x - x^{2}$ , of $x \sin x$ and of $\frac{\sin x}{2 + \cos x}$ .
  Show solution
  AnswerAll three are continuous everywhere.
  The first is a difference and the second a product of continuous functions. In the third, $2 + \cos x \geq 1$ , so the denominator is never zero and the quotient is continuous at every real number.
3. Discuss the continuity of $\sin x$ , $\tan x$ , $cosec x$ and $\cot x$ .
  Show solution
  AnswerEach is continuous on its domain: $\sin$ everywhere, $\tan$ for $x \neq (2 n + 1) \frac{π}{2}$ , $cosec$ and $\cot$ for $x \neq n π$ .
  For sine use $\sin (c + h) = \sin c \cos h + \cos c \sin h \to \sin c$ . Then $\tan = \frac{\sin}{}$ , $cosec = \frac{1}{\sin}$ and $\cot = \frac{\cos}{}$ are quotients of continuous functions, continuous wherever the denominator is non-zero.
10
Short answerHardNCERT Exercise 5.1, Q23–Q25
Limits involving sine.
1. Find the points of discontinuity of $f (x) = \frac{\sin 2 x}{x}$ for $x < 0$ , $f (x) = x + 2$ for $x \geq 0$ .
  Show solution
  AnswerNone.
  $\lim_{x \to 0^{-}} \frac{\sin 2 x}{x} = \lim 2 \cdot \frac{\sin 2 x}{2 x} = 2$ , the right-hand limit is 2 and $f (0) = 2$ . Each piece is continuous on its own interval, so $f$ is continuous everywhere.
2. Is $f (x) = x \sin \frac{1}{x}$ ( $x \neq 0$ ), $f (0) = 0$ a continuous function?
  Show solution
  AnswerYes.
  For $x \neq 0$ , $∣ x \sin \frac{1}{x} ∣ \leq ∣ x ∣$ , which tends to 0. So $\lim_{x \to 0} f (x) = 0 = f (0)$ . Away from 0, $f$ is a product and composite of continuous functions.
3. Examine the continuity of $f (x) = \cos x + \sin x$ ( $x \neq 0$ ), $f (0) = 2$ .
  Show solution
  AnswerDiscontinuous at 0 only.
  $\lim_{x \to 0} (\cos x + \sin x) = 1 + 0 = 1$ , but $f (0) = 2$ . Elsewhere $f$ is a sum of continuous functions.
11
Short answerMediumNCERT Exercise 5.1, Q26–Q30
Find the constants that make each function continuous at the indicated point.
1. $f (x) = \frac{k \sin x}{π - x}$ for $x \neq π$ , $f (π) = 5$ , at $x = π$ . Find $k$ .
  Show solution
  Put $t = π - x$ . Then $\sin x = \sin (π - t) = \sin t$ , and $\frac{\sin t}{t} \to 1$ as $t \to 0$ . The limit is $k$ , so $k = f (π) = 5$ .
2. $f (x) = k x^{2} + 1$ for $x \leq 3$ , $f (x) = 4$ for $x > 3$ , at $x = 3$ . Find $k$ .
  Show solution
  $f (3) = 9 k + 1$ must equal the right-hand limit 4, so $k = \frac{1}{3}$ .
3. $f (x) = k x - 1$ for $x \leq \frac{π}{2}$ , $f (x) = \sin x$ for $x > \frac{π}{2}$ , at $x = \frac{π}{2}$ . Find $k$ .
  Show solution
  $f (\frac{π}{2}) = \frac{k π}{2} - 1$ must equal $\sin \frac{π}{2} = 1$ , so $k = \frac{4}{π}$ .
4. $f (x) = k x + 3$ for $x \leq 4$ , $f (x) = 2 x - 1$ for $x > 4$ , at $x = 4$ . Find $k$ .
  Show solution
  $4 k + 3 = 8 - 1 = 7$ , so $k = 1$ .
5. Find $a$ and $b$ so that $f (x) = 2$ ( $x \leq 1$ ), $a x + b$ ( $1 < x < 4$ ), $14$ ( $x \geq 4$ ) is a continuous function.
  Show solution
  Answer $a = 4$ , $b = - 2$ .
  At 1: $a + b = 2$ . At 4: $4 a + b = 14$ . Subtracting, $3 a = 12$ , so $a = 4$ and $b = - 2$ .
12
Short answerMediumNCERT Exercise 5.1, Q31–Q34
Composites and moduli.
1. Show that $f (x) = \cos (\frac{1}{1 + x^{2}})$ is continuous.
  Show solution
  AnswerContinuous everywhere.
  $u (x) = \frac{1}{1 + x^{2}}$ is rational with a denominator that is never zero, so it is continuous everywhere; cosine is continuous; the composite is continuous by Theorem 2.
2. Show that $f (x) = ∣ \sin 2 x ∣$ is continuous.
  Show solution
  AnswerContinuous everywhere.
  $\sin 2 x$ is a composite of $2 x$ and sine, so it is continuous; the modulus of a continuous function is continuous (Theorem 2 with $m (u) = ∣ u ∣$ ).
3. Show that $f (x) = \sin (∣ x ∣ - 2)$ is continuous.
  Show solution
  AnswerContinuous everywhere.
  $∣ x ∣ - 2$ is continuous and sine is continuous, so the composite is continuous.
4. Find all points of discontinuity of $f (x) = ∣ x - 1 ∣ + ∣ x + 2 ∣$ .
  Show solution
  AnswerNone.
  Each modulus is a composite of a polynomial and $∣ u ∣$ , so it is continuous; their sum is continuous everywhere.

5.3 Differentiability

In Class XI you met the derivative of a real function $f$ at a point $c$ of its domain:

Derivative at c

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

where

$h$: a small step away from $c$ , positive or negative

Geometrically, $f^{'} (c)$ is the slope on which chord slopes settle as the chords shrink. Collecting these slopes at every $x$ where the limit exists builds a new function, the derivative $f^{'}$ . You will meet all its names in this chapter: $f^{'} (x)$ , $\frac{d}{d x} f (x)$ , and $\frac{d y}{d x}$ or $y^{'}$ when $y = f (x)$ ; the whole process is differentiation. Three Class XI rules do most of the work, for functions $u$ and $v$ of $x$ :

$(u \pm v)^{'} = u^{'} \pm v^{'}$
$(u v)^{'} = u^{'} v + u v^{'}$ (the product rule, also called Leibniz's rule)
${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ wherever $v \neq 0$ (the quotient rule)

Table 5.2Standard derivatives from Class XI

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

The formula for $f^{'} (c)$ only works when its limit exists, and at some points it does not. There $f$ simply has no derivative: it is not differentiable at $c$ . To see which way the limit fails, let $h$ approach 0 from each side separately. This gives a left-hand derivative (from $h < 0$ ) and a right-hand derivative (from $h > 0$ ).

Left-hand and 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

Differentiable at a point

We call $f$ differentiable at a point $c$ of its domain when its left-hand and right-hand derivatives at $c$ are both finite and equal; their common value is $f^{'} (c)$ .

Example $f (x) = x^{2}$ at $c = 1$ : both one-sided difference quotients equal $2 + h$ , which tends to 2, so $f^{'} (1) = 2$ .

Differentiability on an open interval $(a, b)$ means differentiability at each of its points. On a closed interval $[a, b]$ we also need the one-sided derivatives at the ends: the right-hand derivative at $a$ and the left-hand derivative at $b$ , just as continuity at an end point uses only one side.

Theorem 3Differentiable implies continuous

Differentiability at a point forces continuity there: if $f^{'} (c)$ exists, then $f$ is continuous at $c$ .

Corollary 1Differentiable functions

Every differentiable function is continuous.

The converse is false: a graph can be unbroken at a point and still have no single slope there. The simplest culprit is a corner.

ExampleNCERT p. 120

A corner at x = 3

Show that $f (x) = ∣ x - 3 ∣$ is continuous but not differentiable at $x = 3$ .

Try it yourself first. Then reveal the solution one step at a time.

At a corner the chords from the two sides settle on two different slopes. Fig. 5.6 shows this for a curved corner, where the chords visibly turn as they shorten.

4/4 $- 2 \neq 2$ : no single tangent, so $f$ is not differentiable at 1.

Fig. 5.6Chords to the corner of

y = ∣ x^{2} - 1 ∣

x = 1

.NCERT p. 120

The graph of y = |x² − 1| for x from about −0.2 to 2 has a sharp corner at the point (1, 0). Step 1 shows only the curve and the corner. Step 2 draws three chords from the corner to points on the left branch at x = 0.2, 0.6 and 0.85; they get steeper towards slope −2, and a dashed line of slope −2 is labelled. Step 3 draws three chords to points on the right branch at x = 1.8, 1.4 and 1.15, settling towards a dashed line of slope 2. Step 4 shows both dashed limiting lines together: two different slopes, so no single tangent.

The zoom lab below makes the same point another way. Zoom in on a point of a smooth graph and the graph soon looks like a straight line, whose slope is the derivative. Zoom in on a corner and it stays a corner however far you go.

InteractiveZoom lab

A graph window centred on a marked point, with five function buttons: x squared at 1, a smooth join at 1, a kinked join at 1, |x − 1| at 1 and the cube root of x − 1 at 1. Zoom in and Zoom out buttons halve or double the window; a readout gives the window width. Two chords join the point to points h to its left and right, with h tied to the window width, and a table lists the left and right chord slopes for h = 0.1, 0.01 and 0.001. A verdict line says whether the two slopes settle on one number (the derivative), on two different numbers (a corner), or grow without bound (a vertical tangent).

Fig. 5.7Pick a function, zoom in on the marked point, and compare the slopes of chords from the left and from the right.

5.3.1 Derivatives of composite functions

How would you differentiate $y = (5 - 2 x)^{2}$ ? One way is to expand first: $y = 25 - 20 x + 4 x^{2}$ , so $\frac{d y}{d x} = - 20 + 8 x = - 4 (5 - 2 x)$ . Now look at the answer differently. The outer operation, squaring, contributes $2 (5 - 2 x)$ ; the inner expression $5 - 2 x$ changes at rate $- 2$ ; and $2 (5 - 2 x) \times (- 2)$ is exactly $- 4 (5 - 2 x)$ .

That pattern is worth having, because expanding $(5 - 2 x)^{40}$ is hopeless, while the pattern gives $40 (5 - 2 x)^{39} \times (- 2) = - 80 (5 - 2 x)^{39}$ at once. The idea behind it is that rates multiply along a chain, as Fig. 5.8 shows.

3/3 $y = u^{2}$ moves from 16 to 18.49: $Δ y \approx 8 Δ u \approx 24 Δ x$ .

Fig. 5.8A change in

x

passes through

u

y

, multiplied at each link.NCERT p. 121

Three horizontal rulers are stacked: the top one for x, the middle one for u = 3x + 1 and the bottom one for y = u squared. Step 1 marks x = 1 moving to 1.1, a change of 0.1. Step 2 shows u moving from 4 to 4.3, a change of 0.3, three times as big, labelled du/dx = 3. Step 3 shows y moving from 16 to 18.49, a change of 2.49, about 8 times the change in u because dy/du = 2u = 8, so y changes about 24 times as fast as x.

Theorem 4The chain rule

Suppose $f (x) = v (u (x))$ , and write $t = u (x)$ for the inner value. Whenever the inner rate $\frac{d t}{d x}$ and the outer rate $\frac{d v}{d t}$ both exist, the composite's rate is their product: $\frac{d f}{d x} = \frac{d v}{d t} \cdot \frac{d t}{d x}$ .

We use the chain rule without proof. In the figure, $\frac{d u}{d x} = 3$ and $\frac{d y}{d u} = 2 u = 8$ at $u = 4$ , so $\frac{d y}{d x} = 24$ ; the actual change 2.49 for a step of 0.1 is close to $24 \times 0.1$ . Longer chains work the same way. If $x \to t = u (x) \to s = v (t) \to y = w (s)$ , then, provided all three derivatives exist,

Chain rule with three links

\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

A cosine of a cube

Differentiate $f (x) = \cos (x^{3})$ .

Try it yourself first. Then reveal the solution one step at a time.

ExampleNCERT p. 121

Three links

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

Try it yourself first. Then reveal the solution one step at a time.

Practice · 3 questions

Practice 5.2: questions with answers

Fresh questions on the skills of NCERT Exercise 5.2.

1
Short answerMediumNCERT Exercise 5.2, Q1–Q8
Differentiate each function with respect to $x$ .
1. $\cos (x^{3} - 2)$
  Show solution
  Answer $- 3 x^{2} \sin (x^{3} - 2)$
  Outer $\cos t$ gives $- \sin t$ ; inner $t = x^{3} - 2$ gives $3 x^{2}$ . So $- 3 x^{2} \sin (x^{3} - 2)$ .
2. $\tan (\sin x)$
  Show solution
  Answer $\cos x \sec^{2} (\sin x)$
  $\frac{d}{d x} \tan t = \sec^{2} t \cdot \frac{d t}{d x}$ with $t = \sin x$ , giving $\sec^{2} (\sin x) \cos x$ .
3. $\cos (4 - 3 x)$
  Show solution
  Answer $3 \sin (4 - 3 x)$
  $- \sin (4 - 3 x) \cdot (- 3) = 3 \sin (4 - 3 x)$ .
4. $\cos (\tan 2 x)$
  Show solution
  Answer $- 2 \sec^{2} 2 x \cdot \sin (\tan 2 x)$
  Three links: $- \sin (\tan 2 x) \cdot \sec^{2} (2 x) \cdot 2$ .
5. $\frac{\cos 2 x}{\sin (x^{2})}$
  Show solution
  Answer $- \frac{2 \sin 2 x}{\sin (x^{2})} - \frac{2 x \cos 2 x \cos (x^{2})}{\sin^{2} (x^{2})}$
  Quotient rule with $u = \cos 2 x$ , $u^{'} = - 2 \sin 2 x$ , $v = \sin (x^{2})$ and $v^{'} = 2 x \cos (x^{2})$ , wherever $\sin (x^{2}) \neq 0$ :
  $\frac{u^{'} v - u v^{'}}{v^{2}} = \frac{- 2 \sin 2 x \sin (x^{2}) - 2 x \cos 2 x \cos (x^{2})}{\sin^{2} (x^{2})}$
6. $\sin (x^{2}) \cos^{3} (2 x)$
  Show solution
  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. $\frac{d}{d x} \sin (x^{2}) = 2 x \cos (x^{2})$ and $\frac{d}{d x} \cos^{3} 2 x = 3 \cos^{2} 2 x \cdot (- \sin 2 x) \cdot 2 = - 6 \cos^{2} 2 x \sin 2 x$ .
7. $\sqrt{\tan 3 x}$ (where $\tan 3 x > 0$ )
  Show solution
  Answer $\frac{3 \sec^{2} 3 x}{2 \sqrt{\tan 3 x}}$
  $\frac{1}{2 \sqrt{t}} \cdot \sec^{2} 3 x \cdot 3$ with $t = \tan 3 x$ .
8. $\cos (\frac{1}{x})$ , $x \neq 0$
  Show solution
  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 that $f (x) = ∣ x + 2 ∣$ is not differentiable at $x = - 2$ .
Show solution
AnswerThe left-hand derivative is −1 and the right-hand derivative is 1.
$\frac{f (- 2 + h) - f (- 2)}{h} = \frac{∣ h ∣ - 0}{h}$ . For $h < 0$ this is $- 1$ ; for $h > 0$ it is $1$ . The one-sided derivatives differ, so $f^{'} (- 2)$ does not exist.
Short answerHardNCERT Exercise 5.2, Q10
Let $f (x) = [2 x]$ for $0 < x < 2$ . Show that $f$ is not differentiable at $x = \frac{1}{2}$ , $x = 1$ and $x = \frac{3}{2}$ .
Show solution
Answer $f$ jumps by 1 at each of these points, so it is not continuous there and cannot be differentiable.
$[2 x]$ jumps when $2 x$ passes an integer, that is at $x = \frac{1}{2}, 1, \frac{3}{2}$ in $(0, 2)$ . For example at $x = 1$ : the left-hand limit is $[2^{-}] = 1$ but $f (1) = [2] = 2$ .
A function that is not continuous at a point cannot be differentiable there (Theorem 3). Directly: for $h < 0$ , $\frac{f (1 + h) - f (1)}{h} = \frac{1 - 2}{h} = - \frac{1}{h}$ , which grows without bound as $h \to 0^{-}$ .

5.3.2 Derivatives of implicit functions

Many curves come as an equation tying $x$ and $y$ together rather than as a formula for $y$ . The relation $3 x + 2 y = 7$ is easily rearranged to $y = \frac{7 - 3 x}{2}$ : here $y$ is an explicit function of $x$ . The relation $y^{5} + y = x$ is different. For each $x$ there is exactly one $y$ , because $y^{5} + y$ increases steadily, yet no formula from algebra writes $y$ in terms of $x$ .

Definition

Implicit function

When a relation between $x$ and $y$ fixes $y$ as a function of $x$ , but is not solved (or not easily solvable) in the form $y = f (x)$ , we say the relation gives $y$ implicitly as a function of $x$ .

Example $y^{5} + y = x$ defines $y$ implicitly.

We do not need the formula to find $\frac{d y}{d x}$ . Differentiate both sides of the relation with respect to $x$ , remembering that $y$ is a function of $x$ , so every term in $y$ brings a factor $\frac{d y}{d x}$ by the chain rule: $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ . Then solve for $\frac{d y}{d x}$ . Fig. 5.9 uses a circle, which is not the graph of a single function, but each half of it is.

3/3 $2 x + 2 y y^{'} = 0$ gives $y^{'} = - \frac{x}{y} = - \frac{3}{4}$ at $(3, 4)$ .

Fig. 5.9Implicit differentiation of

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

.NCERT p. 122

A circle of radius 5 centred at the origin on square axes. Step 1 shows the whole circle with a vertical line x = 3 meeting it twice, at (3, 4) and (3, −4), so the circle is not the graph of one function. Step 2 highlights the upper half, y equals the square root of 25 minus x squared, and dims the lower half. Step 3 adds the tangent line at the point (3, 4), with slope −3/4, and a label giving dy/dx = −x/y.

Check the tangent's slope another way: on the upper half, $y = \sqrt{25 - x^{2}}$ gives $y^{'} = \frac{- x}{\sqrt{25 - x^{2}}} = - \frac{3}{4}$ at $x = 3$ , as before. The implicit route reached it without choosing a half.

Example 22NCERT Example 22 · p. 123

Two routes, one answer

Find $\frac{d y}{d x}$ if $4 x - 2 y = 9$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 23NCERT Example 23 · p. 123

A relation you cannot solve

Find $\frac{d y}{d x}$ if $y^{2} + \sin y = x^{3}$ .

Try it yourself first. Then reveal the solution one step at a time.

5.3.3 Derivatives of inverse trigonometric functions

The inverse trigonometric functions of Chapter 2 are continuous on their domains; we accept this without proof. To differentiate them, turn each one round into a trigonometric equation and differentiate implicitly.

Example 24NCERT Example 24 · p. 124

The derivative of tan⁻¹ x

Find the derivative of $y = \tan^{- 1} x$ , assuming it exists.

Try it yourself first. Then reveal the solution one step at a time.

DerivationDerivatives of sin⁻¹ x and cos⁻¹ x

Let $y = \sin^{- 1} x$ , so $x = \sin y$ with $- \frac{π}{2} \leq y \leq \frac{π}{2}$ . Differentiating, $1 = \cos y \frac{d y}{d x}$ , so $\frac{d y}{d x} = \frac{1}{\cos y}$ . This needs $\cos y \neq 0$ , which rules out $y = \pm \frac{π}{2}$ , that is $x = \pm 1$ .

For $- \frac{π}{2} < y < \frac{π}{2}$ the cosine is positive, so $\cos y = + \sqrt{1 - \sin^{2} y} = \sqrt{1 - x^{2}}$ . Therefore

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

For $y = \cos^{- 1} x$ , $x = \cos y$ with $0 \leq y \leq π$ , and $1 = - \sin y \frac{d y}{d x}$ . On $0 < y < π$ the sine is positive, so $\sin y = \sqrt{1 - x^{2}}$ :

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

A check: $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$ , a constant, so the two derivatives must add to 0, and they do.

Table 5.3Derivatives of the inverse trigonometric functions

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

Why reciprocals? The graph of $\sin^{- 1}$ is the mirror image of a piece of the sine graph in the line $y = x$ . Reflection swaps rise and run, so a tangent of slope $m$ becomes a tangent of slope $\frac{1}{m}$ , as Fig. 5.10 shows.

3/3The reflected tangent has slope $2 = \frac{1}{\sqrt{1 - (\sqrt{3} / 2)^{2}}}$ , the reciprocal of $\frac{1}{2}$ .

Fig. 5.10Mirror images in the dashed line

y = x

have reciprocal slopes at matching points.NCERT p. 124

Square axes with a dashed mirror line y = x. Step 1 shows the piece of y = sin x from −π/2 to π/2 and the point (π/3, √3/2), about (1.05, 0.87), with its tangent of slope cos(π/3) = 1/2. Step 2 adds the reflected curve y = sin⁻¹ x from −1 to 1, and a dotted segment joining the point to its mirror image (√3/2, π/3). Step 3 adds the tangent to the reflected curve at (√3/2, π/3), whose slope is 2, the reciprocal of 1/2.

Quick check

1 of 4

\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})

Show answer

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

Outer:

\cos (5 x^{2})

; inner:

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

; multiply. Option A forgets the inner rate, and option C differentiates the inside twice.

Practice · 2 questions

Practice 5.3: questions with answers

Fresh questions on the skills of NCERT Exercise 5.3.

1
Short answerMediumNCERT Exercise 5.3, Q1–Q8
Find $\frac{d y}{d x}$ for each relation.
1. $5 x - 2 y = \cos x$
  Show solution
  Answer $\frac{5 + \sin x}{2}$
  $5 - 2 y^{'} = - \sin x$ , so $y^{'} = \frac{5 + \sin x}{2}$ .
2. $x + 4 y = \cos y$
  Show solution
  Answer $- \frac{1}{4 + \sin y}$
  $1 + 4 y^{'} = - \sin y y^{'}$ , so $y^{'} (4 + \sin y) = - 1$ .
3. $p x^{2} + q y = \sin y$ ( $p$ , $q$ constants)
  Show solution
  Answer $\frac{2 p x}{\cos y - q}$
  $2 p x + q y^{'} = \cos y y^{'}$ , so $y^{'} (\cos y - q) = 2 p x$ .
4. $x y^{2} + y = \sec x$
  Show solution
  Answer $\frac{\sec x \tan x - y^{2}}{2 x y + 1}$
  $y^{2} + 2 x y y^{'} + y^{'} = \sec x \tan x$ , so $y^{'} (2 x y + 1) = \sec x \tan x - y^{2}$ .
5. $x^{2} - x y + 2 y^{2} = 50$
  Show solution
  Answer $\frac{y - 2 x}{4 y - x}$
  $2 x - (y + x y^{'}) + 4 y y^{'} = 0$ , so $y^{'} (4 y - x) = y - 2 x$ .
6. $x^{3} + y^{3} = 3 x y + 7$
  Show solution
  Answer $\frac{y - x^{2}}{y^{2} - x}$
  $3 x^{2} + 3 y^{2} y^{'} = 3 y + 3 x y^{'}$ , so $y^{'} (y^{2} - x) = y - x^{2}$ .
7. $\cos^{2} y + \sin (x y) = \frac{1}{2}$
  Show solution
  Answer $\frac{y \cos (x y)}{\sin 2 y - x \cos (x y)}$
  $- 2 \cos y \sin y y^{'} + \cos (x y) (y + x y^{'}) = 0$ . With $2 \sin y \cos y = \sin 2 y$ : $y^{'} (x \cos (x y) - \sin 2 y) = - y \cos (x y)$ .
8. $\tan^{2} x + \sec^{2} y = 4$
  Show solution
  Answer $- \frac{\tan x \sec^{2} x}{\tan y \sec^{2} y}$
  $2 \tan x \sec^{2} x + 2 \sec y \cdot \sec y \tan y y^{'} = 0$ , so $y^{'} = - \frac{\tan x \sec^{2} x}{\tan y \sec^{2} y}$ .
2
Short answerHardNCERT Exercise 5.3, Q9–Q15
Simplify with a substitution, then find $\frac{d y}{d x}$ .
1. $y = \tan^{- 1} \frac{2 x}{1 - x^{2}}$ , $- 1 < x < 1$
  Show solution
  Answer $\frac{2}{1 + x^{2}}$
  With $x = \tan θ$ , $- \frac{π}{4} < θ < \frac{π}{4}$ : $\frac{2 \tan θ}{1 - \tan^{2} θ} = \tan 2 θ$ and $2 θ \in (- \frac{π}{2}, \frac{π}{2})$ , so $y = 2 θ = 2 \tan^{- 1} x$ .
2. $y = \sin^{- 1} (3 x - 4 x^{3})$ , $- \frac{1}{2} < x < \frac{1}{2}$
  Show solution
  Answer $\frac{3}{\sqrt{1 - x^{2}}}$
  With $x = \sin θ$ , $- \frac{π}{6} < θ < \frac{π}{6}$ : $3 \sin θ - 4 \sin^{3} θ = \sin 3 θ$ and $3 θ \in (- \frac{π}{2}, \frac{π}{2})$ , so $y = 3 \sin^{- 1} x$ .
3. $y = \sin^{- 1} \frac{1 - x^{2}}{1 + x^{2}}$ , $0 < x < 1$
  Show solution
  Answer $- \frac{2}{1 + x^{2}}$
  With $x = \tan θ$ , $0 < θ < \frac{π}{4}$ : the fraction is $\cos 2 θ = \sin (\frac{π}{2} - 2 θ)$ and $\frac{π}{2} - 2 θ \in (0, \frac{π}{2})$ . So $y = \frac{π}{2} - 2 \tan^{- 1} x$ .
4. $y = \cos^{- 1} (2 x \sqrt{1 - x^{2}})$ , $0 < x < \frac{1}{\sqrt{2}}$
  Show solution
  Answer $- \frac{2}{\sqrt{1 - x^{2}}}$
  With $x = \sin θ$ , $0 < θ < \frac{π}{4}$ : $2 \sin θ \cos θ = \sin 2 θ = \cos (\frac{π}{2} - 2 θ)$ , with $\frac{π}{2} - 2 θ \in (0, \frac{π}{2})$ . So $y = \frac{π}{2} - 2 \sin^{- 1} x$ .
5. $y = \tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}$ , $x \neq 0$
  Show solution
  Answer $\frac{1}{2 (1 + x^{2})}$
  With $x = \tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $θ \neq 0$ : $\frac{\sec θ - 1}{\tan θ} = \frac{1 - \cos θ}{\sin θ} = \tan \frac{θ}{2}$ . So $y = \frac{θ}{2} = \frac{1}{2} \tan^{- 1} x$ .
6. $y = \cos^{- 1} (4 x^{3} - 3 x)$ , $\frac{1}{2} < x < 1$
  Show solution
  Answer $- \frac{3}{\sqrt{1 - x^{2}}}$
  With $x = \cos θ$ , $0 < θ < \frac{π}{3}$ : $4 \cos^{3} θ - 3 \cos θ = \cos 3 θ$ and $3 θ \in (0, π)$ . So $y = 3 \cos^{- 1} x$ .
7. $y = \sec^{- 1} \frac{1}{1 - 2 x^{2}}$ , $0 < x < \frac{1}{\sqrt{2}}$
  Show solution
  Answer $\frac{2}{\sqrt{1 - x^{2}}}$
  With $x = \sin θ$ , $0 < θ < \frac{π}{4}$ : $1 - 2 \sin^{2} θ = \cos 2 θ$ , so the argument is $\sec 2 θ$ with $2 θ \in (0, \frac{π}{2})$ . So $y = 2 \sin^{- 1} x$ .

5.4 Exponential and Logarithmic Functions

Savings that earn interest, bacteria multiplying in milk, a medicine leaving the bloodstream: these change by equal factors over equal times, which no polynomial can describe. The functions that can are the exponentials, and their inverses are the logarithms. We argue for their main properties instead of proving them; the proofs need tools from later courses.

Start with powers. When $x$ goes from 1 to 2, $x^{2}$ goes from 1 to 4, $x^{5}$ from 1 to 32 and $x^{10}$ from 1 to 1024. For $x > 1$ , the higher the power, the faster it grows. Yet a different kind of function, $2^{x}$ , eventually leaves every power behind. Compare $x^{10}$ with $2^{x}$ at $x = 100$ : $100^{10} = 10^{20}$ , while $2^{100} \approx 1.27 \times 10^{30}$ , more than ten billion times larger. From about $x = 59$ onwards, $2^{x}$ stays ahead of $x^{10}$ for good, and a similar crossover happens against any power you choose. Fig. 5.11 shows the race against $x^{3}$ , where the crossover comes early enough to draw.

3/3At $x = 12$ : $4096$ against $1728$ . Each step of 1 doubles $2^{x}$ , but multiplies $x^{3}$ by less and less.

Fig. 5.11

y = x^{3}

against

y = 2^{x}

.NCERT Fig 5.9 · p. 126

Step 1 is a close-up with x from 0 to 6 and y from 0 to 230: the cube curve is far above the exponential, with dots at 125 and 32 above x = 5. Steps 2 and 3 zoom out to x from 0 to 12.6 and y from 0 to 4600. Step 2 marks the crossing near x = 9.94, where both values are close to 980. Step 3 marks the values at x = 12, where 2 to the power 12 is 4096 but 12 cubed is only 1728, and the exponential climbs almost vertically while the cube rises slowly.

Definition

Exponential function

For a real number $b > 1$ , the exponential function with base $b$ is $y = f (x) = b^{x}$ .

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

Sketch $y = 2^{x}$ , $y = 3^{x}$ and $y = 5^{x}$ on one set of axes; you will see the following features in every one of them.

Every graph passes through $(0, 1)$ , because $b^{0} = 1$ for every base.
Each graph rises all the way from left to right: $b^{x}$ is an increasing function.
Far to the left the graph hugs the negative $x$ -axis without ever touching it, because $b^{x}$ shrinks towards 0 but stays positive.
So $b^{x}$ accepts every real input (domain $𝐑$ ) and returns only positive outputs (range $(0, \infty)$ ).

In Class XI you met the number $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ , which lies between 2 and 3 (it is about 2.71828). With $e$ as the base we get the natural exponential function $y = e^{x}$ , the most important exponential in calculus.

Does an exponential function have an inverse? Since $b^{x}$ is increasing, each positive number is $b^{x}$ for exactly one $x$ . That exponent is its logarithm.

Definition

Logarithm

Let $b > 1$ and $a > 0$ . The logarithm of $a$ to base $b$ is the number $x$ with $b^{x} = a$ . We write $\log_{b} a = x$ .

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

More examples: $10^{- 2} = 0.01$ says the same as $\log_{10} 0.01 = - 2$ ; and since $64 = 4^{3} = 8^{2}$ , we have $\log_{4} 64 = 3$ and $\log_{8} 64 = 2$ . Fixing the base, $\log_{b}$ becomes a function from positive reals to all reals: $\log_{b} x = y$ exactly when $b^{y} = x$ . Base 10 gives common logarithms and base $e$ gives natural logarithms, often written $\ln x$ . In this chapter, $\log x$ always means the natural logarithm $\log_{e} x$ .

The graph of $\log_{b} x$ is the graph of $b^{x}$ reflected in the line $y = x$ , as Fig. 5.12 shows for base 2. The other features follow from this.
Reflecting $(0, 1)$ gives $(1, 0)$ : $\log_{b} 1 = 0$ for every base.
Reflection keeps the graph rising, so $\log_{b}$ is increasing.
Inputs and outputs swap roles: the logarithm accepts only positive numbers (domain $(0, \infty)$ ; $\log 0$ and logs of negatives are undefined) and can return any real number (range $𝐑$ ).
As $x$ shrinks towards 0, $\log x$ plunges without bound, hugging the negative $y$ -axis without touching it.

3/3The mirror points lie on $y = \log_{2} x$ , which passes through $(1, 0)$ .

Fig. 5.12Swapping coordinates turns the graph of

2^{x}

into the graph of

\log_{2} x

.NCERT Figs 5.10–5.11 · p. 128

Square axes from −2 to 5 with the dashed line y = x. Step 1 shows y = 2ˣ through the points (−1, 1/2), (0, 1), (1, 2) and (2, 4). Step 2 joins each of these points by a short dotted segment, crossing y = x at right angles, to its mirror point (1/2, −1), (1, 0), (2, 1) and (4, 2). Step 3 draws y = log₂ x through the mirror points, hugging the negative y-axis as x approaches 0.

The laws of logarithms come straight from the laws of exponents. In the derivation below, $p$ and $q$ are positive and the base $b > 1$ is fixed.

DerivationLaws of logarithms

Let $\log_{b} p = β$ and $\log_{b} q = γ$ , so $b^{β} = p$ and $b^{γ} = q$ .

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

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

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

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

Power. Applying the product law $n$ times to $p \cdot p \dots p$ gives $\log_{b} p^{n} = n \log_{b} p$ for every positive integer $n$ . (It holds for every real $n$ , but we do not prove that here.)

Change of base. Let $a > 1$ be another base and $\log_{a} p = α$ , so $a^{α} = p$ . If $\log_{b} a = δ$ , then $a = b^{δ}$ and $p = (b^{δ})^{α} = b^{δ α}$ . Comparing with $p = b^{β}$ gives $β = α δ$ , that is

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

Example 25NCERT Example 25 · p. 129

When does e^(2 log x) equal x²?

Is $e^{2 \log x} = x^{2}$ for every real $x$ ?

Try it yourself first. Then reveal the solution one step at a time.

Theorem 5Derivatives of eˣ and log x

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

We do not prove part (1). It says that at every point the slope of $y = e^{x}$ equals its height, and $e$ is the only base with this property. Part (2) then follows by implicit differentiation, assuming $\log x$ is differentiable: if $y = \log x$ then $x = e^{y}$ , so $1 = e^{y} \frac{d y}{d x}$ and $\frac{d y}{d x} = \frac{1}{e^{y}} = \frac{1}{x}$ .

Example 26NCERT Example 26 · p. 129

Chain rule with exponentials and logarithms

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

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Find

\log_{2} 32

Show answer

Answer 5

2^{5} = 32

, so

\log_{2} 32 = 5

Practice · 2 questions

Practice 5.4: questions with answers

Fresh questions on the skills of NCERT Exercise 5.4.

1
Short answerMediumNCERT Exercise 5.4, Q1–Q5
Differentiate with respect to $x$ .
1. $\frac{e^{x}}{1 + x^{2}}$
  Show solution
  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. $e^{\cos^{- 1} x}$ , $- 1 < x < 1$
  Show solution
  Answer $- \frac{e^{\cos^{- 1} x}}{\sqrt{1 - x^{2}}}$
  $e^{\cos^{- 1} x} \cdot (- \frac{1}{\sqrt{1 - x^{2}}})$ .
3. $e^{\sqrt{x}}$ , $x > 0$
  Show solution
  Answer $\frac{e^{\sqrt{x}}}{2 \sqrt{x}}$
  $e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ .
4. $\sin^{- 1} (e^{- 2 x})$ , $x > 0$
  Show solution
  Answer $- \frac{2 e^{- 2 x}}{\sqrt{1 - e^{- 4 x}}}$
  $\frac{1}{\sqrt{1 - (e^{- 2 x})^{2}}} \cdot (- 2 e^{- 2 x})$ . For $x > 0$ , $0 < e^{- 2 x} < 1$ , so the formula applies.
5. $\log (1 + \sin^{2} x)$
  Show solution
  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
Differentiate with respect to $x$ .
1. $e^{x} + e^{2 x^{2}} + e^{3 x^{3}}$
  Show solution
  Answer $e^{x} + 4 x e^{2 x^{2}} + 9 x^{2} e^{3 x^{3}}$
  Term by term: $e^{x}$ ; $e^{2 x^{2}} \cdot 4 x$ ; $e^{3 x^{3}} \cdot 9 x^{2}$ .
2. $\sqrt[3]{e^{x^{2}}}$
  Show solution
  Answer $\frac{2 x}{3} e^{x^{2} / 3}$
  $\sqrt[3]{e^{x^{2}}} = e^{x^{2} / 3}$ , whose derivative is $e^{x^{2} / 3} \cdot \frac{2 x}{3}$ .
3. $\log (1 + \log x)$ , $x > \frac{1}{e}$
  Show solution
  Answer $\frac{1}{x (1 + \log x)}$
  For $x > \frac{1}{e}$ , $1 + \log x > 0$ . The derivative is $\frac{1}{1 + \log x} \cdot \frac{1}{x}$ .
4. $\frac{\log x}{x}$ , $x > 0$
  Show solution
  Answer $\frac{1 - \log x}{x^{2}}$
  Quotient rule: $\frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^{2}}$ .
5. $\sin (x^{2} + \log x)$ , $x > 0$
  Show solution
  Answer $(2 x + \frac{1}{x}) \cos (x^{2} + \log x)$
  Outer $\sin t$ , inner $t = x^{2} + \log x$ with $t^{'} = 2 x + \frac{1}{x}$ .

5.5 Logarithmic Differentiation

How do you differentiate $y = x^{x}$ ? The power rule needs a fixed exponent and the rule for $a^{x}$ needs a fixed base, but here both vary. Taking logarithms turns the awkward exponent into a product, which the product rule handles. The same trick turns a long product or quotient into a sum. The general case is $y = [u (x)]^{v (x)}$ with $u (x) > 0$ .

DerivationDifferentiating u to the power v

Take natural logarithms of $y = u^{v}$ (both sides are positive):

\log y = v \log u

Differentiate both sides with respect to $x$ , using the chain rule on the left and the product rule on the right:

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

Multiply by $y = u^{v}$ :

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

This is logarithmic differentiation. Its one condition matters: $u (x)$ , and hence $y$ , must be positive, or the logarithms make no sense. For a product or quotient that may be negative, apply the method to $∣ y ∣$ : since $\frac{d}{d x} \log ∣ y ∣ = \frac{y^{'}}{y}$ wherever $y \neq 0$ , you reach the same formula. For $y = x^{x}$ ( $x > 0$ ) the formula gives $\frac{d y}{d x} = x^{x} (1 + \log x)$ , and Fig. 5.13 shows what that says about the graph.

Definition

Logarithmic differentiation

Finding $\frac{d y}{d x}$ by first taking logarithms of both sides of $y = f (x)$ , differentiating $\log y$ , and then multiplying by $y$ ; used for variable powers and for long products and quotients.

3/3The slope is 0 where $\log x = - 1$ , at $x = \frac{1}{e} \approx 0.37$ , $y \approx 0.69$ .

Fig. 5.13

y = x^{x}

and its derivative

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

.NCERT p. 130

Axes show x from 0 to 2 and y from 0 to 4. Step 1 draws y = x^x: it starts near height 1 just right of 0, dips to a low point and then climbs steeply to 4 at x = 2. Step 2 adds the tangent at (1, 1), which has slope 1. Step 3 marks the lowest point at x = 1/e, about 0.37, where y is about 0.69, with a flat tangent, because there 1 + log x = 0.

Example 27NCERT Example 27 · p. 131

A long quotient

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

Try it yourself first. Then reveal the solution one step at a time.

Example 28NCERT Example 28 · p. 131

The derivative of 7ˣ, two ways

Differentiate $y = 7^{x}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 29NCERT Example 29 · p. 132

Variable base, variable exponent

Differentiate $y = x^{\sqrt{x}}$ , $x > 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 30NCERT Example 30 · p. 132

A variable power inside an implicit relation

Find $\frac{d y}{d x}$ if $x^{y} + y^{2} = 5 x$ , where $x > 0$ and $y > 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

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

(for

x > 0

)

x \cdot x^{x - 1}

x^{x} \log x

x^{x}

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

Show answer

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

Logarithmic differentiation gives

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

. The answer

x \cdot x^{x - 1}

uses the power rule as if the exponent were fixed, and

x^{x} \log x

uses the

a^{x}

rule as if the base were fixed; the correct answer is their sum.

Practice · 6 questions

Practice 5.5: questions with answers

Fresh questions on the skills of NCERT Exercise 5.5.

1
Short answerMediumNCERT Exercise 5.5, Q1–Q6
Differentiate with respect to $x$ .
1. $x^{2} e^{x} \sin x$ , $0 < x < π$
  Show solution
  Answer $2 x e^{x} \sin x + x^{2} e^{x} \sin x + x^{2} e^{x} \cos x$
  $\log y = 2 \log x + x + \log \sin x$ , so $\frac{y^{'}}{y} = \frac{2}{x} + 1 + \cot x$ .
2. $\sqrt[3]{\frac{(x + 1) (x + 2)}{(x + 3) (x + 4)}}$ , $x > - 1$
  Show solution
  Answer $\frac{y}{3}$ times $\frac{1}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3} - \frac{1}{x + 4}$ , with $y$ the given function
  $\log y = \frac{1}{3} [\log (x + 1) + \log (x + 2) - \log (x + 3) - \log (x + 4)]$ ; differentiate and multiply by $y$ .
3. $(\tan x)^{x}$ , $0 < x < \frac{π}{2}$
  Show solution
  Answer $(\tan x)^{x} \log \tan x + \frac{2 x (\tan x)^{x}}{\sin 2 x}$
  $\log y = x \log \tan x$ ; $\frac{y^{'}}{y} = \log \tan x + x \frac{\sec^{2} x}{\tan x}$ , and $\frac{\sec^{2} x}{\tan x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2 x}$ .
4. $x^{2 x} + 3^{\cos x}$ , $x > 0$
  Show solution
  Answer $2 x^{2 x} (1 + \log x) - 3^{\cos x} \sin x \log 3$
  For $x^{2 x}$ : $\log u = 2 x \log x$ , $u^{'} = u (2 \log x + 2)$ . For $3^{\cos x}$ : $3^{\cos x} \log 3 \cdot (- \sin x)$ .
5. $(2 x + 1)^{3} (x - 4)^{2} (x + 1)^{5}$
  Show solution
  Answer $y$ times $\frac{6}{2 x + 1} + \frac{2}{x - 4} + \frac{5}{x + 1}$ , with $y$ the given product
  $\log ∣ y ∣ = 3 \log ∣ 2 x + 1 ∣ + 2 \log ∣ x - 4 ∣ + 5 \log ∣ x + 1 ∣$ ; differentiate (wherever $y \neq 0$ ) and multiply by $y$ .
6. $x^{1 / x} + (x + 1)^{x}$ , $x > 0$
  Show solution
  Answer $x^{1 / x} \frac{1 - \log x}{x^{2}} + (x + 1)^{x} \log (x + 1) + x (x + 1)^{x - 1}$
  $\log u = \frac{\log x}{x}$ gives $\frac{u^{'}}{u} = \frac{1 - \log x}{x^{2}}$ ; $\log v = x \log (x + 1)$ gives $\frac{v^{'}}{v} = \log (x + 1) + \frac{x}{x + 1}$ .
2
Short answerHardNCERT Exercise 5.5, Q7–Q11
Differentiate with respect to $x$ .
1. $x^{e^{x}} + (e^{x})^{x}$ , $x > 0$
  Show solution
  Answer $x^{e^{x}} e^{x} (\log x + \frac{1}{x}) + 2 x e^{x^{2}}$
  $\log u = e^{x} \log x$ , so $\frac{u^{'}}{u} = e^{x} \log x + \frac{e^{x}}{x}$ . And $(e^{x})^{x} = e^{x^{2}}$ , with derivative $2 x e^{x^{2}}$ .
2. $(1 + x^{2})^{x} + \tan^{- 1} \sqrt{x}$ , $x > 0$
  Show solution
  Answer $(1 + x^{2})^{x} \log (1 + x^{2}) + 2 x^{2} (1 + x^{2})^{x - 1} + \frac{1}{2 \sqrt{x} (1 + x)}$
  $\log u = x \log (1 + x^{2})$ gives $\frac{u^{'}}{u} = \log (1 + x^{2}) + \frac{2 x^{2}}{1 + x^{2}}$ . And $\frac{d}{d x} \tan^{- 1} \sqrt{x} = \frac{1}{1 + x} \cdot \frac{1}{2 \sqrt{x}}$ .
3. $x^{\tan x} + (\cos x)^{x^{2}}$ , $0 < x < \frac{π}{2}$
  Show solution
  Answer $x^{\tan x} \sec^{2} x \log x + x^{\tan x - 1} \tan x + 2 x (\cos x)^{x^{2}} \cdot \log \cos x - x^{2} (\cos x)^{x^{2}} \cdot \tan x$
  $\log u = \tan x \log x$ and $\log v = x^{2} \log \cos x$ ; differentiate each with the product rule.
4. $x^{x^{2}} + \frac{1 + x}{1 - x}$ , $0 < x < 1$
  Show solution
  Answer $x^{x^{2}} x (1 + 2 \log x) + \frac{2}{(1 - x)^{2}}$
  $\log u = x^{2} \log x$ gives $\frac{u^{'}}{u} = 2 x \log x + x$ . The quotient rule gives $\frac{(1 - x) + (1 + x)}{(1 - x)^{2}}$ .
5. $(x e^{x})^{x} + (\log x)^{1 / x}$ , $x > 1$
  Show solution
  Answer $(x e^{x})^{x} (\log x + 1 + 2 x) + \frac{(\log x)^{1 / x}}{x^{2} \log x} - \frac{(\log x)^{1 / x} \log (\log x)}{x^{2}}$
  $\log u = x \log x + x^{2}$ , so $\frac{u^{'}}{u} = \log x + 1 + 2 x$ . $\log v = \frac{\log (\log x)}{x}$ , so $\frac{v^{'}}{v} = \frac{1}{x^{2} \log x} - \frac{\log (\log x)}{x^{2}}$ .
3
Short answerHardNCERT Exercise 5.5, Q12–Q15
Find $\frac{d y}{d x}$ (all variables positive).
1. $x^{y} \cdot y^{x} = 1$
  Show solution
  Answer $- \frac{\frac{y}{x} + \log y}{\log x + \frac{x}{y}}$
  $y \log x + x \log y = 0$ . Differentiating: $y^{'} \log x + \frac{y}{x} + \log y + \frac{x}{y} y^{'} = 0$ .
2. $y = x^{y}$
  Show solution
  Answer $\frac{y^{2}}{x (1 - y \log x)}$
  $\log y = y \log x$ , so $\frac{y^{'}}{y} = y^{'} \log x + \frac{y}{x}$ , giving $y^{'} (\frac{1}{y} - \log x) = \frac{y}{x}$ .
3. $(1 + x)^{y} = (1 + y)^{x}$
  Show solution
  Answer $\frac{\log (1 + y) - \frac{y}{1 + x}}{\log (1 + x) - \frac{x}{1 + y}}$
  $y \log (1 + x) = x \log (1 + y)$ . Differentiating: $y^{'} \log (1 + x) + \frac{y}{1 + x} = \log (1 + y) + \frac{x}{1 + y} y^{'}$ .
4. $x^{2} y = e^{x + y}$
  Show solution
  Answer $\frac{y (x - 2)}{x (1 - y)}$
  $2 \log x + \log y = x + y$ . Differentiating: $\frac{2}{x} + \frac{y^{'}}{y} = 1 + y^{'}$ , so $y^{'} (\frac{1}{y} - 1) = 1 - \frac{2}{x}$ .
Short answerMediumNCERT Exercise 5.5, Q16
Let $f (x) = (1 + x) (1 + 2 x) (1 + 3 x) (1 + 4 x)$ . Find $f^{'} (x)$ by logarithmic differentiation, and hence find $f^{'} (1)$ .
Show solution
For $x$ where every factor is positive, $\log f = \sum \log (1 + k x)$ , so $f^{'} (x)$ is $f (x)$ times $\frac{1}{1 + x} + \frac{2}{1 + 2 x} + \frac{3}{1 + 3 x} + \frac{4}{1 + 4 x}$ .
At $x = 1$ : $f (1) = 2 \cdot 3 \cdot 4 \cdot 5 = 120$ and the bracket is $\frac{30 + 40 + 45 + 48}{60} = \frac{163}{60}$ . So $f^{'} (1) = 120 \cdot \frac{163}{60} = 326$ .
Short answerMediumNCERT Exercise 5.5, Q17
Differentiate $(x^{2} + 2 x - 1) (x^{3} - 4 x + 6)$ in three ways: (i) the product rule, (ii) expanding first, (iii) logarithmic differentiation. Do the answers agree?
Show solution
AnswerAll three give $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ .
(i) $(2 x + 2) (x^{3} - 4 x + 6) + (x^{2} + 2 x - 1) (3 x^{2} - 4)$ . Expanding each product: $2 x^{4} + 2 x^{3} - 8 x^{2} + 4 x + 12$ and $3 x^{4} + 6 x^{3} - 7 x^{2} - 8 x + 4$ , which add to $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ .
(ii) The product is $x^{5} + 2 x^{4} - 5 x^{3} - 2 x^{2} + 16 x - 6$ , whose derivative is $5 x^{4} + 8 x^{3} - 15 x^{2} - 4 x + 16$ .
(iii) Multiplying the hint by $y$ gives back exactly the expression in (i). So all three agree.
Short answerHardNCERT Exercise 5.5, Q18
If $u$ , $v$ and $w$ are differentiable functions of $x$ with $w \neq 0$ , show in two ways that $\frac{d}{d x} (\frac{u v}{w}) = \frac{u v}{w} (\frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w})$ (where $u, v \neq 0$ ): by the product and quotient rules, and by logarithmic differentiation.
Show solution
AnswerBoth ways give $\frac{u^{'} v w + u v^{'} w - u v w^{'}}{w^{2}}$ .
Rules: $\frac{(u v)^{'} w - u v w^{'}}{w^{2}} = \frac{u^{'} v w + u v^{'} w - u v w^{'}}{w^{2}}$ . Factor out $\frac{u v}{w}$ : this is $\frac{u v}{w} (\frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w})$ .
Logarithms: with $y = \frac{u v}{w}$ , $\log ∣ y ∣ = \log ∣ u ∣ + \log ∣ v ∣ - \log ∣ w ∣$ , so $\frac{y^{'}}{y} = \frac{u^{'}}{u} + \frac{v^{'}}{v} - \frac{w^{'}}{w}$ , and multiplying by $y$ gives the same result.

5.6 Derivatives of Functions in Parametric Forms

A batter lofts a cricket ball. Its horizontal distance $x$ and its height $y$ are each easier to describe in terms of the time $t$ than in terms of each other. Time is the natural link: each of $x$ and $y$ comes as its own formula in $t$ .

Definition

Parameter

If $x = f (t)$ and $y = g (t)$ for a third variable $t$ , the pair of equations gives the relation between $x$ and $y$ in parametric form, and $t$ is called the parameter.

Example $x = 3 \cos t$ , $y = 2 \sin t$ describes an ellipse with parameter $t$ .

To find $\frac{d y}{d x}$ , apply the chain rule to $y$ as a function of $x$ , with $x$ itself a function of $t$ : $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ . Divide, provided $\frac{d x}{d t} \neq 0$ :

Derivative in parametric form

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

In words: the slope of the path is the vertical speed divided by the horizontal speed. Fig. 5.14 shows this for a ball with $x = 20 t$ and $y = 20 t - 5 t^{2}$ (metres, seconds, air resistance ignored, $g$ rounded to $10 m / s^{2}$ ).

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 = vertical speed ÷ horizontal speed.NCERT p. 135

Axes show horizontal distance from 0 to 80 metres and height from 0 to 25 metres. The path is a symmetric arch from (0, 0) up to (40, 20) and down to (80, 0). Step 1 marks the ball at t = 1 s at (20, 15), with a horizontal speed arrow of 20 m/s and an upward arrow of 10 m/s, and the tangent of slope 10/20 = 1/2. Step 2 marks the ball at t = 2 s at the top (40, 20), where the vertical speed is 0 and the tangent is flat. Step 3 marks the ball at t = 3 s at (60, 15), with a downward arrow of 10 m/s and the tangent of slope −1/2.

Example 31NCERT Example 31 · p. 135

An ellipse

Find $\frac{d y}{d x}$ if $x = 3 \cos t$ , $y = 2 \sin t$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 32NCERT Example 32 · p. 135

The cricket ball

For the ball with $x = 20 t$ , $y = 20 t - 5 t^{2}$ , find $\frac{d y}{d x}$ . Where is the path level, and at what angle does the ball land?

Try it yourself first. Then reveal the solution one step at a time.

Example 33NCERT Example 33 · p. 136

A looping curve

Find $\frac{d y}{d x}$ if $x = 2 \cos t - \cos 2 t$ and $y = 2 \sin t - \sin 2 t$ .

Try it yourself first. Then reveal the solution one step at a time.

Notice that the answers above are written in the parameter alone, with no need to find $x$ or $y$ first. A parameter can even be introduced on purpose to tame an implicit relation, as the next example shows.

Example 34NCERT Example 34 · p. 136

Choosing a parameter

Find $\frac{d y}{d x}$ if $\sqrt{x} + \sqrt{y} = \sqrt{a}$ , where $a > 0$ and $x, y > 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

x = t^{2}

and

y = t^{3}

, find

\frac{d y}{d x}

t = 2

Show answer

Answer 3

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

t = 2

Practice · 3 questions

Practice 5.6: questions with answers

Fresh questions on the skills of NCERT Exercise 5.6.

1
Short answerMediumNCERT Exercise 5.6, Q1–Q5
Find $\frac{d y}{d x}$ without eliminating the parameter.
1. $x = 3 t^{2}$ , $y = 2 t^{3}$
  Show solution
  Answer $t$
  $\frac{6 t^{2}}{6 t} = t$ for $t \neq 0$ .
2. $x = 2 \sin θ$ , $y = 5 \sin θ$
  Show solution
  Answer $\frac{5}{2}$
  $\frac{5 \cos θ}{2 \cos θ} = \frac{5}{2}$ where $\cos θ \neq 0$ . (The points lie on the line $y = \frac{5}{2} x$ .)
3. $x = \tan t$ , $y = \cos 2 t$
  Show solution
  Answer $- 2 \sin 2 t \cos^{2} t$
  $\frac{- 2 \sin 2 t}{\sec^{2} t} = - 2 \sin 2 t \cos^{2} t$ .
4. $x = t + \frac{1}{t}$ , $y = t - \frac{1}{t}$
  Show solution
  Answer $\frac{t^{2} + 1}{t^{2} - 1}$
  $\frac{1 + 1 / t^{2}}{1 - 1 / t^{2}} = \frac{t^{2} + 1}{t^{2} - 1}$ for $t \neq \pm 1$ .
5. $x = 2 \cos θ + \cos 2 θ$ , $y = 2 \sin θ - \sin 2 θ$
  Show solution
  Answer $- \tan \frac{θ}{2}$
  $\frac{d y}{d θ} = 2 (\cos θ - \cos 2 θ) = 4 \sin \frac{3 θ}{2} \sin \frac{θ}{2}$ and $\frac{d x}{d θ} = - 2 (\sin θ + \sin 2 θ) = - 4 \sin \frac{3 θ}{2} \cos \frac{θ}{2}$ . The quotient is $- \tan \frac{θ}{2}$ .
2
Short answerMediumNCERT Exercise 5.6, Q6–Q10
Find $\frac{d y}{d x}$ without eliminating the parameter.
1. $x = 2 θ - \sin 2 θ$ , $y = 1 - \cos 2 θ$
  Show solution
  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. $x = \frac{1 - t^{2}}{1 + t^{2}}$ , $y = \frac{2 t}{1 + t^{2}}$
  Show solution
  Answer $\frac{t^{2} - 1}{2 t}$
  $\frac{d x}{d t} = \frac{- 4 t}{(1 + t^{2})^{2}}$ , $\frac{d y}{d t} = \frac{2 (1 - t^{2})}{(1 + t^{2})^{2}}$ . The quotient is $\frac{t^{2} - 1}{2 t}$ , which equals $- \frac{x}{y}$ : these points lie on the unit circle.
3. $x = \log (\sec t + \tan t)$ , $y = \tan t$ , $0 < t < \frac{π}{2}$
  Show solution
  Answer $\sec t$
  $\frac{d x}{d t} = \frac{\sec t \tan t + \sec^{2} t}{\sec t + \tan t} = \sec t$ and $\frac{d y}{d t} = \sec^{2} t$ , so $\frac{d y}{d x} = \sec t$ .
4. $x = 2 \tan θ$ , $y = 3 \sec θ$
  Show solution
  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. $x = e^{θ} (\sin θ + \cos θ)$ , $y = e^{θ} (\sin θ - \cos θ)$
  Show solution
  Answer $\tan θ$
  $\frac{d x}{d θ} = e^{θ} (\sin θ + \cos θ + \cos θ - \sin θ) = 2 e^{θ} \cos θ$ and $\frac{d y}{d θ} = 2 e^{θ} \sin θ$ .
Short answerMediumNCERT Exercise 5.6, Q11
If $x = e^{t} + e^{- t}$ and $y = e^{t} - e^{- t}$ , show that $\frac{d y}{d x} = \frac{x}{y}$ .
Show solution
Answer $\frac{d x}{d t} = y$ and $\frac{d y}{d t} = x$ , so $\frac{d y}{d x} = \frac{x}{y}$ (for $t \neq 0$ ).
$\frac{d x}{d t} = e^{t} - e^{- t} = y$ and $\frac{d y}{d t} = e^{t} + e^{- t} = x$ . Dividing, $\frac{d y}{d x} = \frac{x}{y}$ wherever $y \neq 0$ . (The points satisfy $x^{2} - y^{2} = 4$ , and implicit differentiation gives the same slope.)

5.7 Second Order Derivative

Differentiating $y = x^{4}$ gives $4 x^{3}$ , which is a new function, and nothing stops us differentiating that too: the result is $12 x^{2}$ . Repeating the operation like this is common enough to deserve a name and notation.

Definition

Second order derivative

If $y = f (x)$ and $f^{'}$ is differentiable, then $\frac{d}{d x} (\frac{d y}{d x})$ is the second order derivative of $y$ with respect to $x$ , written $\frac{d^{2} y}{d x^{2}}$ , $f^{''} (x)$ , $D^{2} y$ , $y^{''}$ or $y_{2}$ .

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

In the same notation the first derivative is often written $y_{1}$ , and third and higher derivatives are defined the same way. Two readings make $y^{''}$ concrete. If $s (t)$ is position, $\frac{d s}{d t}$ is velocity and $\frac{d^{2} s}{d t^{2}}$ is acceleration, the rate at which velocity changes. On a graph, $y^{''}$ is the rate at which the slope changes, so its sign tells which way the curve bends (Fig. 5.15).

3/3 $y^{''} = 6 x$ : negative for $x < 0$ (bends down), positive for $x > 0$ (bends up).

Fig. 5.15

y^{''}

is the slope of the slope.NCERT p. 137

The top panel shows y = x³ − 3x for x from −2.2 to 2.2, rising to a peak at (−1, 2), falling through the origin and reaching a valley at (1, −2). Step 1 draws short tangents at x = −1, 0 and 1 with slopes 0, −3 and 0. Step 2 draws the bottom panel, the slope graph y′ = 3x² − 3, a U-shaped curve through (−1, 0), (0, −3) and (1, 0), with dotted links to the matching tangents. Step 3 shades the left half of the top panel as bending down, where y″ = 6x is negative, and the right half as bending up, where y″ is positive.

Example 35NCERT Example 35 · p. 138

Differentiating twice

Find $\frac{d^{2} y}{d x^{2}}$ if $y = x^{4} - \sec x$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 36NCERT Example 36 · p. 138

A function that its second derivative undoes

If $y = A \cos 3 x + B \sin 3 x$ ( $A$ , $B$ constants), show that $\frac{d^{2} y}{d x^{2}} + 9 y = 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 37NCERT Example 37 · p. 138

Two exponentials

If $y = 2 e^{x} - e^{4 x}$ , show that $\frac{d^{2} y}{d x^{2}} - 5 \frac{d y}{d x} + 4 y = 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 38NCERT Example 38 · p. 139

An inverse sine squared

If $y = (\sin^{- 1} x)^{2}$ , $- 1 < x < 1$ , show that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

y = x^{5}

, find

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

x = 1

Show answer

Answer 20

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

and

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

, which is 20 at

x = 1

Practice · 4 questions

Practice 5.7: questions with answers

Fresh questions on the skills of NCERT Exercise 5.7.

1
Short answerEasyNCERT Exercise 5.7, Q1–Q5
Find the second order derivative of each function.
1. $4 x^{3} - 5 x + 7$
  Show solution
  Answer $24 x$
  $y^{'} = 12 x^{2} - 5$ , $y^{''} = 24 x$ .
2. $x^{15}$
  Show solution
  Answer $210 x^{13}$
  $y^{'} = 15 x^{14}$ , $y^{''} = 210 x^{13}$ .
3. $x^{2} \sin x$
  Show solution
  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. $\sqrt{x}$ , $x > 0$
  Show solution
  Answer $- \frac{1}{4 x^{3 / 2}}$
  $y^{'} = \frac{1}{2} x^{- 1 / 2}$ , $y^{''} = - \frac{1}{4} x^{- 3 / 2}$ .
5. $x^{2} \log x$ , $x > 0$
  Show solution
  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
Find the second order derivative of each function.
1. $e^{2 x} \sin x$
  Show solution
  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. $e^{- x} \cos 2 x$
  Show solution
  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. $\sin^{- 1} x$ , $- 1 < x < 1$
  Show solution
  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. $\log (1 + x^{2})$
  Show solution
  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. $\log (\sin x)$ , $0 < x < π$
  Show solution
  Answer $- {cosec}^{2} x$
  $y^{'} = \frac{\cos x}{\sin x} = \cot x$ ; $y^{''} = - {cosec}^{2} x$ .
3
Short answerHardNCERT Exercise 5.7, Q11–Q14
Prove each identity.
1. If $y = 4 \sin 2 x - 3 \cos 2 x$ , show that $y^{''} + 4 y = 0$ .
  Show solution
  Answer $y^{''} = - 4 y$ .
  $y^{'} = 8 \cos 2 x + 6 \sin 2 x$ and $y^{''} = - 16 \sin 2 x + 12 \cos 2 x = - 4 (4 \sin 2 x - 3 \cos 2 x)$ .
2. If $y = \sin^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.
  Show solution
  Answer $\sec^{2} y \tan y$
  $x = \sin y$ gives $y^{'} = \frac{1}{\cos y} = \sec y$ (cosine is positive on $(- \frac{π}{2}, \frac{π}{2})$ ). Then $y^{''} = \frac{d}{d x} \sec y = \sec y \tan y \cdot y^{'} = \sec^{2} y \tan y$ .
3. If $y = x \cos (\log x)$ , $x > 0$ , show that $x^{2} y_{2} - x y_{1} + 2 y = 0$ .
  Show solution
  AnswerSubstitution gives 0.
  $y_{1} = \cos (\log x) - \sin (\log x)$ and $y_{2} = - \frac{\sin (\log x) + \cos (\log x)}{x}$ . Write $s = \sin (\log x)$ and $c = \cos (\log x)$ ; then
  $x^{2} y_{2} - x y_{1} + 2 y = (- x s - x c) - (x c - x s) + 2 x c = 0$
4. If $y = (A + B x) e^{k x}$ , show that $y^{''} - 2 k y^{'} + k^{2} y = 0$ .
  Show solution
  AnswerSubstitution gives 0.
  $y^{'} = B e^{k x} + k y$ and $y^{''} = k B e^{k x} + k y^{'} = 2 k B e^{k x} + k^{2} y$ . So $y^{''} - 2 k y^{'} + k^{2} y = 2 k B e^{k x} + k^{2} y - 2 k B e^{k x} - 2 k^{2} y + k^{2} y = 0$ .
4
Short answerHardNCERT Exercise 5.7, Q15–Q17
Prove each identity.
1. If $y = 4 e^{- 3 x} + 7 e^{3 x}$ , show that $\frac{d^{2} y}{d x^{2}} = 9 y$ .
  Show solution
  Answer $y^{''} = 36 e^{- 3 x} + 63 e^{3 x} = 9 y$ .
  $y^{'} = - 12 e^{- 3 x} + 21 e^{3 x}$ and $y^{''} = 36 e^{- 3 x} + 63 e^{3 x} = 9 (4 e^{- 3 x} + 7 e^{3 x})$ .
2. If $y = \log (\cos x)$ , $- \frac{π}{2} < x < \frac{π}{2}$ , show that $y^{''} + (y^{'})^{2} + 1 = 0$ .
  Show solution
  Answer $y^{''} = - \sec^{2} x = - (1 + (y^{'})^{2})$ .
  $y^{'} = - \tan x$ and $y^{''} = - \sec^{2} x = - (1 + \tan^{2} x) = - (1 + (y^{'})^{2})$ .
3. If $y = e^{\tan^{- 1} x}$ , show that $(1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0$ .
  Show solution
  AnswerDifferentiate $(1 + x^{2}) y_{1} = y$ .
  $y_{1} = \frac{y}{1 + x^{2}}$ , so $(1 + x^{2}) y_{1} = y$ . Differentiating, $2 x y_{1} + (1 + x^{2}) y_{2} = y_{1}$ , which rearranges to the result.

Miscellaneous examples

These examples combine the techniques of the chapter. Two habits help: state where the function is defined before differentiating, and simplify first when an identity is available, checking that it holds on the whole interval you need.

Example 39NCERT Example 39 · p. 140

Roots and a change of base

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

Try it yourself first. Then reveal the solution one step at a time.

Example 40NCERT Example 40 · p. 141

Simplify first, but mind the interval

Differentiate (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) $y = \cos^{- 1} \frac{1 - 9^{x}}{1 + 9^{x}}$ .

Try it yourself first. Then reveal the solution one step at a time.

2/2Beyond $[0, π]$ the graph turns: slopes alternate between $1$ and $- 1$ .

Fig. 5.16

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

equals

x

only on

[0, π]

.NCERT p. 141

Axes show x from −π to 3π and y from 0 to π. Step 1 draws the line y = x dashed and the graph of y = cos⁻¹(cos x) only for x from 0 to π, where the two coincide. Step 2 draws the whole graph: a zigzag of straight segments rising with slope 1 from 0 to π and falling with slope −1 from π to 2π, repeating, with peaks of height π at x = −π, π and 3π. The line y = x leaves the zigzag at x = π.

Example 41NCERT Example 41 · p. 143

The compound-interest function

Find $f^{'} (x)$ if $f (x) = {(1 + \frac{1}{x})}^{x}$ , $x > 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 42NCERT Example 42 · p. 143

A parametric pair with variable powers

Find $\frac{d y}{d x}$ if $x = (t^{2} + 1)^{3}$ and $y = 3^{t^{2}}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 43NCERT Example 43 · p. 144

One function with respect to another

Differentiate $\tan x$ with respect to $\sin x$ , for $- \frac{π}{2} < x < \frac{π}{2}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

On which interval is

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

[- \frac{π}{2}, \frac{π}{2}]

[0, π]

[0, 2 π]

every real

x

Show answer

Answer (B) $[0, π]$

\cos^{- 1}

returns values in

[0, π]

, so the identity can hold only there.

[- \frac{π}{2}, \frac{π}{2}]

is the range of

\sin^{- 1}

, the usual mix-up.

Practice · 13 questions

Practice questions with answers (NCERT “Miscellaneous exercises”)

Fresh questions on the skills of NCERT's Miscellaneous Exercise on Chapter 5, plus three quick self-checks.

1
Short answerMediumNCERT Miscellaneous Exercise 5, Q1–Q6
Differentiate with respect to $x$ .
1. $(2 x^{3} - x + 4)^{7}$
  Show solution
  Answer $7 (2 x^{3} - x + 4)^{6} (6 x^{2} - 1)$
  Chain rule: $7 u^{6} \cdot u^{'}$ with $u = 2 x^{3} - x + 4$ .
2. $\sin^{4} x + \cos^{4} x$
  Show solution
  Answer $- \sin 4 x$
  The derivative is $4 \sin^{3} x \cos x - 4 \cos^{3} x \sin x$ . Factorise: $- 4 \sin x \cos x \cdot (\cos^{2} x - \sin^{2} x)$ , which is $- 2 \sin 2 x \cos 2 x = - \sin 4 x$ .
3. $(x^{2})^{\sin 3 x}$ , $x > 0$
  Show solution
  Answer $3 \cos 3 x \log (x^{2}) \cdot (x^{2})^{\sin 3 x} + \frac{2 \sin 3 x}{x} \cdot (x^{2})^{\sin 3 x}$
  $\log y = \sin 3 x \cdot 2 \log x$ , so $\frac{y^{'}}{y} = 3 \cos 3 x \cdot 2 \log x + \frac{2 \sin 3 x}{x}$ , and $2 \log x = \log (x^{2})$ .
4. $\tan^{- 1} (x \sqrt{x})$ , $x \geq 0$
  Show solution
  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. $\frac{\sin^{- 1} (x / 3)}{\sqrt{x + 4}}$ , $- 3 < x < 3$
  Show solution
  Answer $\frac{1}{\sqrt{9 - x^{2}} \sqrt{x + 4}} - \frac{\sin^{- 1} (x / 3)}{2 (x + 4)^{3 / 2}}$
  $\frac{d}{d x} \sin^{- 1} \frac{x}{3} = \frac{1}{\sqrt{1 - x^{2} / 9}} \cdot \frac{1}{3} = \frac{1}{\sqrt{9 - x^{2}}}$ . Then the quotient rule with $v = (x + 4)^{1 / 2}$ , $v^{'} = \frac{1}{2} (x + 4)^{- 1 / 2}$ .
6. $\tan^{- 1} \frac{\cos x}{1 + \sin x}$ , $- \frac{π}{2} < x < \frac{π}{2}$
  Show solution
  Answer $- \frac{1}{2}$
  $\frac{\cos x}{1 + \sin x} = \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2})^{2}} = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \tan (\frac{π}{4} - \frac{x}{2})$ , with $\frac{π}{4} - \frac{x}{2} \in (0, \frac{π}{2})$ . So $y = \frac{π}{4} - \frac{x}{2}$ .
2
Short answerHardNCERT Miscellaneous Exercise 5, Q7–Q11
Differentiate with respect to $x$ .
1. $(\sin x)^{\log x}$ , $0 < x < π$
  Show solution
  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)
  Show solution
  Answer $(a \cos x - b \sin x) e^{a \sin x + b \cos x}$
  Chain rule with inner derivative $a \cos x - b \sin x$ .
3. $(x^{2} + 1)^{x^{2} + 1}$
  Show solution
  Answer $2 x (x^{2} + 1)^{x^{2} + 1} \cdot [\log (x^{2} + 1) + 1]$
  $\log y = (x^{2} + 1) \log (x^{2} + 1)$ , so $\frac{y^{'}}{y} = 2 x \log (x^{2} + 1) + (x^{2} + 1) \cdot \frac{2 x}{x^{2} + 1}$ .
4. $x^{x} - x^{e} + e^{x} - e^{e}$ , $x > 0$
  Show solution
  Answer $x^{x} (1 + \log x) - e x^{e - 1} + e^{x}$
  Four different rules: variable base and exponent; a constant power ( $e$ is a number); the natural exponential; a constant.
5. $x^{x^{2} + 1} + (x - 1)^{x}$ , $x > 1$
  Show solution
  Answer $2 x \log x \cdot x^{x^{2} + 1} + (x^{2} + 1) x^{x^{2}} + (x - 1)^{x} \log (x - 1) + x (x - 1)^{x - 1}$
  $\log u = (x^{2} + 1) \log x$ and $\log v = x \log (x - 1)$ ; differentiate each and multiply back.
3
Short answerMediumNCERT Miscellaneous Exercise 5, Q12–Q14
Parametric and implicit slopes.
1. Find $\frac{d y}{d x}$ if $x = 6 (t + \sin t)$ , $y = 4 (1 - \cos t)$ , $- π < t < π$ .
  Show solution
  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. Find $\frac{d y}{d x}$ if $y = \tan^{- 1} x + \tan^{- 1} \frac{1}{x}$ , $x > 0$ .
  Show solution
  Answer0
  Directly: the two derivatives cancel. Or note that for $x > 0$ the sum is the constant $\frac{π}{2}$ .
3. If $y \sqrt{x^{2} + 1} = \log (\sqrt{x^{2} + 1} - x)$ , show that $(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ .
  Show solution
  AnswerDifferentiate both sides and multiply by $\sqrt{x^{2} + 1}$ .
  Right side: $\frac{\frac{x}{\sqrt{x^{2} + 1}} - 1}{\sqrt{x^{2} + 1} - x} = - \frac{1}{\sqrt{x^{2} + 1}}$ . Left side: $y^{'} \sqrt{x^{2} + 1} + \frac{x y}{\sqrt{x^{2} + 1}}$ . Equate and multiply by $\sqrt{x^{2} + 1}$ .
Short answerHardNCERT Miscellaneous Exercise 5, Q15
A circle has centre $(a, b)$ and radius $c$ : $x = a + c \cos t$ , $y = b + c \sin t$ . Show that $\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{∣ \frac{d^{2} y}{d x^{2}} ∣} = c$ , a constant that does not depend on $a$ and $b$ .
Show solution
Answer $\frac{d y}{d x} = - \cot t$ and $\frac{d^{2} y}{d x^{2}} = - \frac{1}{c \sin^{3} t}$ , so the ratio is $c$ .
$\frac{d y}{d x} = \frac{c \cos t}{- c \sin t} = - \cot t$ . Then $\frac{d^{2} y}{d x^{2}} = \frac{{cosec}^{2} t}{- c \sin t} = - \frac{1}{c \sin^{3} t}$ .
Numerator: $(1 + \cot^{2} t)^{3 / 2} = ({cosec}^{2} t)^{3 / 2} = \frac{1}{∣ \sin t ∣^{3}}$ . Dividing by $\frac{1}{c ∣ \sin t ∣^{3}}$ leaves $c$ . (This ratio is the radius of curvature; for a circle it is the radius.)
Short answerHardNCERT Miscellaneous Exercise 5, Q16
If $\sin y = x \sin (a + y)$ , where $\sin a \neq 0$ , prove that $\frac{d y}{d x} = \frac{\sin^{2} (a + y)}{\sin a}$ .
Show solution
Answer $\frac{d x}{d y} = \frac{\sin a}{\sin^{2} (a + y)}$ .
$x = \frac{\sin y}{\sin (a + y)}$ , so $\frac{d x}{d y} = \frac{\cos y \sin (a + y) - \sin y \cos (a + y)}{\sin^{2} (a + y)} = \frac{\sin ((a + y) - y)}{\sin^{2} (a + y)} = \frac{\sin a}{\sin^{2} (a + y)}$ .
Since $\frac{d y}{d x} = 1 \div \frac{d x}{d y}$ , the result follows.
6
Short answerHardNCERT Miscellaneous Exercise 5, Q17–Q18
Second derivatives in unusual settings.
1. If $x = e^{t} \cos t$ and $y = e^{t} \sin t$ , find $\frac{d^{2} y}{d x^{2}}$ .
  Show solution
  Answer $\frac{2}{e^{t} (\cos t - \sin t)^{3}}$
  $\frac{d y}{d x} = \frac{\sin t + \cos t}{\cos t - \sin t}$ . By the quotient rule its $t$ -derivative is
  $\frac{(\cos t - \sin t)^{2} + (\sin t + \cos t)^{2}}{(\cos t - \sin t)^{2}} = \frac{2}{(\cos t - \sin t)^{2}}$
  Divide this by $\frac{d x}{d t} = e^{t} (\cos t - \sin t)$ .
2. Let $f (x) = x ∣ x ∣$ . Show that $f^{'} (x)$ exists for every real $x$ , but $f^{''} (0)$ does not exist.
  Show solution
  Answer $f^{'} (x) = 2 ∣ x ∣$ ; $f^{''}$ is $- 2$ for $x < 0$ and $2$ for $x > 0$ .
  $f (x) = x^{2}$ for $x \geq 0$ and $- x^{2}$ for $x < 0$ . Away from 0, $f^{'} (x) = 2 x$ or $- 2 x$ , that is $2 ∣ x ∣$ . At 0, $\frac{f (h) - f (0)}{h} = ∣ h ∣ \to 0$ , so $f^{'} (0) = 0$ and $f^{'} (x) = 2 ∣ x ∣$ everywhere. But $2 ∣ x ∣$ has a corner at 0 (one-sided derivatives $- 2$ and $2$ ), so $f^{''} (0)$ does not exist.
Short answerMediumNCERT Miscellaneous Exercise 5, Q19
Starting from $\cos (x + B) = \cos x \cos B - \sin x \sin B$ , where $B$ is a constant, differentiate with respect to $x$ to obtain the addition formula for $\sin (x + B)$ .
Show solution
Answer $\sin (x + B) = \sin x \cos B + \cos x \sin B$ .
Differentiating, $- \sin (x + B) = - \sin x \cos B - \cos x \sin B$ . Multiply by $- 1$ .
Short answerMediumNCERT Miscellaneous Exercise 5, Q20
Give a function that is continuous at every real number but not differentiable at exactly three points, $x = - 1$ , $0$ and $2$ . Justify your answer.
Show solution
AnswerFor example $f (x) = ∣ x + 1 ∣ + ∣ x ∣ + ∣ x - 2 ∣$ .
Each term is continuous, so $f$ is continuous. Away from $- 1, 0, 2$ , $f$ is linear near each point and so differentiable.
At $x = 0$ the other two terms are differentiable, while $∣ x ∣$ has one-sided derivatives $- 1$ and $1$ . So the one-sided derivatives of $f$ differ by 2 and $f^{'} (0)$ does not exist. The same argument works at $- 1$ and at $2$ .
Short answerHardNCERT Miscellaneous Exercise 5, Q21
Let $y = ∣ \begin{array}{cc} x^{2} & \sin x \\ e^{x} & x \end{array} ∣$ . Find $\frac{d y}{d x}$ by expanding, and check that it equals $∣ \begin{array}{cc} 2 x & \cos x \\ e^{x} & x \end{array} ∣ + ∣ \begin{array}{cc} x^{2} & \sin x \\ e^{x} & 1 \end{array} ∣$ (differentiate one row at a time).
Show solution
Answer $3 x^{2} - e^{x} (\sin x + \cos x)$ both ways.
$y = x^{3} - e^{x} \sin x$ , so $y^{'} = 3 x^{2} - e^{x} \sin x - e^{x} \cos x$ .
The first determinant is $2 x^{2} - e^{x} \cos x$ and the second is $x^{2} - e^{x} \sin x$ . Their sum is $3 x^{2} - e^{x} \cos x - e^{x} \sin x$ , the same.
Short answerHardNCERT Miscellaneous Exercise 5, Q22
If $y = \sin (m \sin^{- 1} x)$ , $- 1 < x < 1$ , show that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ .
Show solution
AnswerDifferentiating $\sqrt{1 - x^{2}} y_{1} = m \cos (m \sin^{- 1} x)$ gives the result.
$y_{1} = \cos (m \sin^{- 1} x) \cdot \frac{m}{\sqrt{1 - x^{2}}}$ , so $\sqrt{1 - x^{2}} y_{1} = m \cos (m \sin^{- 1} x)$ .
Differentiate: $\sqrt{1 - x^{2}} y_{2} - \frac{x}{\sqrt{1 - x^{2}}} y_{1} = - m \sin (m \sin^{- 1} x) \cdot \frac{m}{\sqrt{1 - x^{2}}} = - \frac{m^{2} y}{\sqrt{1 - x^{2}}}$ . Multiply by $\sqrt{1 - x^{2}}$ .
Quick answerEasy
Which of these functions is continuous at $x = 0$ but not differentiable there?
$x^{2}$ $[x]$ $∣ \sin x ∣$ $\frac{1}{x}$
Show answer
Answer (C) $∣ \sin x ∣$
$∣ \sin x ∣$ has a corner at 0 (one-sided derivatives $- 1$ and $1$ ). $x^{2}$ is differentiable, $[x]$ jumps at 0, and $\frac{1}{x}$ is not defined at 0.
Quick answerEasy
If $x = 2 t^{2}$ and $y = t^{4}$ , find $\frac{d y}{d x}$ when $t = 3$ .
Show answer
Answer 9
$\frac{d y}{d x} = t^{2} = 9$ .
Quick answerEasy
True or false: if $f$ is differentiable at $c$ , then $\lim_{x \to c} f (x)$ exists.
TrueFalse
Show answer
Answer True
Differentiability implies continuity (Theorem 3), so the limit exists and equals $f (c)$ .

Recap

Continuity and Differentiability: key points and summary

A real function $f$ is continuous at a point $c$ of its domain if $\lim_{x \to c} f (x) = f (c)$ : both one-sided limits exist, agree, and equal $f (c)$ (Section 5.2).
A continuous function has no point of discontinuity in its domain; at the end of a closed interval only the one-sided limit is used (continuous functions).
Sums, differences, products and quotients (where the denominator is non-zero) of continuous functions are continuous, and so are composites (Section 5.2.1).
Polynomials, rational functions, trigonometric functions, $∣ x ∣$ , exponentials and logarithms are continuous on their domains; $[x]$ jumps at every integer (Example 15).
$f$ is differentiable at $c$ when its left-hand and right-hand derivatives at $c$ are finite and equal. Differentiability implies continuity, not the other way round: $∣ x - 3 ∣$ is unbroken but has a corner at 3 (Section 5.3).
Chain rule: if $t = u (x)$ and $f = v \circ u$ , then $\frac{d f}{d x} = \frac{d v}{d t} \cdot \frac{d t}{d x}$ , and longer chains multiply more rates (Section 5.3.1).
For an implicit relation, differentiate both sides with respect to $x$ , attaching $\frac{d y}{d x}$ to every term in $y$ (Section 5.3.2).
$\frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}$ on $(- 1, 1)$ , and $\frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}}$ (Section 5.3.3).
$\log_{b} a = x$ means $b^{x} = a$ ; $\log$ turns products into sums, quotients into differences and powers into multiples, and $\log_{b} x$ is the mirror image of $b^{x}$ in $y = x$ (Section 5.4).
$\frac{d}{d x} e^{x} = e^{x}$ , $\frac{d}{d x} \log x = \frac{1}{x}$ and $\frac{d}{d x} a^{x} = a^{x} \log a$ (Theorem 5).
Logarithmic differentiation handles $[u (x)]^{v (x)}$ and long products: $\frac{d y}{d x} = u^{v} [\frac{v}{u} u^{'} + v^{'} \log u]$ , valid when $u > 0$ (Section 5.5).
For $x = f (t)$ , $y = g (t)$ : $\frac{d y}{d x} = \frac{g^{'} (t)}{f^{'} (t)}$ wherever $f^{'} (t) \neq 0$ (Section 5.6).
The second derivative $\frac{d^{2} y}{d x^{2}}$ is the derivative of $\frac{d y}{d x}$ : the rate of change of the slope (Section 5.7).

Flash cards · 10

Derivatives to know

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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 on Continuity and Differentiability

5.2 Continuity4 questions
5.2.1 Algebra of continuous functions3 questions
5.3 Differentiability4 questions
5.3.3 Derivatives of inverse trigonometric functions4 questions
5.4 Exponential and Logarithmic Functions3 questions
5.5 Logarithmic Differentiation3 questions
5.6 Derivatives of Functions in Parametric Forms3 questions
5.7 Second Order Derivative3 questions
Miscellaneous examples3 questions

Glossary · 13

Key terms (English–Hindi)

chain ruleशृंखला नियमshrinkhala niyam: The derivative of a composite is the product of the rates along the chain: $\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}$ .First seen in the chapter →
continuous (at a point)संततsantat: $f$ is continuous at a point $c$ of its domain when $\lim_{x \to c} f (x)$ exists and equals $f (c)$ .First seen in the chapter →
continuous functionसंतत फलनsantat phalan: A function that is continuous at every point of its domain.

differentiableअवकलनीयavakalaneeya: $f$ is differentiable at $c$ when its left-hand and right-hand derivatives at $c$ are finite and equal.First seen in the chapter →

exponential functionचरघातांकी फलनcharghaataanki phalan: $y = b^{x}$ for a fixed base $b > 1$ ; with base $e$ it is the natural exponential $e^{x}$ .

greatest integer functionमहत्तम पूर्णांक फलनmahattam poornaank phalan: $[x]$ , the largest integer less than or equal to $x$ ; for example $[- 1.2] = - 2$ .First seen in the chapter →

implicit functionअस्पष्ट फलनaspasht phalan: A function of $x$ fixed by a relation between $x$ and $y$ that is not written in the form $y = f (x)$ .

logarithmलघुगणकlaghuganak: $\log_{b} a = x$ means $b^{x} = a$ . In this chapter $\log x$ is the natural logarithm, base $e$ .
logarithmic differentiationलघुगणकीय अवकलनlaghuganakiya avakalan: Differentiating $\log y$ instead of $y$ , then multiplying by $y$ ; used for $[u (x)]^{v (x)}$ and long products.

one-sided limitएकपक्षीय सीमाekpakshiya seema: The value $f (x)$ approaches as $x \to c$ from the left only ( $x < c$ ) or from the right only ( $x > c$ ).First seen in the chapter →

parameterप्राचलpraachal: A third variable $t$ through which $x = f (t)$ and $y = g (t)$ are both given.
point of discontinuityअसांतत्य का बिंदुasaantatya ka bindu: A point of the domain at which the function is not continuous.First seen in the chapter →

second order derivativeद्वितीय कोटि का अवकलजdvitiya koti ka avakalaj: The derivative of $\frac{d y}{d x}$ , written $\frac{d^{2} y}{d x^{2}}$ or $y^{''}$ .

5.1 Introduction

5.2 Continuity

5.2.1 Algebra of continuous functions

5.3 Differentiability

5.3.1 Derivatives of composite functions

5.3.2 Derivatives of implicit functions

5.3.3 Derivatives of inverse trigonometric functions

5.4 Exponential and Logarithmic Functions

5.5 Logarithmic Differentiation

5.6 Derivatives of Functions in Parametric Forms

5.7 Second Order Derivative

Miscellaneous examples

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Quick check: MCQs on Continuity and Differentiability

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