Limits and Derivatives: Class 11 Maths · Chapter 12

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

Published 22 Oct 2026 · Updated 22 Oct 2026

95 min
1 simulation
52 practice questions
23 MCQs

What you will learn in Limits and Derivatives

Estimate an instantaneous rate of change from average rates over shrinking intervals and interpret it as the slope of a tangent.
Find left-hand and right-hand limits from tables, graphs and formulas, and decide whether a limit exists.
Evaluate limits with the algebra of limits, substitution, cancelling common factors and the standard limit of (xⁿ − aⁿ)/(x − a).
Prove cos x < sin x/x < 1 by comparing areas, use the sandwich theorem to find the limits of sin x/x and (1 − cos x)/x, and evaluate trigonometric limits.
Define the derivative at a point and as a function, find derivatives from the first principle, and interpret them as slopes of tangents.
Differentiate sums, differences, products and quotients, powers of x, polynomials and the six trigonometric functions.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 12, pp. 217–256. Not affiliated with or endorsed by NCERT.

12.1 Introduction

A car's speedometer shows one number: how fast the car is going at this instant. A doctor checking a fever wants to know how quickly the temperature is falling now, not its average fall overnight. An engineer at a river gauge in the monsoon watches how fast the water is rising this hour. Each is a rate of change at an instant, and the mathematics built to measure such rates is called calculus (कलन).

In calculus we ask how fast the output of a function changes when its input changes, and we answer with two new ideas: the limit and the derivative. We first meet speed at an instant informally, then study limits with tables, graphs and algebra, and finally define the derivative and find it for polynomials and trigonometric functions. Class 12 builds continuity, more rules and integration on this base.

12.2 Intuitive Idea of Derivatives

A metro train pulls out of a station and speeds up steadily for the first 20 seconds. Its distance $s$ (in metres) from the starting point after $t$ seconds is $s = 0.5 t^{2}$ . The driver's speedometer shows the speed at every instant. Can we work out its reading at $t = 10$ s from the distance formula alone?

The difficulty is that during a single instant the train covers no distance at all, so 'distance divided by time' gives $\frac{0}{0}$ . What we can compute is the average speed over a time interval $[t_{1}, t_{2}]$ :

average speed over an interval

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

(12.1)

Now let the intervals shrink towards $t = 10$ s, first from the left (intervals ending at 10 s) and then from the right (intervals starting at 10 s). Since $s (10) = 50$ m, the interval from 9.9 s to 10 s covers $50 - 49.005 = 0.995$ m in 0.1 s, an average of 9.95 m/s. Table 12.1 lists eight such averages; the last column is distance divided by the length of the interval.

Table 12.1Average speeds of the train over intervals that end or start at

t = 10

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

Read the last column downwards from the top: the averages climb 9, 9.5, 9.95, 9.995 as the intervals close in on 10 s from the left. Read it upwards from the bottom: they fall 11, 10.5, 10.05, 10.005 as the intervals close in from the right. Both lists home in on the same number, so the speedometer at $t = 10$ s must read 10 m/s, that is 36 km/h. No single interval produces this value. It is the number the averages approach, the instantaneous speed (तात्क्षणिक चाल) of the train at $t = 10$ s.

The same idea has a picture. On the graph of $s = 0.5 t^{2}$ mark the point $A (10, 50)$ . The average speed over $[10, 10 + h]$ is the rise over the run of the chord from $A$ to the point $Q$ of the graph above $t = 10 + h$ , so it is the slope of the chord $A Q$ . As $h$ shrinks, the chord swings round and settles on the line that just touches the curve at $A$ : the tangent at $A$ (Fig. 12.1).

4/4The tangent at $A$ has slope 10 m/s, the instantaneous speed.

Fig. 12.1Chords from

A

swing towards the tangent as the time interval shrinks.NCERT Fig. 12.1 · p. 219

A distance–time graph: the curve s = 0.5t² rises from the origin and passes through A(10, 50). Step 1 draws the dashed chord from A to Q at t = 14, with slope 12 m/s shown in a panel. Step 2 moves Q to t = 12; the chord is less steep, slope 11 m/s. Step 3 moves Q to t = 10.5; the chord almost hugs the curve, slope 10.25 m/s. Step 4 replaces the chord by the solid tangent at A, slope 10 m/s, which meets the time axis at t = 5.

So the instantaneous speed at $t = 10$ s equals the slope of the tangent to the distance–time graph at that instant. In the language of Section 12.5, 10 m/s is the derivative of $s = 0.5 t^{2}$ at $t = 10$ . To make 'the number the averages approach' precise, we now study limits.

Quick check

1 of 3

For the train,

s = 0.5 t^{2}

. What is its average speed between

t = 4

s and

t = 6

m/s

Show answer

Answer 5 m/s

s (6) = 18

m and

s (4) = 8

m, so the train covers 10 m in 2 s: 5 m/s. Dividing 18 m by 6 s gives 3 m/s, which is the average speed from the start, not over this interval.

12.3 Limits

The speed calculation followed a pattern we will use again and again: we were not allowed to put $t = 10$ into the formula for average speed, so we watched what happened as $t$ crept close to 10. That process is called taking a limit.

Try it on $g (x) = \frac{x^{3} - 8}{x - 2}$ . At $x = 2$ the formula gives $\frac{0}{0}$ , so $g (2)$ is not defined. Nothing stops us from evaluating $g$ near 2, though:

Table 12.2Values of

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

near

x = 2

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

From both sides the values crowd around 12, and they stay as close to 12 as we like once $x$ is close enough to 2. We write $\lim_{x \to 2} g (x) = 12$ and read it as 'the limit of $g (x)$ as $x$ tends to 2 is 12'. The graph explains why. For $x \neq 2$ we may cancel $x - 2$ , since $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ , so $g$ is the parabola $y = x^{2} + 2 x + 4$ with one point punched out (Fig. 12.2).

Definition

Limit

We write $\lim_{x \to a} f (x) = L$ when the values $f (x)$ can be made as close to $L$ as we please by taking $x$ close enough to $a$ , with $x \neq a$ . The number $L$ is called the limit of $f$ at $a$ .

Example $\lim_{x \to 2} \frac{x^{3} - 8}{x - 2} = 12$ , although the function is not defined at 2.

4/4Set $g (2) = 5$ : the limit is still 12.

Fig. 12.2A missing or misplaced point does not change the limit.NCERT Fig. 12.2 · p. 220

The graph of g(x) = (x³ − 8)/(x − 2) is the parabola y = x² + 2x + 4 with an open circle at (2, 12); dashed guides run from the hole to x = 2 and y = 12. Step 1 shows the curve and notes that g(2) is undefined. Step 2 adds three dots on the left branch moving towards the hole, with g(x) → 12 as x → 2 from the left. Step 3 does the same from the right. Step 4 defines g(2) = 5: a filled dot appears at (2, 5), well below the hole, yet the limit is still 12.

Two features of this definition need attention. First, the value $f (a)$ plays no part in it. The function may be undefined at $a$ , or defined there with a different value, as in the last step of Fig. 12.2, and the limit is unchanged. Second, $x$ may approach $a$ from either side, and the definition needs both sides to agree.

Left-hand and right-hand limits

A courier counter in Pune charges by weight: ₹60 for a parcel up to and including 2 kg, and ₹95 for a parcel over 2 kg (up to 5 kg). Let $C (w)$ be the charge, in rupees, for a parcel of $w$ kg.

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

Let the weight approach 2 kg from below: 1.9 kg, 1.99 kg, 1.999 kg. The charge is ₹60 every time, so as $w$ comes towards 2 from the left, $C (w)$ approaches 60. This is the left-hand limit (बाएँ पक्ष की सीमा) of $C$ at 2, written $\lim_{w \to 2^{-}} C (w) = 60$ . From above (2.1 kg, 2.01 kg, 2.001 kg) the charge is always ₹95, so the right-hand limit is $\lim_{w \to 2^{+}} C (w) = 95$ (Fig. 12.3).

3/3The one-sided limits differ, so $\lim_{w \to 2} C (w)$ does not exist.

Fig. 12.3The two one-sided limits at

w = 2

disagree.NCERT Fig. 12.3 · p. 221

The charge C in rupees is plotted against the weight w in kilograms. A solid horizontal segment at height 60 runs from w = 0 to w = 2, ending in a filled dot at (2, 60). A dashed segment at height 95 starts with an open circle at (2, 95) and runs to w = 5. Step 1 adds an arrow along the lower segment towards w = 2: the left-hand limit is 60. Step 2 adds an arrow along the upper segment towards w = 2 from the right: the right-hand limit is 95. Step 3 shows both arrows and a double arrow marking the gap between 60 and 95: there is no limit at 2.

The two one-sided limits disagree, so there is no single number that $C (w)$ approaches as $w \to 2$ . The limit of $C$ at 2 does not exist, even though $C (2) = 60$ is perfectly well defined.

Definition

Left- and right-hand limits

The left-hand limit $\lim_{x \to a^{-}} f (x)$ is the number that $f (x)$ approaches as $x$ comes close to $a$ through values less than $a$ . The right-hand limit $\lim_{x \to a^{+}} f (x)$ uses values greater than $a$ . The limit $\lim_{x \to a} f (x)$ exists exactly when both one-sided limits exist and are equal, and it is then their common value.

ExampleFor the courier charge, the left-hand limit at 2 is 60 and the right-hand limit is 95, so there is no limit at 2.

Limits from tables and graphs

A table of values is a dependable first tool. Table 12.3 evaluates four more functions 0.01 to the left and 0.01 to the right of a chosen point $a$ .

Table 12.3Values of

f

just left of

a

(at

a - 0.01

) and just right of

a

(at

a + 0.01

)

$f (x)$ and $a$	Left	Right	Limit
$5 - 2 x$ , $3$	−0.98	−1.02	−1
$x^{2} - 3 x$ , $- 1$	4.0501	3.9501	4
$\tan x$ , $\frac{π}{4}$	0.9802	1.0202	1
$\frac{1}{(x - 1)^{2}}$ , $1$	$10^{4}$	$10^{4}$	none

The first three rows behave politely. The values close in on one number from both sides, and that number is just $f (a)$ : $5 - 2 (3) = - 1$ , $(- 1)^{2} - 3 (- 1) = 4$ and $\tan \frac{π}{4} = 1$ . For polynomials and the trigonometric functions, at points where the formula makes sense, the limit is the value. Section 12.3.2 proves this for polynomials.

The last row is different. Closer to 1 the values explode: $10^{6}$ at $x = 0.999$ and $10^{8}$ at $x = 0.9999$ , beyond any bound you care to name. No real number is being approached, so the limit does not exist. This behaviour is sometimes written $\lim_{x \to 1} \frac{1}{(x - 1)^{2}} = + \infty$ , but $+ \infty$ only signals values growing without bound; it is not a number. Such limits lie outside this course.

InteractiveLimit and slope lab

Two modes. Approach: choose one of six functions (the parabola with a hole, the courier charge, tan x near π/4, 1/(x − 1)² near 1, sin x / x near 0 and (1 − cos x)/x near 0); a slider sets how far x is from the point, a graph marks the two points just left and right of it, and a table lists f(x) at distances 0.1, 0.01, 0.001 and 0.0001 on each side with a verdict on whether the limit exists. Chord to tangent: choose f(x) = 0.5x², x³/3 − x, 6/x or sin x, set the point a and the step h (positive or negative); the graph shows the chord PQ, the panel shows the difference quotient, and a button reveals the tangent and the value of the derivative.

Fig. 12.4In Approach mode, pick a function and slide

x

towards the point from either side while the table fills in. In Chord to tangent mode, shrink

h

and watch the chord slope settle on the derivative.

Here is what the tables and graphs have taught us:

The limit at $a$ depends only on values near $a$ , never on $f (a)$ itself.
A limit exists only when the left-hand and right-hand limits exist and agree.
$f (a)$ may be undefined while the limit exists (Fig. 12.2), defined while the limit does not exist (Fig. 12.3), or defined and different from the limit.
Values that grow without bound have no limit.

Quick check

1 of 3

Let

f (x) = x + 1

for

x < 2

and

f (x) = 7 - x

for

x \geq 2

. What is

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

354, the average of 3 and 5It does not exist.

Show answer

Answer (D) It does not exist.

From the left

f (x) \to 3

and from the right

f (x) \to 5

. The one-sided limits differ, so there is no limit. The value

f (2) = 5

does not decide it, and averaging the two one-sided limits is not allowed.

12.3.1 Algebra of limits

Tables are slow, and luckily we seldom need them. If two functions have limits at $a$ , so do their sum, difference, product and quotient, and the limits combine in the obvious way. The one catch: we may not divide by a limit of 0. We take these rules on trust here, since proving them needs an exact definition of a limit, which comes in later courses.

Theorem 12.1Algebra of limits

Suppose $\lim_{x \to a} f (x) = l$ and $\lim_{x \to a} g (x) = m$ both exist. Then, as $x \to a$ : (i) $f (x) + g (x) \to l + m$ ; (ii) $f (x) - g (x) \to l - m$ ; (iii) $f (x) g (x) \to l m$ ; (iv) $\frac{f (x)}{g (x)} \to \frac{l}{m}$ , provided $m \neq 0$ .

If $g$ is the constant function $c$ , rule (iii) says $\lim_{x \to a} c f (x) = c l$ : constants come out of limits. For example, (iii) gives $\lim_{x \to - 1} x^{2} = (- 1) (- 1) = 1$ , and then (ii) gives $\lim_{x \to - 1} (x^{2} - 3 x) = 1 - 3 (- 1) = 4$ , as the table predicted. Rule (iv) has a real condition: $\lim_{x \to 0} \frac{\sin x}{x}$ cannot be split into $\frac{\lim \sin x}{\lim x}$ , because the bottom limit is 0. Limits of that kind need Section 12.4.

12.3.2 Limits of polynomials and rational functions

Two limits are obvious: a constant function $c$ stays at $c$ , so its limit is $c$ , and the function $x$ approaches $a$ as $x$ does. Every polynomial is built from these two by multiplying and adding. Since $x^{2} = x \cdot x$ , rule (iii) gives $\lim_{x \to a} x^{2} = a^{2}$ ; repeating the step (induction on $n$ ) gives $\lim_{x \to a} x^{n} = a^{n}$ for every positive integer $n$ .

DerivationLimit of a polynomial

Let $p (x) = c_{0} + c_{1} x + \dots + c_{n} x^{n}$ and treat each term as a function. The sum rule lets us take the limit term by term, the constant rule turns the limit of $c_{k} x^{k}$ into $c_{k} \lim_{x \to a} x^{k}$ , and that is $c_{k} a^{k}$ . Adding up the terms gives $c_{0} + c_{1} a + \dots + c_{n} a^{n}$ , which is $p (a)$ .

limit of a polynomial

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

(12.3)

A rational function is a quotient $f (x) = \frac{g (x)}{h (x)}$ of two polynomials, defined wherever $h (x) \neq 0$ . If $h (a) \neq 0$ , the quotient rule and (12.3) give

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

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

(12.4)

Trouble starts when $h (a) = 0$ . If $g (a) \neq 0$ , the top stays near $g (a)$ while the bottom shrinks to 0, so the quotient grows without bound and there is no limit. If $g (a) = 0$ as well, we have the form $\frac{0}{0}$ , which by itself tells us nothing. Then $x - a$ divides both polynomials (factor theorem), and we can pull out every factor of $x - a$ :

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

Here $g_{1} (a) \neq 0$ and $h_{1} (a) \neq 0$ . Cancelling for $x \neq a$ leaves three outcomes (Fig. 12.5):

$k > l$ : the factor $(x - a)^{k - l}$ stays on top, so $f (x) \to 0$ .
$k = l$ : everything cancels, so $f (x) \to \frac{g_{1} (a)}{h_{1} (a)}$ .
$k < l$ : a factor $(x - a)^{l - k}$ stays below, the values grow without bound, and there is no limit.

2/3The factors cancel: the limit is $\frac{g_{1} (a)}{h_{1} (a)}$ .

Fig. 12.5Counting factors of

x - a

decides a

\frac{0}{0}

limit.

One set of axes with a dashed vertical guide at x = 1. Step 1 (k = 2, l = 1): the graph of (x − 1)²/(x − 1) is the line y = x − 1 with a hole at (1, 0), so the limit is 0. Step 2 (k = l = 1): the graph of (x² − 1)/(x − 1) is the line y = x + 1 with a hole at (1, 2), so the limit is 2. Step 3 (k = 1, l = 2): the graph of (x − 1)/(x − 1)² is the hyperbola y = 1/(x − 1), whose two branches run off to minus and plus infinity beside x = 1, so there is no limit.

Cancelling is legitimate because a limit only uses values with $x \neq a$ , and there $x - a$ is not zero.

Example 12.1NCERT Example 1 · p. 230

Polynomials: substitute

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

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

Example 12.2NCERT Example 2 · p. 230

Rational functions

Evaluate each limit, or show that it does not exist.

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

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

Theorem 12.2A standard limit

For every positive integer $n$ and every real number $a$ , $\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$ .

Long division of $x^{n} - a^{n}$ by $x - a$ leaves no remainder: $x^{n} - a^{n} = (x - a) S (x)$ , where $S (x)$ is the sum of the $n$ terms $x^{n - 1}$ , $x^{n - 2} a$ , …, $x a^{n - 2}$ , $a^{n - 1}$ . (Multiplying out $(x - a) S (x)$ checks this: everything cancels except $x^{n} - a^{n}$ .)

For $x \neq a$ we cancel $x - a$ . What remains is a polynomial with $n$ terms, and at $x = a$ every term equals $a^{n - 1}$ . By (12.3) the limit is $n$ copies of $a^{n - 1}$ , that is $n a^{n - 1}$ .

difference of powers

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

(12.5)

The same formula holds when $n$ is any rational number and $a > 0$ ; we use this without proof. For example, with $n = \frac{1}{2}$ it says that $\frac{\sqrt{x} - \sqrt{a}}{x - a} \to \frac{1}{2 \sqrt{a}}$ as $x \to a$ .

Example 12.3NCERT Example 3 · p. 233

Using the difference-of-powers limit

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

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

12.4 Limits of Trigonometric Functions

The quotient $\frac{\sin x}{x}$ gives $\frac{0}{0}$ at $x = 0$ , but there is no polynomial factor to cancel. We need two facts about inequalities and limits, and then a picture.

Theorem 12.3Limits keep inequalities

Suppose $f (x) \leq g (x)$ for all $x$ near $a$ (except possibly at $a$ ), and that $\lim_{x \to a} f (x)$ and $\lim_{x \to a} g (x)$ both exist. Then $\lim_{x \to a} f (x) \leq \lim_{x \to a} g (x)$ .

The conclusion has $\leq$ even when $f (x) < g (x)$ strictly: $x^{2} < 2 x^{2}$ for every $x \neq 0$ , yet both tend to 0 as $x \to 0$ .

Theorem 12.4Sandwich theorem

Suppose $f (x) \leq g (x) \leq h (x)$ whenever $x$ is near $a$ (except possibly at $a$ itself), and $\lim_{x \to a} f (x) = l = \lim_{x \to a} h (x)$ . Then $\lim_{x \to a} g (x)$ exists and equals $l$ .

The floor $f$ and the ceiling $h$ close in on $l$ , and $g$ is trapped between them, so it has nowhere else to go. This sandwich theorem settles limits that tables cannot. In Fig. 12.6, $g (x) = x^{2} \sin \frac{1}{x}$ wobbles infinitely often near 0. Since $- 1 \leq \sin \frac{1}{x} \leq 1$ , we have $- x^{2} \leq g (x) \leq x^{2}$ , and both bounds tend to 0. Hence $\lim_{x \to 0} x^{2} \sin \frac{1}{x} = 0$ .

2/3 $g (x) = x^{2} \sin \frac{1}{x}$ always lies between them.

Fig. 12.6The two bounds squeeze

g

to 0 at the origin.NCERT Fig. 12.9 · p. 234

Axes from −1 to 1. Step 1 draws two dashed parabolas, y = x² opening upwards and y = −x² opening downwards; they meet only at the origin. Step 2 adds the solid curve g(x) = x² sin(1/x), which oscillates faster and faster near 0 but always stays between the two parabolas. Step 3 circles the origin and adds arrows: both bounds tend to 0, so g(x) is forced to tend to 0 as well.

Now the key inequality. For $0 < ∣ x ∣ < \frac{π}{2}$ , with $x$ measured in radians,

for $0 < |x| < \frac{\pi}{2}$

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

(12.6)

DerivationProof of (12.6) by comparing areas

Take $0 < x < \frac{π}{2}$ first. On the unit circle centred at the origin $O$ , mark $A (1, 0)$ and $P (\cos x, \sin x)$ , so that $∠ A O P = x$ . The tangent to the circle at $A$ is the vertical line through $A$ ; it meets the ray $O P$ at $T (1, \tan x)$ (Fig. 12.7).

Three regions stand on the same base $O A = 1$ , each inside the next: triangle $O A P$ with height $\sin x$ , the sector $O A P$ with angle $x$ (area $\frac{x}{2}$ , since the radius is 1), and triangle $O A T$ with height $\tan x$ . So their areas, $\frac{1}{2} \sin x$ , $\frac{x}{2}$ and $\frac{1}{2} \tan x$ , increase in that order. Doubling them,

\sin x < x < \tan x

Divide through by the positive number $\sin x$ :

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

Now turn each (positive) quantity upside down, which reverses the signs:

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

For $- \frac{π}{2} < x < 0$ nothing new is needed: $\cos x$ and $\frac{\sin x}{x}$ take the same values at $x$ and at $- x$ .

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

Fig. 12.7Three nested regions give

\sin x < x < \tan x

.NCERT Fig. 12.10 · p. 235

A quarter of a unit circle with centre O and A = (1, 0) on the horizontal radius. The radius OP makes an angle x with OA, and the dashed tangent at A meets OP extended at T. Step 1 shows the set-up with P = (cos x, sin x) and AT = tan x. Step 2 shades triangle OAP, of area ½ · 1 · sin x. Step 3 shades the sector OAP, of area ½ x. Step 4 shades the larger triangle OAT, of area ½ tan x, and states sin x < x < tan x.

Theorem 12.5Two standard trigonometric limits

With $x$ in radians, (i) $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and (ii) $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ .

(i) In Theorem 12.4 take the floor $\cos x$ , the ceiling 1 and $g (x) = \frac{\sin x}{x}$ ; (12.6) puts $g$ between them near 0. As $x \to 0$ , $\cos x \to 1$ , and the ceiling is 1 already, so $g (x) \to 1$ .

(ii) Multiply the top and the bottom by $1 + \cos x$ , which is close to 2 near 0, and use $1 - \cos^{2} x = \sin^{2} x$ . The quotient becomes the product of $\frac{\sin x}{x}$ and $\frac{\sin x}{1 + \cos x}$ . As $x \to 0$ the first factor tends to 1 and the second to $\frac{0}{2} = 0$ , so the product tends to 0.

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

(12.7)

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

(12.8)

There is a neat way to see (12.7). On a circle of radius $r$ , an angle $θ$ at the centre cuts off an arc of length $r θ$ and a chord of length $2 r \sin \frac{θ}{2}$ . Their ratio is $\frac{\sin (θ / 2)}{θ / 2}$ , which (12.7) says tends to 1: for small angles the chord and the arc are almost the same length. Try it in Fig. 12.8.

SimulationArc and chord as the angle shrinks

A circle with centre O and two draggable points A and B on it, joined to O by radii; they start 120 degrees apart. The panel shows the angle AOB in degrees, the length of the arc AB and the length of the chord AB in centimetres, and the radius, which can be changed with a handle. As the points are dragged closer, the arc and chord lengths get closer until they agree to one decimal place.

Fig. 12.8Drag

A

and

B

towards each other and compare the arc

A B

with the chord

A B

. Their ratio is

\frac{\sin (θ / 2)}{θ / 2}

, which tends to 1.

Try this

At 120° on a 5 cm circle the arc is about 10.5 cm and the chord about 8.7 cm. Shrink the angle to 30°. How far apart are the two readouts now?
Make the radius as large as you can and shrink the angle again. Does the chord-to-arc ratio depend on the radius?

Example 12.4NCERT Example 4 · p. 236

Building the standard pattern

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

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

Practice · 22 questions

Practice 12.1: questions with answers

1
Short answerEasyNCERT Exercise 12.1, Q1–Q3
Evaluate each limit.
1. $\lim_{x \to 4} (2 x - 5)$
  Show solution
  Answer3
  A polynomial, so substitute: $2 (4) - 5 = 3$ .
2. $\lim_{x \to π / 2} (\frac{2 x}{π} + 1)$
  Show solution
  Answer2
  Substitute: $\frac{2}{π} \cdot \frac{π}{2} + 1 = 2$ .
3. $\lim_{r \to 3} \frac{4}{3} π r^{3}$ (the volume of a ball as its radius tends to 3)
  Show solution
  Answer $36 π$
  Substitute: $\frac{4}{3} π \cdot 27 = 36 π$ .
2
Short answerEasyNCERT Exercise 12.1, Q4–Q5
Evaluate each limit.
1. $\lim_{x \to 3} \frac{5 x - 1}{x + 1}$
  Show solution
  Answer $\frac{7}{2}$
  At $x = 3$ the denominator is $4 \neq 0$ , so by (12.4) the limit is $\frac{14}{4} = \frac{7}{2}$ .
2. $\lim_{x \to - 1} \frac{x^{6} - x^{3} + 1}{x - 2}$
  Show solution
  Answer−1
  The denominator is $- 3 \neq 0$ . The numerator is $1 + 1 + 1 = 3$ , so the limit is $\frac{3}{- 3} = - 1$ .
Short answerMediumNCERT Exercise 12.1, Q6
Evaluate $\lim_{x \to 0} \frac{(x + 2)^{4} - 16}{x}$ .
Show solution
Answer32
Put $y = x + 2$ ; then $y \to 2$ as $x \to 0$ , and $x = y - 2$ :
$\frac{(x + 2)^{4} - 16}{x} = \frac{y^{4} - 2^{4}}{y - 2}$
By (12.5) the limit is $4 \cdot 2^{3} = 32$ . (Expanding $(x + 2)^{4}$ and cancelling $x$ gives the same answer.)
4
Short answerMediumNCERT Exercise 12.1, Q7–Q8
Evaluate each limit.
1. $\lim_{x \to 3} \frac{2 x^{2} - 5 x - 3}{x^{2} - 9}$
  Show solution
  Answer $\frac{7}{6}$
  $2 x^{2} - 5 x - 3 = (2 x + 1) (x - 3)$ and $x^{2} - 9 = (x - 3) (x + 3)$ . Cancelling $x - 3$ leaves $\frac{2 x + 1}{x + 3}$ , which tends to $\frac{7}{6}$ .
2. $\lim_{x \to 2} \frac{x^{4} - 16}{3 x^{2} - 4 x - 4}$
  Show solution
  Answer4
  $x^{4} - 16 = (x - 2) (x + 2) (x^{2} + 4)$ and $3 x^{2} - 4 x - 4 = (3 x + 2) (x - 2)$ . Cancelling $x - 2$ leaves $\frac{(x + 2) (x^{2} + 4)}{3 x + 2}$ , which tends to $\frac{4 \cdot 8}{8} = 4$ .
5
Short answerMediumNCERT Exercise 12.1, Q9, Q11
Evaluate each limit; the letters are constants.
1. $\lim_{x \to 0} \frac{a x^{2} + b x + c}{p x + q}$ , where $q \neq 0$
  Show solution
  Answer $\frac{c}{q}$
  The denominator tends to $q \neq 0$ , so substitute: the limit is $\frac{c}{q}$ .
2. $\lim_{x \to 2} \frac{x^{2} + a x}{x^{2} + b x}$ , where $b \neq - 2$
  Show solution
  Answer $\frac{2 + a}{2 + b}$
  At $x = 2$ the denominator is $4 + 2 b = 2 (2 + b) \neq 0$ . Substituting gives $\frac{4 + 2 a}{4 + 2 b} = \frac{2 + a}{2 + b}$ .
Short answerMediumNCERT Exercise 12.1, Q10
Evaluate $\lim_{x \to 16} \frac{x^{1 / 4} - 2}{x^{1 / 2} - 4}$ .
Show solution
Answer $\frac{1}{4}$
Divide the top and the bottom by $x - 16$ and use (12.5) with $a = 16$ :
$\begin{aligned} \frac{x^{1 / 4} - 16^{1 / 4}}{x^{1 / 2} - 16^{1 / 2}} & \to \frac{\frac{1}{4} \cdot 16^{- 3 / 4}}{\frac{1}{2} \cdot 16^{- 1 / 2}} \\ = \frac{1 / 32}{1 / 8} = \frac{1}{4} \end{aligned}$
Check: with $t = x^{1 / 4}$ the quotient is $\frac{t - 2}{t^{2} - 4} = \frac{1}{t + 2}$ , which tends to $\frac{1}{4}$ as $t \to 2$ .
Short answerMediumNCERT Exercise 12.1, Q12
Evaluate $\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ .
Show solution
Answer $- \frac{1}{9}$
The numerator is $\frac{3 - x}{3 x} = - \frac{x - 3}{3 x}$ . Dividing by $x - 3$ (allowed for $x \neq 3$ ) leaves $- \frac{1}{3 x}$ , which tends to $- \frac{1}{9}$ .
8
Short answerMediumNCERT Exercise 12.1, Q13–Q14
Evaluate each limit.
1. $\lim_{x \to 0} \frac{\sin π x}{2 x}$
  Show solution
  Answer $\frac{π}{2}$
  $\frac{\sin π x}{2 x} = \frac{π}{2} \cdot \frac{\sin π x}{π x} \to \frac{π}{2}$ .
2. $\lim_{x \to 0} \frac{\sin a x}{\tan b x}$ , where $a, b \neq 0$
  Show solution
  Answer $\frac{a}{b}$
  Write it as $\frac{a}{b} \cdot \frac{\sin a x}{a x} \cdot \frac{b x}{\tan b x}$ . Both fractions tend to 1 (Example 12.4), so the limit is $\frac{a}{b}$ .
9
Short answerMediumNCERT Exercise 12.1, Q15–Q16
Evaluate each limit.
1. $\lim_{x \to π / 2} \frac{\cos x}{\frac{π}{2} - x}$
  Show solution
  Answer1
  Put $y = \frac{π}{2} - x$ , so $y \to 0$ and $\cos x = \sin y$ . The quotient is $\frac{\sin y}{y} \to 1$ .
2. $\lim_{x \to 0} \frac{1 + \cos x}{π + x}$
  Show solution
  Answer $\frac{2}{π}$
  Nothing is 0 here: the denominator tends to $π$ . Substitute: $\frac{1 + 1}{π} = \frac{2}{π}$ . Not every trigonometric limit needs (12.7).
10
Short answerHardNCERT Exercise 12.1, Q17–Q18
Evaluate each limit.
1. $\lim_{x \to 0} \frac{1 - \cos 6 x}{1 - \cos 2 x}$
  Show solution
  Answer9
  $1 - \cos 6 x = 2 \sin^{2} 3 x$ and $1 - \cos 2 x = 2 \sin^{2} x$ , so the quotient is ${(\frac{\sin 3 x}{\sin x})}^{2}$ . As in Example 12.4, $\frac{\sin 3 x}{\sin x} \to 3$ , so the limit is 9.
2. $\lim_{x \to 0} \frac{2 x + x \cos 2 x}{3 \sin x}$
  Show solution
  Answer1
  Take out $x$ : the quotient is $\frac{2 + \cos 2 x}{3} \cdot \frac{x}{\sin x}$ . This tends to $\frac{3}{3} \cdot 1 = 1$ .
11
Short answerMediumNCERT Exercise 12.1, Q19–Q21
Evaluate each limit.
1. $\lim_{x \to π} x \cos x$
  Show solution
  Answer $- π$
  No $\frac{0}{0}$ here: substitute to get $π \cos π = - π$ .
2. $\lim_{x \to 0} \frac{\sin 2 x + 4 x}{x + \sin 3 x}$
  Show solution
  Answer $\frac{3}{2}$
  Divide the top and the bottom by $x$ : $\frac{\frac{\sin 2 x}{x} + 4}{1 + \frac{\sin 3 x}{x}}$ . Since $\frac{\sin 2 x}{x} \to 2$ and $\frac{\sin 3 x}{x} \to 3$ , the limit is $\frac{2 + 4}{1 + 3} = \frac{3}{2}$ .
3. $\lim_{x \to 0} \frac{\sec x - 1}{x}$
  Show solution
  Answer0
  $\sec x - 1 = \frac{1 - \cos x}{\cos x}$ , so the quotient is $\frac{1 - \cos x}{x} \cdot \frac{1}{\cos x} \to 0 \cdot 1 = 0$ by (12.8).
Short answerHardNCERT Exercise 12.1, Q22
Evaluate $\lim_{x \to π / 4} \frac{1 - \tan x}{x - \frac{π}{4}}$ .
Show solution
Answer−2
Put $y = x - \frac{π}{4}$ , so $y \to 0$ . Then $\tan x = \frac{1 + \tan y}{1 - \tan y}$ and
$1 - \tan x = \frac{- 2 \tan y}{1 - \tan y}$
The quotient becomes $\frac{- 2}{1 - \tan y} \cdot \frac{\tan y}{y} \to \frac{- 2}{1} \cdot 1 = - 2$ .
13
Short answerMediumNCERT Exercise 12.1, Q23–Q24
Find the limits of these functions defined in pieces.
1. $f (x) = 4 - x^{2}$ for $x < 1$ and $f (x) = 2 x + 1$ for $x \geq 1$ . Find $\lim_{x \to 0} f (x)$ and $\lim_{x \to 1} f (x)$ .
  Show solution
  Answer4 and 3
  Near 0 only the first piece is used, so the limit is $4 - 0 = 4$ .
  At 1: from the left $4 - 1 = 3$ , from the right $2 + 1 = 3$ . They agree, so $\lim_{x \to 1} f (x) = 3$ .
2. $g (x) = x^{2} + 1$ for $x \leq 2$ and $g (x) = 3 x - 2$ for $x > 2$ . Does $\lim_{x \to 2} g (x)$ exist?
  Show solution
  AnswerNo
  From the left $g (x) \to 5$ ; from the right $g (x) \to 4$ . They differ, so the limit does not exist.
14
Short answerMediumNCERT Exercise 12.1, Q25–Q27
Find each limit, or show that it does not exist.
1. $\lim_{x \to 3} \frac{∣ x - 3 ∣}{x - 3}$
  Show solution
  Answerdoes not exist
  For $x > 3$ the quotient is $1$ ; for $x < 3$ it is $- 1$ . The one-sided limits are $- 1$ and $1$ , so there is no limit.
2. $\lim_{x \to - 4} (∣ x ∣ + 2 x)$
  Show solution
  Answer−4
  Near $- 4$ , $∣ x ∣ = - x$ , so the function is $x$ , which tends to $- 4$ . (Or substitute: $4 - 8 = - 4$ .)
3. $\lim_{x \to 0} \frac{∣ x ∣ + x^{2}}{x}$
  Show solution
  Answerdoes not exist
  For $x > 0$ it is $1 + x \to 1$ ; for $x < 0$ it is $- 1 + x \to - 1$ . The one-sided limits differ, so there is no limit.
Short answerMediumNCERT Exercise 12.1, Q28
Let $f (x) = a x + b$ for $x < 2$ , $f (2) = 6$ , and $f (x) = a x^{2} - b$ for $x > 2$ . Choose $a$ and $b$ so that the limit of $f$ at 2 equals $f (2)$ .
Show solution
Answer $a = 2$ , $b = 2$
Left-hand limit $2 a + b$ ; right-hand limit $4 a - b$ . Both must equal 6.
Adding $2 a + b = 6$ and $4 a - b = 6$ gives $6 a = 12$ , so $a = 2$ and $b = 2$ .
16
Short answerMediumNCERT Exercise 12.1, Q29
Let $f (x) = (x - 1) (x - 2) (x - 3) (x - 4)$ .
1. Find $\lim_{x \to 3} f (x)$ .
  Show solution
  Answer0
  Substitute: one factor is $3 - 3 = 0$ , so the limit is 0.
2. Find $\lim_{x \to 5} f (x)$ .
  Show solution
  Answer24
  $4 \cdot 3 \cdot 2 \cdot 1 = 24$ .
3. Find $\lim_{x \to 3} \frac{f (x)}{x - 3}$ .
  Show solution
  Answer−2
  Cancel $x - 3$ : the quotient is $(x - 1) (x - 2) (x - 4) \to 2 \cdot 1 \cdot (- 1) = - 2$ .
Short answerMediumNCERT Exercise 12.1, Q30
Let $f (x) = x + 2$ for $x < 1$ , $f (1) = 0$ , and $f (x) = 4 - x$ for $x > 1$ . At which points $a$ does $f$ have a limit?
Show solution
Answerevery real a
For $a \neq 1$ , $f$ is given by a single polynomial near $a$ , so the limit exists.
At 1 the left-hand limit is $1 + 2 = 3$ and the right-hand limit is $4 - 1 = 3$ . They agree, so the limit exists there too and equals 3, even though $f (1) = 0$ .
The limit exists for every real $a$ .
18
Short answerMediumNCERT Exercise 12.1, Q31
A function satisfies $\lim_{x \to 2} \frac{f (x) - 5}{x - 2} = 3$ .
1. Find $\lim_{x \to 2} f (x)$ .
  Show solution
  Answer5
  $f (x) - 5 = \frac{f (x) - 5}{x - 2} \cdot (x - 2) \to 3 \cdot 0 = 0$ , so $f (x) \to 5$ .
2. Find $\lim_{x \to 2} \frac{f (x) - 5}{x^{2} - 4}$ .
  Show solution
  Answer $\frac{3}{4}$
  $\frac{f (x) - 5}{x^{2} - 4} = \frac{f (x) - 5}{x - 2} \cdot \frac{1}{x + 2} \to 3 \cdot \frac{1}{4} = \frac{3}{4}$ .
Short answerHardNCERT Exercise 12.1, Q32
Let $f (x) = m x + n$ for $x < 0$ , $f (x) = n x^{2} + 3$ for $0 \leq x \leq 1$ , and $f (x) = m x + 1$ for $x > 1$ . Find $m$ and $n$ if both $\lim_{x \to 0} f (x)$ and $\lim_{x \to 1} f (x)$ exist.
Show solution
Answer $m = 5$ , $n = 3$
At 0: the left-hand limit is $n$ and the right-hand limit is $3$ , so $n = 3$ .
At 1: the left-hand limit is $n + 3 = 6$ and the right-hand limit is $m + 1$ , so $m = 5$ .
Quick answerMedium
(Entrance-style) $\lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}}$ equals
0 $\frac{1}{2}$ 1 $\frac{3}{2}$
Show answer
Answer (B) $\frac{1}{2}$
$\tan x - \sin x = \tan x (1 - \cos x)$ , so the quotient is $\frac{\tan x}{x} \cdot \frac{1 - \cos x}{x^{2}} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ . Splitting it as $\frac{\tan x}{x^{3}} - \frac{\sin x}{x^{3}}$ fails because each piece grows without bound.
Quick answerMedium
(Assertion–reason) A: $\lim_{x \to 0} \frac{∣ x ∣}{x}$ does not exist. R: The left-hand and right-hand limits of $\frac{∣ x ∣}{x}$ at 0 are $- 1$ and $1$ .
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
For $x < 0$ the quotient is $- 1$ and for $x > 0$ it is $1$ , so R is true. Unequal one-sided limits are exactly why the limit fails to exist, so R explains A.
Quick answerHard
(Entrance-style) Evaluate $\lim_{x \to 1} \frac{x + x^{2} + x^{3} + \dots + x^{10} - 10}{x - 1}$ .
Show answer
Answer 55
Split the quotient into $\sum_{k = 1}^{10} \frac{x^{k} - 1}{x - 1}$ . By (12.5) each term tends to $k$ , so the limit is $1 + 2 + \dots + 10 = 55$ .

12.5 Derivatives

In Section 12.2 the speed of a train at one instant came out as a limit of average speeds. The question how fast is this quantity changing right now? turns up everywhere: how quickly a patient's fever falls an hour after a tablet, how fast a cricket ball is moving as it leaves the bowler's hand, how much a factory's cost grows when it makes one more unit. Each is answered by the same kind of limit.

Definition

Derivative at a point

Let $f$ be a real function and $a$ a point of its domain. When the limit $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ exists, we call it the derivative of $f$ at $a$ and write it $f^{'} (a)$ .

ExampleFor $s = 0.5 t^{2}$ the limit at $t = 10$ is 10, the speed of the train.

derivative at $a$

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

(12.9)

The fraction inside the limit is the average rate of change of $f$ between $a$ and $a + h$ , called the difference quotient. Positive $h$ looks to the right of $a$ and negative $h$ to the left, so the definition needs both one-sided limits to agree. We read $f^{'} (a)$ as ' $f$ dash $a$ ' or ' $f$ prime of $a$ '.

Example 12.5NCERT Example 5 · p. 240

A straight line

Find the derivative of $f (x) = 5 - 2 x$ at $x = 4$ .

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

Example 12.6NCERT Example 6 · p. 240

A quadratic at two points

For $f (x) = 3 x^{2} - 4 x + 1$ , find $f^{'} (2)$ and $f^{'} (- 1)$ , and show that $5 f^{'} (2) + 4 f^{'} (- 1) = 0$ .

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

Example 12.7NCERT Example 7 · p. 241

Cosine at zero

Find the derivative of $\cos x$ at $x = 0$ .

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

Example 12.8NCERT Example 8 · p. 241

A constant function

Find the derivative of $f (x) = - 7$ at $x = 5$ and at $x = - 2$ .

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

Derivative as the slope of a tangent

In Chapter 9 the slope of a line through two points was rise over run. For the chord joining $P (a, f (a))$ to $Q (a + h, f (a + h))$ on the graph of $y = f (x)$ , the run is $h$ and the rise is $f (a + h) - f (a)$ : they are the legs $P M$ and $M Q$ of the right triangle in Fig. 12.9. So the difference quotient is exactly the slope of the chord $P Q$ .

Now let $h$ shrink. $Q$ slides along the curve to $P$ , and the chord turns about $P$ until it lies along the tangent at $P$ . The slopes follow, so the derivative is the slope of the tangent:

f^{'} (a) = \tan ψ

where

$ψ$: the angle the tangent at $P$ makes with the positive $x$ -axis

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

Fig. 12.9As

Q

slides to

P

, the chord

P Q

becomes the tangent at

P

.NCERT Fig. 12.11 · p. 241

The decreasing curve y = 6/x for x from about 1 to 6.5, with P(2, 3) marked. In steps 1 to 3 a dashed chord joins P to Q at x = 5, 3.5 and 2.5, with the right triangle PMQ drawn under it: the horizontal side PM has length h and the vertical side MQ goes down. The panel gives the chord slopes −0.6, −0.86 and −1.2. In step 4 the chord is replaced by the solid tangent at P, which crosses the x-axis at x = 4; an arc marks its angle ψ with the positive x-axis, and the panel gives the slope −1.5 = tan ψ.

A falling curve has a negative derivative, and $ψ$ is then obtuse. This links the two pictures of Section 12.2: a rate of change and the slope of a tangent are the same number.

The derivative function

Nothing forces us to keep $a$ fixed. Finding $f^{'}$ at every point where the limit exists produces a new function.

Definition

Derivative (first principle)

For a real function $f$ , the derivative is the function $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ , defined at every $x$ where this limit exists. Working straight from this limit is called finding the derivative from the first principle.

first principle

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

(12.10)

Several notations are in use. If $y = f (x)$ , its derivative may be written $f^{'} (x)$ , $\frac{d}{d x} f (x)$ , $\frac{d y}{d x}$ or $D (f (x))$ , read 'the derivative of $f (x)$ , or of $y$ , with respect to $x$ '. Its value at $x = a$ is written $f^{'} (a)$ , ${\frac{d f}{d x} ∣}_{x = a}$ or ${(\frac{d y}{d x})}_{x = a}$ . Finding a derivative is called differentiation (अवकलन).

Fig. 12.10 builds $f^{'}$ point by point for $f (x) = \frac{x^{3}}{3} - x$ . Where $f$ climbs, $f^{'}$ is positive. At the two level spots, $x = \pm 1$ , $f^{'}$ is 0, and where $f$ falls, $f^{'}$ is negative. After Section 12.5.2 you can check that $f^{'} (x) = x^{2} - 1$ .

4/4The slopes trace the parabola $y = x^{2} - 1$ .

Fig. 12.10Each tangent slope of

f

becomes one point on the graph of

f^{'}

Two panels share the x-axis. The top panel shows y = f(x) = x³/3 − x, rising to a peak at x = −1, falling through the origin to a trough at x = 1, then rising. The bottom panel will hold y = f′(x). Step 1 draws the level tangent at x = −1 and plots a diamond at height 0 below it. Step 2 draws the tangent at x = 0 with slope −1 and plots a diamond at −1. Step 3 draws the steep tangent at x = 2 with slope 3 and plots a diamond at 3. Step 4 adds diamonds at x = −2 and x = 1 and joins all five with the dashed parabola y = x² − 1.

Example 12.9NCERT Example 9 · p. 242

A linear function

Find the derivative of $f (x) = 4 x - 9$ .

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

Example 12.10NCERT Example 10 · p. 243

A quadratic from the first principle

Find the derivative of $f (x) = 2 x^{2} - x$ from the first principle.

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

Example 12.11NCERT Example 11 · p. 243

Constants have derivative 0

Show that the constant function $f (x) = c$ has derivative 0, and use this to differentiate $g (x) = \sin^{2} x + \cos^{2} x$ .

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

Example 12.12NCERT Example 12 · p. 243

A reciprocal

Find the derivative of $f (x) = \frac{6}{x}$ , $x \neq 0$ , and check the tangent slope in Fig. 12.9.

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

Quick check

1 of 3

Use the definition to find

f^{'} (4)

for

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

Show answer

Answer 5

f (4) = 4

and

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

, so the difference quotient is

5 + h \to 5

. The value

f (4) = 4

is the height of the graph, not its slope.

12.5.1 Algebra of derivative of functions

Every derivative is a limit of difference quotients, so Theorem 12.1 can be put to work on derivatives too. Below, $u$ and $v$ are functions of $x$ with derivatives $u^{'}$ and $v^{'}$ on a common domain.

Theorem 12.6Rules of differentiation

(i) $(u + v)^{'} = u^{'} + v^{'}$ ; (ii) $(u - v)^{'} = u^{'} - v^{'}$ ; (iii) $(u v)^{'} = u^{'} v + u v^{'}$ , the product rule (also called the Leibniz rule); (iv) ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ wherever $v \neq 0$ , the quotient rule.

(i) and (ii) follow at once from the sum and difference rules for limits. For (iii), write $u$ , $v$ for $u (x)$ , $v (x)$ and $u_{h}$ , $v_{h}$ for $u (x + h)$ , $v (x + h)$ . Adding and subtracting $u_{h} v$ in the numerator of the difference quotient of $u v$ gives

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

As $h \to 0$ the two fractions tend to $v^{'}$ and $u^{'}$ . Also $u_{h} \to u$ , because $u_{h} - u = h \cdot \frac{u_{h} - u}{h} \to 0 \cdot u^{'} = 0$ . So the limit is $u v^{'} + u^{'} v$ .

Rule (iv) can be proved in the same way, by adding and subtracting a suitable term.

product rule

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

(12.11)

quotient rule

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

(12.12)

2/3The new area is two strips plus a corner.

Fig. 12.11When

u

and

v

grow a little, the area

u v

gains two strips and a tiny corner.

A rectangle of width u and height v has area uv. Step 1 shows just this rectangle. Step 2 lets u grow by Δu and v by Δv: a strip of area v Δu appears on the right, a dashed strip of area u Δv on top, and a small corner of area Δu Δv. Step 3 shows thinner strips for a smaller change; after dividing by h and letting h → 0 the strips give v u′ + u v′, while the corner term Δu · v′ tends to 0.

In words: the derivative of the first times the second, plus the first times the derivative of the second. In the quotient rule keep the order, $u^{'} v$ first, and square the denominator. A tempting slip is $(u v)^{'} = u^{'} v^{'}$ ; Fig. 12.11 shows why it cannot be right, since the area gains two strips, not one corner.

Some first consequences. The function $x$ has derivative 1, because $\frac{(x + h) - x}{h} = 1$ . For a constant $c$ the product rule gives $(c u)^{'} = 0 \cdot u + c u^{'} = c u^{'}$ : constants come out of derivatives. Applied to $x^{2} = x \cdot x$ the product rule gives $1 \cdot x + x \cdot 1 = 2 x$ , and applied to $x^{3} = x^{2} \cdot x$ it gives $2 x \cdot x + x^{2} \cdot 1 = 3 x^{2}$ . The pattern continues.

Theorem 12.7Power rule

For every positive integer $n$ , $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ .

power rule

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

(12.13)

The power rule holds for every real power $n$ ; we use this without proof. With $n = - 1$ it gives $\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x^{2}}$ , which agrees with Example 12.12, and with $n = \frac{1}{2}$ it gives $\frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}}$ .

12.5.2 Derivative of polynomials and trigonometric functions

Theorem 12.8Derivative of a polynomial

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

Example 12.13NCERT Example 13 · p. 246

Term by term

Differentiate $4 x^{25} - 3 x^{10} + 7 x - 2$ .

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

Example 12.14NCERT Example 14 · p. 247

A long polynomial at two points

For $f (x) = x + x^{2} + x^{3} + \dots + x^{20}$ , find $f^{'} (1)$ and $f^{'} (- 1)$ .

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

Example 12.15NCERT Example 15 · p. 247

The quotient rule

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

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

Example 12.16NCERT Example 16 · p. 247

The derivative of sin x

Show from the first principle that $\frac{d}{d x} (\sin x) = \cos x$ .

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

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

(12.14)

The same method, starting from $\cos (x + h) = \cos x \cos h - \sin x \sin h$ , proves the companion result (Practice 12.2, Q9):

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

(12.15)

4/4Plotted slopes form $y = \cos x$ .

Fig. 12.12The slopes of the sine curve trace out the cosine curve.NCERT Example 16 · p. 247

The curve y = sin x from 0 to 2π. Step 1 draws the tangent at the origin, with slope 1, and plots a diamond at height 1 above x = 0. Step 2 draws the level tangent at the crest x = π/2 and plots a diamond at height 0. Step 3 draws the tangent at x = π, with slope −1, and plots a diamond at −1. Step 4 adds diamonds at 3π/2 (height 0) and 2π (height 1) and joins all the diamonds with a dashed curve, y = cos x.

Example 12.17NCERT Example 17 · p. 247

The derivative of tan x

Show from the first principle that $\frac{d}{d x} (\tan x) = \sec^{2} x$ wherever $\cos x \neq 0$ .

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

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

(12.16)

Example 12.18NCERT Example 18 · p. 248

Product rule with a sine

Differentiate $f (x) = x^{2} \sin x$ , and find $f^{'} (π)$ .

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

Table 12.4 collects the standard derivatives of this chapter. The last three come from the quotient rule, in Example 12.21 and Practice 12.2.

Table 12.4Standard derivatives

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

Quick check

1 of 4

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

- 3 x^{- 2}

- 3 x^{- 4}

3 x^{- 4}

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

Show answer

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

Power rule with

n = - 3

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

. Lowering the power by 1 takes

- 3

- 4

, not to

- 2

Practice · 14 questions

Practice 12.2: questions with answers

1
Short answerEasyNCERT Exercise 12.2, Q1–Q3
Use the definition (12.9) to find each derivative.
1. $f (x) = x^{2} + 5 x$ at $x = - 3$
  Show solution
  Answer−1
  $f (- 3 + h) - f (- 3) = (9 - 6 h + h^{2} - 15 + 5 h) - (- 6) = h^{2} - h$ . Dividing by $h$ gives $h - 1 \to - 1$ .
2. $f (x) = x$ at $x = - 4$
  Show solution
  Answer1
  $\frac{(- 4 + h) - (- 4)}{h} = 1$ for every $h \neq 0$ , so $f^{'} (- 4) = 1$ .
3. $f (x) = 25 x$ at $x = 40$
  Show solution
  Answer25
  $\frac{25 (40 + h) - 25 \cdot 40}{h} = 25$ , so $f^{'} (40) = 25$ . Note that the point 40 plays no part.
2
Short answerMediumNCERT Exercise 12.2, Q4
Use the first principle (12.10) to differentiate each function.
1. $4 - x^{3}$
  Show solution
  Answer $- 3 x^{2}$
  $(4 - (x + h)^{3}) - (4 - x^{3}) = - (3 x^{2} h + 3 x h^{2} + h^{3})$ . Divide by $h$ and let $h \to 0$ : $- 3 x^{2}$ .
2. $(x + 3) (x - 1)$
  Show solution
  Answer $2 x + 2$
  The function is $x^{2} + 2 x - 3$ . The difference is $2 x h + h^{2} + 2 h$ ; divide by $h$ and let $h \to 0$ : $2 x + 2$ .
3. $\frac{1}{x^{3}}$
  Show solution
  Answer $- \frac{3}{x^{4}}$
  $\frac{1}{(x + h)^{3}} - \frac{1}{x^{3}} = \frac{x^{3} - (x + h)^{3}}{x^{3} (x + h)^{3}} = \frac{- h (3 x^{2} + 3 x h + h^{2})}{x^{3} (x + h)^{3}}$ . Divide by $h$ and let $h \to 0$ : $\frac{- 3 x^{2}}{x^{6}} = - \frac{3}{x^{4}}$ .
4. $\frac{2 x + 1}{x + 1}$
  Show solution
  Answer $\frac{1}{(x + 1)^{2}}$
  Over the common denominator $(x + h + 1) (x + 1)$ the numerator is $(2 x + 2 h + 1) (x + 1) - (2 x + 1) (x + h + 1) = h$ . So the quotient is $\frac{1}{(x + h + 1) (x + 1)} \to \frac{1}{(x + 1)^{2}}$ .
Short answerMediumNCERT Exercise 12.2, Q5
Let $f (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{60}}{60}$ . Show that $f^{'} (1) = 60 f^{'} (0)$ , and find $f^{'} (- 1)$ .
Show solution
Answer $f^{'} (- 1) = 0$
$f^{'} (x) = 1 + x + x^{2} + \dots + x^{59}$ , a sum of 60 terms.
$f^{'} (1) = 60$ and $f^{'} (0) = 1$ , so $f^{'} (1) = 60 f^{'} (0)$ .
$f^{'} (- 1) = 1 - 1 + 1 - \dots - 1 = 0$ , since the 60 terms cancel in pairs.
Short answerMediumNCERT Exercise 12.2, Q6
For a fixed real number $a$ , differentiate $g (x) = x^{5} + a x^{4} + a^{2} x^{3} + a^{3} x^{2} + a^{4} x + a^{5}$ , and find $g^{'} (a)$ .
Show solution
Answer $g^{'} (a) = 15 a^{4}$
$\begin{aligned} g^{'} (x) = 5 x^{4} & + 4 a x^{3} + 3 a^{2} x^{2} \\ + 2 a^{3} x + a^{4} \end{aligned}$
At $x = a$ every term is a multiple of $a^{4}$ : $g^{'} (a) = (5 + 4 + 3 + 2 + 1) a^{4} = 15 a^{4}$ .
5
Short answerMediumNCERT Exercise 12.2, Q7
For constants $p$ and $q$ , differentiate each function.
1. $(x + p) (2 x - q)$
  Show solution
  Answer $4 x + 2 p - q$
  Product rule: $1 \cdot (2 x - q) + (x + p) \cdot 2 = 4 x + 2 p - q$ .
2. $(p x^{3} - q)^{2}$
  Show solution
  Answer $6 p^{2} x^{5} - 6 p q x^{2}$
  Expand: $p^{2} x^{6} - 2 p q x^{3} + q^{2}$ . The derivative is $6 p^{2} x^{5} - 6 p q x^{2}$ .
3. $\frac{x + p}{x + q}$
  Show solution
  Answer $\frac{q - p}{(x + q)^{2}}$
  Quotient rule: $\frac{1 \cdot (x + q) - (x + p) \cdot 1}{(x + q)^{2}} = \frac{q - p}{(x + q)^{2}}$ .
Short answerMediumNCERT Exercise 12.2, Q8
For a constant $a$ , differentiate $\frac{x^{4} - a^{4}}{x - a}$ in two ways: by first simplifying, and by the quotient rule.
Show solution
Answer $3 x^{2} + 2 a x + a^{2}$
Simplified, the function is $x^{3} + a x^{2} + a^{2} x + a^{3}$ (for $x \neq a$ ), with derivative $3 x^{2} + 2 a x + a^{2}$ .
By the quotient rule the derivative is $\frac{4 x^{3} (x - a) - (x^{4} - a^{4})}{(x - a)^{2}} = \frac{3 x^{4} - 4 a x^{3} + a^{4}}{(x - a)^{2}}$ .
The two agree: multiplying out $(3 x^{2} + 2 a x + a^{2}) (x - a)^{2}$ gives $3 x^{4} - 4 a x^{3} + a^{4}$ .
7
Short answerEasyNCERT Exercise 12.2, Q9 (i)–(iii)
Differentiate each function.
1. $3 x + \frac{2}{5}$
  Show solution
  Answer3
  The constant $\frac{2}{5}$ has derivative 0, so the answer is 3.
2. $(2 x^{2} - x + 4) (x + 3)$
  Show solution
  Answer $6 x^{2} + 10 x + 1$
  Product rule: $(4 x - 1) (x + 3) + (2 x^{2} - x + 4) = 6 x^{2} + 10 x + 1$ .
3. $x^{- 2} (4 + 5 x)$
  Show solution
  Answer $- 8 x^{- 3} - 5 x^{- 2}$
  The function is $4 x^{- 2} + 5 x^{- 1}$ , so the derivative is $- 8 x^{- 3} - 5 x^{- 2}$ .
8
Short answerMediumNCERT Exercise 12.2, Q9 (iv)–(vi)
Differentiate each function.
1. $x^{4} (2 - 3 x^{- 6})$
  Show solution
  Answer $8 x^{3} + 6 x^{- 3}$
  The function is $2 x^{4} - 3 x^{- 2}$ , so the derivative is $8 x^{3} + 6 x^{- 3}$ .
2. $x^{- 3} (5 - 2 x^{- 4})$
  Show solution
  Answer $- 15 x^{- 4} + 14 x^{- 8}$
  The function is $5 x^{- 3} - 2 x^{- 7}$ , so the derivative is $- 15 x^{- 4} + 14 x^{- 8}$ .
3. $\frac{3}{x - 2} - \frac{x^{2}}{2 x + 1}$
  Show solution
  Answersee solution
  First term: $- \frac{3}{(x - 2)^{2}}$ . Second term, by the quotient rule: $\frac{2 x (2 x + 1) - 2 x^{2}}{(2 x + 1)^{2}} = \frac{2 x^{2} + 2 x}{(2 x + 1)^{2}}$ .
  The derivative is $- \frac{3}{(x - 2)^{2}} - \frac{2 x^{2} + 2 x}{(2 x + 1)^{2}}$ .
ProofMediumNCERT Exercise 12.2, Q10
Prove (12.15) from the first principle, then find the slope of $y = \cos x$ at $x = \frac{π}{6}$ .
Show solution
Answer $- \frac{1}{2}$
$\cos (x + h) - \cos x = \cos x (\cos h - 1) - \sin x \sin h$ . Dividing by $h$ :
$\begin{aligned} - \cos x \cdot \frac{1 - \cos h}{h} \\ - \sin x \cdot \frac{\sin h}{h} \end{aligned}$
As $h \to 0$ this tends to $- \cos x \cdot 0 - \sin x \cdot 1 = - \sin x$ .
At $x = \frac{π}{6}$ the slope is $- \sin \frac{π}{6} = - \frac{1}{2}$ .
10
Short answerMediumNCERT Exercise 12.2, Q11 (i)–(iv)
Use the quotient or product rule to differentiate each function.
1. $\sec x$
  Show solution
  Answer $\sec x \tan x$
  $\frac{0 \cdot \cos x - 1 \cdot (- \sin x)}{\cos^{2} x} = \frac{\sin x}{\cos^{2} x} = \sec x \tan x$ .
2. $cosec x$
  Show solution
  Answer $- cosec x \cot x$
  $\frac{0 - \cos x}{\sin^{2} x} = - cosec x \cot x$ .
3. $\sin x \tan x$
  Show solution
  Answer $\sin x (1 + \sec^{2} x)$
  Product rule: $\cos x \tan x + \sin x \sec^{2} x = \sin x + \sin x \sec^{2} x = \sin x (1 + \sec^{2} x)$ .
11
Short answerEasyNCERT Exercise 12.2, Q11 (v)–(vii)
Differentiate each function.
1. $4 \sec x - 3 \cos x$
  Show solution
  Answer $4 \sec x \tan x + 3 \sin x$
  $4 \sec x \tan x + 3 \sin x$ .
2. $2 \cot x - 3 cosec x$
  Show solution
  Answersee solution
  $- 2 {cosec}^{2} x + 3 cosec x \cot x$ .
3. $3 \cos x - 4 \sin x + 9$
  Show solution
  Answer $- 3 \sin x - 4 \cos x$
  $- 3 \sin x - 4 \cos x$ ; the constant 9 contributes 0.
4. $5 \tan x + 2 \sec x$
  Show solution
  Answer $5 \sec^{2} x + 2 \sec x \tan x$
  $5 \sec^{2} x + 2 \sec x \tan x$ .
Quick answerMedium
(Entrance-style) If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$ , then $\frac{d y}{d x}$ at $x = 0$ is
01415
Show answer
Answer (B) 1
Multiplying out, $y = 1 + x + x^{2} + \dots + x^{15}$ , so $y^{'} = 1 + 2 x + \dots + 15 x^{14}$ and $y^{'} (0) = 1$ . The answer 15 is $y^{'}$ 's top coefficient, not its value at 0.
Quick answerEasy
(Assertion–reason) A: The derivative of $\sin^{2} x + \cos^{2} x$ is 0 for every $x$ . R: The derivative of a constant function is 0.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (A) Both A and R are true, and R explains A
$\sin^{2} x + \cos^{2} x = 1$ , a constant, so A follows directly from R. (Differentiating term by term gives $2 \sin x \cos x - 2 \cos x \sin x = 0$ as well.)
Quick answerMedium
(Entrance-style) For $f (x) = x^{3} - 6 x^{2} + 9 x$ , the equation $f^{'} (x) = 0$ has two roots. What is their sum?
Show answer
Answer 4
$f^{'} (x) = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ , so the roots are 1 and 3, with sum 4. These are where the graph of $f$ has level tangents.

Miscellaneous Examples

These examples mix the first principle with the rules. When a question says 'from the first principle', work from (12.10); otherwise use whichever rule is quickest.

Example 12.19NCERT Example 19 · p. 249

Two derivatives from the first principle

From the first principle, differentiate (i) $f (x) = \frac{3 x - 1}{x + 2}$ , $x \neq - 2$ , and (ii) $g (x) = \sqrt{x}$ , $x > 0$ .

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

Fig. 12.13 shows what the formula $g^{'} (x) = \frac{1}{2 \sqrt{x}}$ means for the graph: the tangents to $y = \sqrt{x}$ get steeper and steeper as the point of contact slides towards the origin, and at 0 itself the difference quotient $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$ grows without bound, so there is no derivative there.

3/4At $x = \frac{1}{4}$ the slope is 1.

Fig. 12.13The slope

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

grows without bound near 0.

The graph of y = √x rises from the origin, steeply at first and then more gently. Step 1 draws the tangent at x = 4, slope 1/4. Step 2 draws the tangent at x = 1, slope 1/2. Step 3 draws the tangent at x = 1/4, slope 1. Step 4 draws the steep tangent at x = 0.04, slope 2.5: the slope 1/(2√x) grows without bound as x approaches 0.

Example 12.20NCERT Example 20 · p. 250

Trigonometric functions from the first principle

From the first principle, differentiate (i) $\sin 2 x$ and (ii) $x \cos x$ .

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

Example 12.21NCERT Example 21 · p. 251

Using the rules

Differentiate (i) $\cos 2 x$ , using $\cos 2 x = \cos^{2} x - \sin^{2} x$ , and (ii) $\cot x$ .

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

Example 12.22NCERT Example 22 · p. 252

Quotients with trigonometric parts

Differentiate (i) $\frac{x^{3} + \sin x}{\cos x}$ and (ii) $\frac{1 + \sin x}{x + \cos x}$ , wherever they are defined.

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

Quick check

1 of 2

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

for

x \neq 0

\frac{\cos x}{1}

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

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

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

Show answer

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

Quotient rule:

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

. Option B has the terms of the numerator in the wrong order, which flips the sign, and option A divides the two derivatives.

Practice · 16 questions

Practice questions with answers (NCERT “Miscellaneous exercises on Chapter 12”)

1
Short answerMediumNCERT Miscellaneous Exercise, Q1
Work from the definition (12.10) to differentiate each function.
1. $3 - 2 x$
  Show solution
  Answer−2
  $\frac{(3 - 2 x - 2 h) - (3 - 2 x)}{h} = - 2$ .
2. $(2 x)^{- 1}$
  Show solution
  Answer $- \frac{1}{2 x^{2}}$
  $\frac{1}{2 (x + h)} - \frac{1}{2 x} = \frac{- h}{2 x (x + h)}$ . Divide by $h$ and let $h \to 0$ : $- \frac{1}{2 x^{2}}$ .
3. $\cos (x + 2)$
  Show solution
  Answer $- \sin (x + 2)$
  With $u = x + 2$ , the difference is $\cos (u + h) - \cos u$ , so as in Practice 12.2 Q9 the limit is $- \sin u = - \sin (x + 2)$ .
4. $\sin (x - \frac{π}{6})$
  Show solution
  Answer $\cos (x - \frac{π}{6})$
  With $u = x - \frac{π}{6}$ , the steps of Example 12.16 give $\cos u = \cos (x - \frac{π}{6})$ .
2
Short answerMediumNCERT Miscellaneous Exercise, Q2–Q4
Differentiate; all letters other than $x$ are non-zero constants.
1. $\frac{2 x + a}{3}$
  Show solution
  Answer $\frac{2}{3}$
  $\frac{2}{3} x + \frac{a}{3}$ has derivative $\frac{2}{3}$ .
2. $(a x + b) (\frac{c}{x} + d)$
  Show solution
  Answer $a d - \frac{b c}{x^{2}}$
  Expand: $a c + a d x + \frac{b c}{x} + b d$ . The derivative is $a d - \frac{b c}{x^{2}}$ .
3. $(p x - q)^{2} (r x + s)$
  Show solution
  Answersee solution
  Product rule, with $\frac{d}{d x} (p x - q)^{2} = 2 p (p x - q)$ : $2 p (p x - q) (r x + s) + r (p x - q)^{2}$ .
  Taking out $p x - q$ : $(p x - q) (3 p r x + 2 p s - q r)$ .
3
Short answerMediumNCERT Miscellaneous Exercise, Q5–Q7
Differentiate; letters other than $x$ are constants.
1. $\frac{p x + q}{r x + s}$
  Show solution
  Answer $\frac{p s - q r}{(r x + s)^{2}}$
  $\frac{p (r x + s) - r (p x + q)}{(r x + s)^{2}} = \frac{p s - q r}{(r x + s)^{2}}$ .
2. $\frac{x - \frac{1}{x}}{x + \frac{1}{x}}$
  Show solution
  Answer $\frac{4 x}{(x^{2} + 1)^{2}}$
  Multiply the top and the bottom by $x$ : the function is $\frac{x^{2} - 1}{x^{2} + 1}$ .
  Quotient rule: $\frac{2 x (x^{2} + 1) - 2 x (x^{2} - 1)}{(x^{2} + 1)^{2}} = \frac{4 x}{(x^{2} + 1)^{2}}$ .
3. $\frac{1}{p x^{2} + q}$
  Show solution
  Answersee solution
  $\frac{0 - 2 p x}{(p x^{2} + q)^{2}} = - \frac{2 p x}{(p x^{2} + q)^{2}}$ .
4
Short answerMediumNCERT Miscellaneous Exercise, Q8–Q9
Differentiate; $c$ , $d$ , $p$ , $q$ , $r$ are constants.
1. $\frac{x + c}{x^{2} + d}$
  Show solution
  Answersee solution
  $\frac{(x^{2} + d) - (x + c) (2 x)}{(x^{2} + d)^{2}} = \frac{d - x^{2} - 2 c x}{(x^{2} + d)^{2}}$ .
2. $\frac{x^{2} + p x + q}{x + r}$
  Show solution
  Answersee solution
  $\frac{(2 x + p) (x + r) - (x^{2} + p x + q)}{(x + r)^{2}} = \frac{x^{2} + 2 r x + p r - q}{(x + r)^{2}}$ .
5
Short answerMediumNCERT Miscellaneous Exercise, Q10–Q12
Differentiate; $p$ and $q$ are constants and $n$ is a positive integer.
1. $\frac{p}{x^{3}} + \frac{q}{x} + \sin x$
  Show solution
  Answersee solution
  $- \frac{3 p}{x^{4}} - \frac{q}{x^{2}} + \cos x$ .
2. $3 \sqrt{x} + 5$
  Show solution
  Answer $\frac{3}{2 \sqrt{x}}$
  $3 \cdot \frac{1}{2} x^{- 1 / 2} = \frac{3}{2 \sqrt{x}}$ .
3. $(2 x + 1)^{n}$
  Show solution
  Answer $2 n (2 x + 1)^{n - 1}$
  Claim: the derivative is $2 n (2 x + 1)^{n - 1}$ . For $n = 1$ it is 2. If it holds for $n - 1$ , the product rule on $(2 x + 1)^{n - 1} (2 x + 1)$ gives $2 (n - 1) (2 x + 1)^{n - 1} + 2 (2 x + 1)^{n - 1} = 2 n (2 x + 1)^{n - 1}$ .
Short answerHardNCERT Miscellaneous Exercise, Q13
Differentiate $(x + 1)^{4} (x - 1)^{3}$ and factorise the answer.
Show solution
Answer $(x + 1)^{3} (x - 1)^{2} (7 x - 1)$
Product rule: $4 (x + 1)^{3} (x - 1)^{3} + 3 (x + 1)^{4} (x - 1)^{2}$ .
Take out $(x + 1)^{3} (x - 1)^{2}$ : the bracket is $4 (x - 1) + 3 (x + 1) = 7 x - 1$ .
$(x + 1)^{3} (x - 1)^{2} (7 x - 1)$
7
Short answerMediumNCERT Miscellaneous Exercise, Q14–Q16
Differentiate each function.
1. $\cos (x + a)$ , $a$ a constant
  Show solution
  Answer $- \sin (x + a)$
  $\cos x \cos a - \sin x \sin a$ has derivative $- \sin x \cos a - \cos x \sin a = - \sin (x + a)$ .
2. $\sec x \tan x$
  Show solution
  Answersee solution
  Product rule: $\sec x \tan x \cdot \tan x + \sec x \cdot \sec^{2} x = \sec x \tan^{2} x + \sec^{3} x$ .
3. $\frac{\sin x}{1 + \cos x}$
  Show solution
  Answer $\frac{1}{1 + \cos x}$
  $\frac{\cos x (1 + \cos x) + \sin^{2} x}{(1 + \cos x)^{2}} = \frac{1 + \cos x}{(1 + \cos x)^{2}} = \frac{1}{1 + \cos x}$ .
8
Short answerMediumNCERT Miscellaneous Exercise, Q17–Q18
Differentiate each function.
1. $\frac{1 + \tan x}{1 - \tan x}$
  Show solution
  Answersee solution
  $\frac{\sec^{2} x (1 - \tan x) + (1 + \tan x) \sec^{2} x}{(1 - \tan x)^{2}} = \frac{2 \sec^{2} x}{(1 - \tan x)^{2}}$ .
2. $\frac{1 - \cos x}{1 + \cos x}$
  Show solution
  Answersee solution
  $\frac{\sin x (1 + \cos x) + (1 - \cos x) \sin x}{(1 + \cos x)^{2}} = \frac{2 \sin x}{(1 + \cos x)^{2}}$ .
ProofHardNCERT Miscellaneous Exercise, Q19
Show that $\frac{d}{d x} (\cos^{n} x) = - n \cos^{n - 1} x \sin x$ for every positive integer $n$ .
Show solution
Answer $- n \cos^{n - 1} x \sin x$
For $n = 1$ this is (12.15). Suppose it holds for $n - 1$ . Write $\cos^{n} x = \cos^{n - 1} x \cdot \cos x$ and use the product rule:
$\begin{aligned} - (n - 1) \cos^{n - 2} x \sin x \cdot \cos x \\ + \cos^{n - 1} x \cdot (- \sin x) \end{aligned}$
This is $- n \cos^{n - 1} x \sin x$ , so the rule holds for $n$ as well.
10
Short answerHardNCERT Miscellaneous Exercise, Q20–Q21
Differentiate; $a$ , $p$ , $q$ , $r$ , $s$ are constants.
1. $\frac{p + q \cos x}{r + s \sin x}$
  Show solution
  Answersee solution
  The numerator is $- q \sin x (r + s \sin x) - (p + q \cos x) s \cos x$ .
  With $\sin^{2} x + \cos^{2} x = 1$ it becomes $- (q r \sin x + p s \cos x + q s)$ ; divide by $(r + s \sin x)^{2}$ .
2. $\frac{\cos (x + a)}{\sin x}$
  Show solution
  Answer $- \frac{\cos a}{\sin^{2} x}$
  The numerator is $- \sin (x + a) \sin x - \cos (x + a) \cos x = - \cos ((x + a) - x) = - \cos a$ .
  The derivative is $- \frac{\cos a}{\sin^{2} x}$ .
11
Short answerMediumNCERT Miscellaneous Exercise, Q22–Q24
Differentiate each function; $a$ and $b$ are constants.
1. $x^{3} (2 \cos x + 3 \sin x)$
  Show solution
  Answersee solution
  $3 x^{2} (2 \cos x + 3 \sin x) + x^{3} (3 \cos x - 2 \sin x)$ .
2. $(x^{2} - 2) \sin x$
  Show solution
  Answersee solution
  $2 x \sin x + (x^{2} - 2) \cos x$ .
3. $(x^{2} + \cos x) (a + b \sin x)$
  Show solution
  Answersee solution
  $(2 x - \sin x) (a + b \sin x) + b \cos x (x^{2} + \cos x)$ .
12
Short answerMediumNCERT Miscellaneous Exercise, Q25, Q29
Differentiate each product.
1. $(x + \sin x) (x + \cot x)$
  Show solution
  Answersee solution
  $(1 + \cos x) (x + \cot x) + (x + \sin x) (1 - {cosec}^{2} x)$ .
2. $(x + \tan x) (1 - \sec x)$
  Show solution
  Answersee solution
  $(1 + \sec^{2} x) (1 - \sec x) - (x + \tan x) \sec x \tan x$ .
13
Short answerHardNCERT Miscellaneous Exercise, Q26–Q28, Q30
Differentiate each quotient; $n$ is a positive integer.
1. $\frac{2 x - 3 \cos x}{5 x + 4 \sin x}$
  Show solution
  Answersee solution
  The numerator of the quotient rule is $(2 + 3 \sin x) (5 x + 4 \sin x) - (2 x - 3 \cos x) (5 + 4 \cos x)$ .
  Expanding and using $\sin^{2} x + \cos^{2} x = 1$ , it is $15 x \sin x - 8 x \cos x + 8 \sin x + 15 \cos x + 12$ ; divide by $(5 x + 4 \sin x)^{2}$ .
2. $\frac{x^{3} \sin \frac{π}{3}}{\cos x}$
  Show solution
  Answersee solution
  $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ is a constant. The derivative is $\frac{\sqrt{3}}{2} \cdot \frac{3 x^{2} \cos x + x^{3} \sin x}{\cos^{2} x}$ .
3. $\frac{x}{1 + \cot x}$
  Show solution
  Answersee solution
  $\frac{(1 + \cot x) - x (- {cosec}^{2} x)}{(1 + \cot x)^{2}} = \frac{1 + \cot x + x {cosec}^{2} x}{(1 + \cot x)^{2}}$ .
4. $\frac{x}{\cos^{n} x}$
  Show solution
  Answersee solution
  By Q9, the quotient rule gives $\frac{\cos^{n} x + n x \cos^{n - 1} x \sin x}{\cos^{2 n} x}$ .
  Cancel $\cos^{n - 1} x$ : $\frac{\cos x + n x \sin x}{\cos^{n + 1} x}$ .
Quick answerMedium
(Entrance-style) If $y = (\sin x + \cos x)^{2}$ , then $\frac{d y}{d x}$ at $x = \frac{π}{4}$ is
012 $\sqrt{2}$
Show answer
Answer (A) 0
$y = 1 + \sin 2 x$ , so $y^{'} = 2 \cos 2 x$ , and $\cos \frac{π}{2} = 0$ . The answer 2 is the value of $y$ itself at $\frac{π}{4}$ .
Quick answerMedium
(Assertion–reason) A: The derivative of $x \sin x$ at $x = 0$ is 0. R: The derivative of a product is the product of the derivatives.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true, but R is falseA is false, but R is true
Show answer
Answer (C) A is true, but R is false
By the product rule the derivative is $\sin x + x \cos x$ , which is 0 at $x = 0$ , so A is true. R is false: $(u v)^{'} = u^{'} v + u v^{'}$ , not $u^{'} v^{'}$ .
Quick answerMedium
(Entrance-style) For $f (x) = (x + 1) (x + 2) (x + 3)$ , find $f^{'} (0)$ .
Show answer
Answer 11
$f^{'} (x) = (x + 2) (x + 3) + (x + 1) (x + 3) + (x + 1) (x + 2)$ , so $f^{'} (0) = 6 + 3 + 2 = 11$ . (Or expand: $f (x) = x^{3} + 6 x^{2} + 11 x + 6$ .)

Recap

Limits and Derivatives: key points and summary

Instantaneous speed is the number that average speeds over shrinking intervals approach; it is the slope of the tangent to the distance–time graph (Section 12.2).
$\lim_{x \to a} f (x) = L$ means $f (x)$ gets as close to $L$ as we like for $x$ close enough to $a$ , $x \neq a$ ; the value $f (a)$ plays no part (Section 12.3).
The limit exists exactly when the left-hand and right-hand limits exist and are equal (Left-hand and right-hand limits).
Limits of sums, differences, products and quotients (with a non-zero bottom limit) are the sums, differences, products and quotients of the limits (Theorem 12.1).
For a polynomial, $\lim_{x \to a} p (x) = p (a)$ ; for a rational function with a $\frac{0}{0}$ form, cancel the factors $x - a$ (Section 12.3.2).
$\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$ ((12.5)).
For $0 < ∣ x ∣ < \frac{π}{2}$ , $\cos x < \frac{\sin x}{x} < 1$ ; with the sandwich theorem this gives $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ (Section 12.4).
$f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ is the slope of the tangent at $(a, f (a))$ ((12.9)).
The derivative function is $f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ , also written $\frac{d y}{d x}$ ((12.10)).
$(u \pm v)^{'} = u^{'} \pm v^{'}$ , $(u v)^{'} = u^{'} v + u v^{'}$ and ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ (Section 12.5.1).
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$ ; polynomials are differentiated term by term ((12.13)).
$\frac{d}{d x} (\sin x) = \cos x$ , $\frac{d}{d x} (\cos x) = - \sin x$ and $\frac{d}{d x} (\tan x) = \sec^{2} x$ (Section 12.5.2).

From historyNCERT Historical Note · p. 255

Who invented calculus?

Calculus has two inventors who worked separately. Isaac Newton (1642–1727) called his rates of change 'fluxions'; Gottfried Wilhelm Leibniz (1646–1716) invented the symbols $d x$ and $d y$ that we still write. Their main discoveries fall between the 1660s and the 1680s.

For over a century the foundations stayed shaky, because nobody could say exactly what an 'infinitely small' change was. Joseph-Louis Lagrange gave us the word 'derived function' and the notation $f^{'} (x)$ . Augustin-Louis Cauchy rebuilt the subject on limits and defined the derivative as the limit of the difference quotient, as we did. Karl Weierstrass later made the definition of a limit completely exact. In 1901 the engineer and teacher John Perry argued, in an address at Glasgow, that calculus should be part of school mathematics, which is one reason you meet it in Class 11.

Long before Newton, mathematicians of the Kerala school, beginning with Mādhava of Saṅgamagrāma (about 1340–1425), found infinite series for sine and cosine. Class 12 returns to their work.

Timeline

From fluxions to limits

1665–1666
Newton develops his method of fluxions
1684
Leibniz publishes his differential calculus, with $d x$ and $d y$
1797
Lagrange's derived functions and the notation $f^{'} (x)$
1823
Cauchy defines the derivative as a limit
1860s
Weierstrass gives the exact definition of a limit
1901
Perry's Glasgow address argues for calculus in schools

Find your NCERT exercise question

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

Quick check: MCQs on Limits and Derivatives

12.2 Intuitive Idea of Derivatives3 questions
Limits from tables and graphs3 questions
12.3.2 Limits of polynomials and rational functions4 questions
12.4 Limits of Trigonometric Functions4 questions
The derivative function3 questions
12.5.2 Derivative of polynomials and trigonometric functions4 questions
Miscellaneous Examples2 questions

Glossary · 12

Key terms (English–Hindi)

derivativeअवकलजavakalaj: $f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ , the instantaneous rate of change of $f$ at $a$ and the slope of the tangent there.
difference quotientअंतर भागफलantar bhagphal: $\frac{f (a + h) - f (a)}{h}$ , the average rate of change of $f$ between $a$ and $a + h$ ; the slope of a chord.First seen in the chapter →

first principleप्रथम सिद्धांतpratham siddhant: Finding a derivative directly from its definition as a limit of difference quotients.

instantaneous rate of changeतात्क्षणिक परिवर्तन दरtatkshanik parivartan dar: The rate at which a quantity is changing at one instant: the number that average rates over shrinking intervals approach.First seen in the chapter →

left-hand limitबाएँ पक्ष की सीमाbaayein paksh ki seema: The number $f (x)$ approaches as $x \to a$ through values less than $a$ , written $\lim_{x \to a^{-}} f (x)$ .First seen in the chapter →
limitसीमाseema: $\lim_{x \to a} f (x) = L$ : the values $f (x)$ get as close to $L$ as we like when $x$ is close enough to $a$ , with $x \neq a$ .

product ruleगुणनफल नियमgunanphal niyam: $(u v)^{'} = u^{'} v + u v^{'}$ , also called the Leibniz rule.First seen in the chapter →

quotient ruleभागफल नियमbhagphal niyam: ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ wherever $v \neq 0$ .First seen in the chapter →

rational functionपरिमेय फलनparimey phalan: A quotient $\frac{g (x)}{h (x)}$ of two polynomials, defined where $h (x) \neq 0$ .First seen in the chapter →
right-hand limitदाएँ पक्ष की सीमाdaayein paksh ki seema: The number $f (x)$ approaches as $x \to a$ through values greater than $a$ , written $\lim_{x \to a^{+}} f (x)$ .First seen in the chapter →

sandwich theoremसैंडविच प्रमेयsandwich pramey: If $f \leq g \leq h$ near $a$ and $f$ and $h$ both tend to $l$ , then $g$ also tends to $l$ .First seen in the chapter →

tangentस्पर्श रेखाsparsh rekha: The line that just touches a curve at a point; the limiting position of the chords through that point.First seen in the chapter →

12.1 Introduction

12.2 Intuitive Idea of Derivatives

12.3 Limits

Left-hand and right-hand limits

Limits from tables and graphs

12.3.1 Algebra of limits

12.3.2 Limits of polynomials and rational functions

12.4 Limits of Trigonometric Functions

12.5 Derivatives

Derivative as the slope of a tangent

The derivative function

12.5.1 Algebra of derivative of functions

12.5.2 Derivative of polynomials and trigonometric functions

Miscellaneous Examples

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Quick check: MCQs on Limits and Derivatives

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