NCERT Class 11 Mathematics — Limits and Derivatives, Miscellaneous Exercise on Chapter 12. All 30 questions solved.

Thirty derivatives, almost all of them applications of two rules:

$$(uv)’ = u’v + uv’, \qquad \left(\frac uv\right)’ = \frac{u’v – uv’}{v^2}$$

together with the standard list

$$\left(x^n\right)’ = nx^{n-1}, \quad (\sin x)’ = \cos x, \quad (\cos x)’ = -\sin x$$ $$(\tan x)’ = \sec^2 x, \quad (\sec x)’ = \sec x\tan x, \quad (\operatorname{cosec} x)’ = -\operatorname{cosec} x\cot x, \quad (\cot x)’ = -\operatorname{cosec}^2 x$$

Key insight. Decide before differentiating whether to expand or to use a rule. A product of two short factors — question 4’s $(ax+b)(cx+d)^2$ — is usually faster expanded; a quotient is almost always faster by rule. And where the answer is wanted in factored form, as in questions 24, 25 and 29, leaving the product rule unexpanded is not laziness but the expected form.

Question 1

Find the derivative of the following functions from first principles:

(i) $-x$    (ii) $(-x)^{-1}$    (iii) $\sin(x+1)$    (iv) $\cos\left(x – \dfrac\pi8\right)$

Solution.

(i) $f(x) = -x$:

$$f'(x) = \lim_{h\to0}\frac{-(x+h) – (-x)}{h} = \lim_{h\to0}\frac{-h}{h} = -1$$

(ii) $f(x) = (-x)^{-1} = -\dfrac1x$:

$$f(x+h) – f(x) = -\frac{1}{x+h} + \frac1x = \frac{-x + (x+h)}{x(x+h)} = \frac{h}{x(x+h)}$$

$$f'(x) = \lim_{h\to0}\frac{1}{x(x+h)} = \frac{1}{x^2}$$

(iii) $f(x) = \sin(x+1)$. Convert the difference to a product with $\sin \mathrm A – \sin \mathrm B = 2\cos\dfrac{\mathrm A+\mathrm B}{2}\sin\dfrac{\mathrm A – \mathrm B}{2}$:

$$\sin(x+h+1) – \sin(x+1) = 2\cos\left(x + 1 + \frac h2\right)\sin\frac h2$$

$$f'(x) = \lim_{h\to0}\left[\cos\left(x+1+\frac h2\right) \cdot \frac{\sin\frac h2}{\frac h2}\right] = \cos(x+1)$$

(iv) The same working with $x + 1$ replaced by $x – \tfrac\pi8$ and the cosine difference formula:

$$f'(x) = -\sin\left(x – \frac\pi8\right)$$

In questions 2 to 30, find the derivative of the given function. Throughout, $a$, $b$, $c$, $d$, $p$, $q$, $r$, $s$ are fixed non-zero constants and $m$, $n$ are integers.

Question 2

$x + a$

Solution. The derivative of $x$ is $1$ and of the constant $a$ is $0$:

$$1$$

Question 3

$(px + q)\left(\dfrac rx + s\right)$

Solution. Expand rather than use the product rule:

$$(px+q)\left(\frac rx + s\right) = pr + psx + \frac{qr}{x} + qs$$

Differentiating, with $pr$ and $qs$ constant and $\tfrac{qr}{x} = qr\,x^{-1}$:

$$ps – \frac{qr}{x^2}$$

Question 4

$(ax+b)(cx+d)^2$

Solution. Product rule with $u = ax + b$ and $v = (cx+d)^2$. Here $u’ = a$ and $v’ = 2c(cx+d)$:

$$\frac{d}{dx} = a(cx+d)^2 + (ax+b) \cdot 2c(cx+d)$$

$$= 2c(ax+b)(cx+d) + a(cx+d)^2$$

Question 5

$\dfrac{ax+b}{cx+d}$

Solution. Quotient rule with $u’ = a$, $v’ = c$:

$$\frac{a(cx+d) – (ax+b)c}{(cx+d)^2} = \frac{acx + ad – acx – bc}{(cx+d)^2} = \frac{ad – bc}{(cx+d)^2}$$

The $x$ terms cancel, leaving a constant numerator — a feature of every such “linear over linear” function.

Question 6

$\dfrac{1 + \frac1x}{1 – \frac1x}$

Solution. Simplify before differentiating. Multiplying above and below by $x$:

$$\frac{x+1}{x-1}$$

Now the quotient rule:

$$\frac{1 \cdot (x-1) – (x+1) \cdot 1}{(x-1)^2} = \frac{-2}{(x-1)^2}, \qquad x \ne 0, 1$$

The exclusion $x \ne 0$ survives from the original form, where $\tfrac1x$ appears.

Question 7

$\dfrac{1}{ax^2 + bx + c}$

Solution. Quotient rule with $u = 1$ (so $u’ = 0$):

$$\frac{0 – 1 \cdot (2ax + b)}{\left(ax^2+bx+c\right)^2} = \frac{-(2ax+b)}{\left(ax^2+bx+c\right)^2}$$

Question 8

$\dfrac{ax+b}{px^2 + qx + r}$

Solution. Quotient rule with $u’ = a$ and $v’ = 2px + q$:

$$\frac{a\left(px^2+qx+r\right) – (ax+b)(2px+q)}{\left(px^2+qx+r\right)^2}$$

Expanding the numerator:

$$apx^2 + aqx + ar – \left(2apx^2 + aqx + 2bpx + bq\right) = -apx^2 – 2bpx + ar – bq$$

Question 9

$\dfrac{px^2 + qx + r}{ax + b}$

Solution. This is question 8 with numerator and denominator swapped, so the numerator of the derivative is the negative of the previous one:

$$\frac{(2px+q)(ax+b) – \left(px^2+qx+r\right)a}{(ax+b)^2} = \frac{apx^2 + 2bpx + bq – ar}{(ax+b)^2}$$

Question 10

$\dfrac{a}{x^4} – \dfrac{b}{x^2} + \cos x$

Solution. Rewrite the powers in index form: $ax^{-4} – bx^{-2} + \cos x$.

$$-4ax^{-5} + 2bx^{-3} – \sin x = -\frac{4a}{x^5} + \frac{2b}{x^3} – \sin x$$

Note the middle sign: differentiating $-bx^{-2}$ gives $+2bx^{-3}$, because the two minus signs multiply.

Question 11

$4\sqrt x – 2$

Solution. Write $\sqrt x = x^{1/2}$:

$$4 \cdot \frac12 x^{-1/2} = 2x^{-1/2} = \frac{2}{\sqrt x}$$

Question 12

$(ax+b)^n$

Solution. By the chain rule (or by expanding and differentiating term by term):

$$n(ax+b)^{n-1} \cdot a = na(ax+b)^{n-1}$$

Question 13

$(ax+b)^n(cx+d)^m$

Solution. Product rule, using question 12 for each factor:

$$na(ax+b)^{n-1}(cx+d)^m + mc(ax+b)^n(cx+d)^{m-1}$$

Take out the common factors $(ax+b)^{n-1}$ and $(cx+d)^{m-1}$:

$$(ax+b)^{n-1}(cx+d)^{m-1}\left[mc(ax+b) + na(cx+d)\right]$$

Question 14

$\sin(x+a)$

Solution. Expand using the compound-angle formula and treat $a$ as constant:

$$\sin x\cos a + \cos x\sin a$$

Differentiating:

$$\cos x\cos a – \sin x\sin a = \cos(x+a)$$

Question 15

$\operatorname{cosec} x\cot x$

Solution. Product rule with $u = \operatorname{cosec} x$ and $v = \cot x$:

$$\frac{d}{dx} = \left(-\operatorname{cosec} x\cot x\right)\cot x + \operatorname{cosec} x\left(-\operatorname{cosec}^2 x\right)$$

$$= -\operatorname{cosec} x\cot^2 x – \operatorname{cosec}^3 x$$

Question 16

$\dfrac{\cos x}{1 + \sin x}$

Solution. Quotient rule:

$$\frac{-\sin x(1+\sin x) – \cos x\cos x}{(1+\sin x)^2} = \frac{-\sin x – \sin^2 x – \cos^2 x}{(1+\sin x)^2}$$

Using $\sin^2 x + \cos^2 x = 1$:

$$= \frac{-\sin x – 1}{(1+\sin x)^2} = \frac{-(1 + \sin x)}{(1+\sin x)^2} = \frac{-1}{1 + \sin x}$$

The Pythagorean identity is what collapses the answer to a single fraction.

Question 17

$\dfrac{\sin x + \cos x}{\sin x – \cos x}$

Solution. Quotient rule with $u’ = \cos x – \sin x$ and $v’ = \cos x + \sin x$:

$$\frac{(\cos x – \sin x)(\sin x – \cos x) – (\sin x + \cos x)(\cos x + \sin x)}{(\sin x – \cos x)^2}$$

The first product is $-(\sin x – \cos x)^2$ and the second is $(\sin x + \cos x)^2$:

$$= \frac{-\left[(\sin x – \cos x)^2 + (\sin x + \cos x)^2\right]}{(\sin x – \cos x)^2}$$

Expanding the bracket, the cross terms cancel and each square contributes $1$:

$$(\sin x – \cos x)^2 + (\sin x + \cos x)^2 = 2\left(\sin^2 x + \cos^2 x\right) = 2$$

$$= \frac{-2}{(\sin x – \cos x)^2}$$

Question 18

$\dfrac{\sec x – 1}{\sec x + 1}$

Solution. Quotient rule with $u’ = v’ = \sec x\tan x$:

$$\frac{\sec x\tan x(\sec x + 1) – (\sec x – 1)\sec x\tan x}{(\sec x+1)^2}$$

Factor out $\sec x\tan x$ from the numerator:

$$= \frac{\sec x\tan x\left[(\sec x + 1) – (\sec x – 1)\right]}{(\sec x+1)^2} = \frac{2\sec x\tan x}{(\sec x+1)^2}$$

Question 19

$\sin^n x$

Solution. Chain rule: the outer function is the $n$th power and the inner is $\sin x$:

$$n\sin^{n-1}x \cdot \cos x$$

Question 20

$\dfrac{a + b\sin x}{c + d\cos x}$

Solution. Quotient rule with $u’ = b\cos x$ and $v’ = -d\sin x$:

$$\frac{b\cos x(c + d\cos x) + d\sin x(a + b\sin x)}{(c+d\cos x)^2}$$

Expanding the numerator:

$$bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x = bc\cos x + ad\sin x + bd$$

using $\sin^2 x + \cos^2 x = 1$ on the two $bd$ terms.

Question 21

$\dfrac{\sin(x+a)}{\cos x}$

Solution. Quotient rule, using question 14 for the numerator’s derivative:

$$\frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x}$$

The numerator is $\cos\left[(x+a) – x\right] = \cos a$:

$$= \frac{\cos a}{\cos^2 x}$$

Recognising the compound-angle formula turns a messy expression into a constant over $\cos^2 x$.

Question 22

$x^4\left(5\sin x – 3\cos x\right)$

Solution. Product rule:

$$4x^3\left(5\sin x – 3\cos x\right) + x^4\left(5\cos x + 3\sin x\right)$$

Taking out $x^3$:

$$x^3\left[20\sin x – 12\cos x + 5x\cos x + 3x\sin x\right]$$

Question 23

$\left(x^2 + 1\right)\cos x$

Solution. Product rule:

$$2x\cos x + \left(x^2+1\right)(-\sin x) = -x^2\sin x – \sin x + 2x\cos x$$

Question 24

$\left(ax^2 + \sin x\right)\left(p + q\cos x\right)$

Solution. Product rule, leaving the answer factored:

$$(2ax + \cos x)(p + q\cos x) + \left(ax^2 + \sin x\right)(-q\sin x)$$

$$= -q\sin x\left(ax^2 + \sin x\right) + (p + q\cos x)(2ax + \cos x)$$

Question 25

$(x + \cos x)(x – \tan x)$

Solution. Product rule with $u’ = 1 – \sin x$ and $v’ = 1 – \sec^2 x = -\tan^2 x$:

$$(1 – \sin x)(x – \tan x) + (x + \cos x)\left(-\tan^2 x\right)$$

$$= -\tan^2 x\,(x + \cos x) + (x – \tan x)(1 – \sin x)$$

The simplification $1 – \sec^2 x = -\tan^2 x$ is what makes the answer compact.

Question 26

$\dfrac{4x + 5\sin x}{3x + 7\cos x}$

Solution. Quotient rule with $u’ = 4 + 5\cos x$ and $v’ = 3 – 7\sin x$:

$$\frac{(4 + 5\cos x)(3x + 7\cos x) – (4x + 5\sin x)(3 – 7\sin x)}{(3x + 7\cos x)^2}$$

Expanding the numerator:

$$\left(12x + 28\cos x + 15x\cos x + 35\cos^2 x\right) – \left(12x – 28x\sin x + 15\sin x – 35\sin^2 x\right)$$

The $12x$ terms cancel, and $35\cos^2 x + 35\sin^2 x = 35$:

$$= 35 + 15x\cos x + 28\cos x + 28x\sin x – 15\sin x$$

Question 27

$\dfrac{x^2\cos\frac\pi4}{\sin x}$

Solution. $\cos\tfrac\pi4$ is a constant, so it comes straight out of the differentiation:

$$\cos\frac\pi4 \cdot \frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \cos\frac\pi4 \cdot \frac{2x\sin x – x^2\cos x}{\sin^2 x}$$

$$= \frac{x\cos\frac\pi4\left(2\sin x – x\cos x\right)}{\sin^2 x}$$

Question 28

$\dfrac{x}{1 + \tan x}$

Solution. Quotient rule with $u’ = 1$ and $v’ = \sec^2 x$:

$$\frac{1 \cdot (1 + \tan x) – x\sec^2 x}{(1 + \tan x)^2} = \frac{1 + \tan x – x\sec^2 x}{(1+\tan x)^2}$$

Question 29

$(x + \sec x)(x – \tan x)$

Solution. Product rule with $u’ = 1 + \sec x\tan x$ and $v’ = 1 – \sec^2 x$:

$$(x + \sec x)\left(1 – \sec^2 x\right) + (x – \tan x)\left(1 + \sec x\tan x\right)$$

Question 30

$\dfrac{x}{\sin^n x}$

Solution. Quotient rule with $u = x$ and $v = \sin^n x$, using question 19 for $v’$:

$$\frac{1 \cdot \sin^n x – x \cdot n\sin^{n-1}x\cos x}{\sin^{2n}x}$$

Cancel a factor of $\sin^{n-1}x$ from top and bottom:

$$= \frac{\sin^{n-1}x\left(\sin x – nx\cos x\right)}{\sin^{2n}x} = \frac{\sin x – nx\cos x}{\sin^{n+1}x}$$

Common mistakes

• Question 1, quoting standard derivatives. “From first principles” requires the limit of the difference quotient. For (iii) and (iv), the sum-to-product identities are what create the standard limit $\tfrac{\sin\theta}{\theta}$.
• Question 6, differentiating the compound fraction as it stands. Multiplying above and below by $x$ first turns it into $\tfrac{x+1}{x-1}$ and saves a page. The domain exclusion $x \ne 0$ still has to be carried over.
• Question 10, mis-signing the middle term. $-bx^{-2}$ differentiates to $+2bx^{-3}$; the two minus signs cancel.
• Question 15, forgetting one of the two minus signs. Both $\operatorname{cosec}$ and $\cot$ have negative derivatives, so both terms of the product rule come out negative.
• Questions 16, 20 and 26, stopping before simplifying. In each case $\sin^2 x + \cos^2 x = 1$ collapses two terms into a constant. Leaving them separate is not wrong, but the printed answer assumes the simplification.
• Question 21, expanding $\sin(x+a)$ instead of recognising the result. The numerator after differentiating is $\cos\left[(x+a)-x\right] = \cos a$ — a compound angle read backwards.
• Question 27, differentiating $\cos\tfrac\pi4$. It is the number $\tfrac{1}{\sqrt2}$, not a function of $x$, so its derivative is zero and it simply multiplies through.
• Questions 24, 25 and 29, expanding the product rule output. The expected answer is the factored form; expanding it is correct but does not match the key and buries the structure.

Practise next

• Exercise 13.1 — the next chapter, Statistics, which changes topic entirely.
• Exercise 12.2 — worth working again alongside questions 15 to 21, since every trigonometric derivative used here is established there.