NCERT Class 11 Mathematics — Limits and Derivatives, Exercise 12.2. All 11 questions solved.

The derivative of $f$ at $x$ is defined by the limit

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

and computing it straight from this definition is what “first principles” means.

Once established, the standard results do the routine work:

$$\frac{d}{dx}\left(x^n\right) = nx^{n-1}, \qquad \frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(\cos x) = -\sin x$$

Product rule: $(uv)’ = u’v + uv’$    Quotient rule: $\left(\dfrac uv\right)’ = \dfrac{u’v – uv’}{v^2}$

Key insight. The power rule $\tfrac{d}{dx}x^n = nx^{n-1}$ holds for negative $n$ as well as positive. So $\tfrac{1}{x^2} = x^{-2}$ differentiates to $-2x^{-3} = -\tfrac{2}{x^3}$ in one step. Rewriting every negative power and root in index form before differentiating — as in questions 9(iii) to 9(v) — turns awkward-looking expressions into routine ones.

Question 1

Find the derivative of $x^2 – 2$ at $x = 10$.

Solution. From first principles, with $f(x) = x^2 – 2$:

$$f'(x) = \lim_{h\to0}\frac{\left[(x+h)^2 – 2\right] – \left[x^2 – 2\right]}{h} = \lim_{h\to0}\frac{2xh + h^2}{h} = \lim_{h\to0}(2x + h) = 2x$$

At $x = 10$:

$$f'(10) = 20$$

Question 2

Find the derivative of $x$ at $x = 1$.

Solution. With $f(x) = x$:

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

The derivative is $1$ everywhere, so in particular $f'(1) = 1$ — as it must be, since $y = x$ is a line of slope $1$.

Question 3

Find the derivative of $99x$ at $x = 100$.

Solution. $f(x) = 99x$ is a straight line of slope $99$, so its derivative is the constant $99$ at every point, including $x = 100$.

Question 4

Find the derivative of the following functions from first principles.

(i) $x^3 – 27$    (ii) $(x-1)(x-2)$    (iii) $\dfrac{1}{x^2}$    (iv) $\dfrac{x+1}{x-1}$

Solution.

(i) Expanding $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$, the constant $-27$ cancels:

$$f'(x) = \lim_{h\to0}\frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h\to0}\left(3x^2 + 3xh + h^2\right) = 3x^2$$

(ii) Expand first: $f(x) = x^2 – 3x + 2$. Then

$$f(x+h) – f(x) = \left[(x+h)^2 – 3(x+h)\right] – \left[x^2 – 3x\right] = 2xh + h^2 – 3h$$

$$f'(x) = \lim_{h\to0}(2x + h – 3) = 2x – 3$$

(iii) With $f(x) = \dfrac{1}{x^2}$:

$$f(x+h) – f(x) = \frac{1}{(x+h)^2} – \frac{1}{x^2} = \frac{x^2 – (x+h)^2}{x^2(x+h)^2} = \frac{-2xh – h^2}{x^2(x+h)^2}$$

Dividing by $h$ and letting $h \to 0$:

$$f'(x) = \lim_{h\to0}\frac{-2x – h}{x^2(x+h)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$

(iv) With $f(x) = \dfrac{x+1}{x-1}$:

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

Over the common denominator $(x+h-1)(x-1)$, the numerator is

$$(x+h+1)(x-1) – (x+1)(x+h-1)$$

$$= \left(x^2 + hx + x – x – h – 1\right) – \left(x^2 + hx – x + x + h – 1\right) = -2h$$

So

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

Question 5

For the function

$$f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \cdots + \frac{x^2}{2} + x + 1$$

prove that $f'(1) = 100\,f'(0)$.

Solution. Differentiate term by term. The general term $\dfrac{x^n}{n}$ gives

$$\frac{d}{dx}\left(\frac{x^n}{n}\right) = \frac{nx^{n-1}}{n} = x^{n-1}$$

so every coefficient collapses to $1$ — which is exactly why the denominators were chosen. The constant $1$ differentiates to $0$:

$$f'(x) = x^{99} + x^{98} + \cdots + x + 1$$

At $x = 1$: every one of the $100$ terms equals $1$:

$$f'(1) = 100$$

At $x = 0$: every term with a positive power vanishes, leaving only the constant:

$$f'(0) = 1$$

Therefore

$$f'(1) = 100 = 100 \times 1 = 100\,f'(0) \qquad \blacksquare$$

Question 6

Find the derivative of $x^n + ax^{n-1} + a^2x^{n-2} + \cdots + a^{n-1}x + a^n$ for some fixed real number $a$.

Solution. Differentiate term by term, treating $a$ as a constant. The term $a^k x^{n-k}$ gives $a^k(n-k)x^{n-k-1}$, and the last term $a^n$ is constant so contributes nothing:

$$nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \cdots + a^{n-1}$$

The coefficients descend $n, n-1, n-2, \ldots, 1$ while the powers of $a$ ascend.

Question 7

For some constants $a$ and $b$, find the derivative of

(i) $(x-a)(x-b)$    (ii) $\left(ax^2 + b\right)^2$    (iii) $\dfrac{x-a}{x-b}$

Solution.

(i) Expand rather than use the product rule — it is quicker here:

$$(x-a)(x-b) = x^2 – (a+b)x + ab$$

$$\frac{d}{dx} = 2x – (a+b) = 2x – a – b$$

(ii) Expand the square:

$$\left(ax^2+b\right)^2 = a^2x^4 + 2abx^2 + b^2$$

$$\frac{d}{dx} = 4a^2x^3 + 4abx = 4ax\left(ax^2 + b\right)$$

(iii) Quotient rule with $u = x – a$, $v = x – b$, so $u’ = v’ = 1$:

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

Question 8

Find the derivative of $\dfrac{x^n – a^n}{x – a}$ for some constant $a$.

Solution. Quotient rule with $u = x^n – a^n$ and $v = x – a$, so $u’ = nx^{n-1}$ and $v’ = 1$:

$$\frac{nx^{n-1}(x-a) – \left(x^n – a^n\right)(1)}{(x-a)^2}$$

Expanding the numerator:

$$\frac{nx^n – anx^{n-1} – x^n + a^n}{(x-a)^2}$$

Question 9

Find the derivative of

(i) $2x – \dfrac34$    (ii) $\left(5x^3 + 3x – 1\right)(x-1)$ (iii) $x^{-3}(5 + 3x)$    (iv) $x^5\left(3 – 6x^{-9}\right)$ (v) $x^{-4}\left(3 – 4x^{-5}\right)$    (vi) $\dfrac{2}{x+1} – \dfrac{x^2}{3x-1}$

Solution.

(i) The constant $\tfrac34$ vanishes:

$$\frac{d}{dx}\left(2x – \frac34\right) = 2$$

(ii) Expand first:

$$\left(5x^3+3x-1\right)(x-1) = 5x^4 – 5x^3 + 3x^2 – 3x – x + 1 = 5x^4 – 5x^3 + 3x^2 – 4x + 1$$

$$\frac{d}{dx} = 20x^3 – 15x^2 + 6x – 4$$

(iii) Multiply out into pure powers:

$$x^{-3}(5+3x) = 5x^{-3} + 3x^{-2}$$

$$\frac{d}{dx} = -15x^{-4} – 6x^{-3} = -\frac{15}{x^4} – \frac{6}{x^3} = -\frac{3}{x^4}\left(5 + 2x\right)$$

(iv)

$$x^5\left(3 – 6x^{-9}\right) = 3x^5 – 6x^{-4}$$

$$\frac{d}{dx} = 15x^4 + 24x^{-5} = 15x^4 + \frac{24}{x^5}$$

(v)

$$x^{-4}\left(3 – 4x^{-5}\right) = 3x^{-4} – 4x^{-9}$$

$$\frac{d}{dx} = -12x^{-5} + 36x^{-10} = -\frac{12}{x^5} + \frac{36}{x^{10}}$$

(vi) Two quotients, differentiated separately.

For $\dfrac{2}{x+1} = 2(x+1)^{-1}$:

$$\frac{d}{dx} = \frac{0 \cdot (x+1) – 2 \cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2}$$

For $\dfrac{x^2}{3x-1}$, by the quotient rule:

$$\frac{2x(3x-1) – x^2(3)}{(3x-1)^2} = \frac{6x^2 – 2x – 3x^2}{(3x-1)^2} = \frac{3x^2 – 2x}{(3x-1)^2} = \frac{x(3x-2)}{(3x-1)^2}$$

Subtracting:

$$\frac{-2}{(x+1)^2} – \frac{x(3x-2)}{(3x-1)^2}$$

Question 10

Find the derivative of $\cos x$ from first principles.

Solution. With $f(x) = \cos x$:

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

Convert the difference into a product using $\cos \mathrm A – \cos \mathrm B = -2\sin\dfrac{\mathrm A + \mathrm B}{2}\sin\dfrac{\mathrm A – \mathrm B}{2}$, with $\mathrm A = x + h$ and $\mathrm B = x$:

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

So

$$f'(x) = \lim_{h\to0}\frac{-2\sin\left(x+\frac h2\right)\sin\frac h2}{h} = \lim_{h\to0}\left[-\sin\left(x + \frac h2\right) \cdot \frac{\sin\frac h2}{\frac h2}\right]$$

As $h \to 0$, the first factor tends to $-\sin x$ and the second to $1$:

$$f'(x) = -\sin x$$

The manoeuvre of writing $h$ as $2 \cdot \tfrac h2$ is what produces the standard form $\tfrac{\sin\theta}{\theta}$ exactly.

Question 11

Find the derivative of the following functions:

(i) $\sin x\cos x$   (ii) $\sec x$   (iii) $5\sec x + 4\cos x$ (iv) $\operatorname{cosec} x$   (v) $3\cot x + 5\operatorname{cosec} x$ (vi) $5\sin x – 6\cos x + 7$   (vii) $2\tan x – 7\sec x$

Solution.

(i) By the product rule:

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

(ii) Write $\sec x = \dfrac{1}{\cos x}$ and use the quotient rule:

$$\frac{0 \cdot \cos x – 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x\tan x$$

(iii) Term by term, using (ii):

$$5\sec x\tan x – 4\sin x$$

(iv) $\operatorname{cosec} x = \dfrac{1}{\sin x}$:

$$\frac{0 \cdot \sin x – 1 \cdot \cos x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\operatorname{cosec} x\cot x$$

(v) Using $\dfrac{d}{dx}\cot x = -\operatorname{cosec}^2 x$ and part (iv):

$$-3\operatorname{cosec}^2 x – 5\operatorname{cosec} x\cot x$$

(vi)

$$5\cos x + 6\sin x$$

(The $+7$ is constant and contributes nothing.)

(vii) Using $\dfrac{d}{dx}\tan x = \sec^2 x$ and part (ii):

$$2\sec^2 x – 7\sec x\tan x$$

Common mistakes

• Question 4, differentiating instead of using the definition. “From first principles” means computing the limit of the difference quotient; quoting $\tfrac{d}{dx}x^3 = 3x^2$ answers a different question.
• Question 4(iii), differentiating $\tfrac{1}{x^2}$ as $\tfrac{1}{2x}$. In index form it is $x^{-2}$, whose derivative is $-2x^{-3}$. Rewriting before differentiating removes the temptation.
• Question 9(iii) to (v), applying the product rule. Multiplying out first turns each into a sum of pure powers and makes the differentiation a single line.
• Question 9(vi), forgetting the minus in front of the second quotient. The derivative of a difference is the difference of the derivatives — the whole second result must be subtracted, brackets and all.
• Question 10, using $\cos \mathrm A – \cos \mathrm B = 2\sin\ldots$ The identity carries a leading minus; dropping it makes the derivative come out as $+\sin x$.
• Question 11(iv) and (v), sign errors on the co-functions. The derivatives of $\cos$, $\cot$ and $\operatorname{cosec}$ — the “co-” functions — all carry a minus sign; those of $\sin$, $\tan$ and $\sec$ do not.
• Question 11(vi), differentiating the constant to $7$. The derivative of any constant is $0$.

Practise next

• Miscellaneous Exercise on Chapter 12 — thirty derivatives applying the product and quotient rules to messier expressions.
• Exercise 12.1 — worth revising alongside questions 4 and 10, since every first principles computation ends in a limit from that exercise.