NCERT Class 12 Mathematics — Continuity and Differentiability, Miscellaneous Exercise on Chapter 5. All 22 questions solved.

Every technique in the chapter, with no labels. Before starting a question, decide which it is:

What you see	Technique
A composite function	Chain rule (Exercise 5.2)
An equation in $x$ and $y$	Implicit differentiation (Exercise 5.3)
A variable in the exponent, or a long product	Logarithmic differentiation (Exercise 5.5)
Both variables given via a parameter	$\tfrac{dy/dt}{dx/dt}$ (Exercise 5.6)

Key insight. Several questions here reward a simplification before any differentiation. Question 6 looks fearsome and its answer is $\tfrac12$, because the argument of the $\cot^{-1}$ collapses to $\cot\tfrac{x}{2}$. Question 13’s answer is $0$, because $\sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$ is the constant $\tfrac{\pi}{2}$ on the stated interval.

Whenever an inverse trigonometric function has a complicated argument, spend thirty seconds trying to write that argument as a single ratio of a single angle. It usually is one.

Differentiate w.r.t. $x$ the function in Exercises 1 to 11.

Question 1

$\left(3x^2 – 9x + 5\right)^9$

Solution. Chain rule:

$$\frac{dy}{dx} = 9\left(3x^2-9x+5\right)^8(6x – 9) = 27\left(3x^2-9x+5\right)^8(2x-3)$$

Question 2

$\sin^3x + \cos^6x$

Solution.

$$\frac{dy}{dx} = 3\sin^2x\cos x – 6\cos^5x\sin x = 3\sin x\cos x\left(\sin x – 2\cos^4x\right)$$

Question 3

$(5x)^{3\cos 2x}$

Solution. Variable base and variable exponent, so take logarithms:

$$\log y = 3\cos 2x\log(5x)$$

$$\frac{1}{y}\frac{dy}{dx} = 3\cos 2x \cdot \frac1x – 6\sin 2x\log(5x)$$

using $\tfrac{d}{dx}\log(5x) = \tfrac1x$ (the $5$ disappears because $\log 5x = \log 5 + \log x$).

Question 4

$\sin^{-1}\left(x\sqrt{x}\right)$, $0 \le x \le 1$

Solution. Write $x\sqrt{x} = x^{3/2}$:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1 – x^3}} \cdot \frac32 x^{1/2} = \frac32\sqrt{\frac{x}{1-x^3}}$$

Question 5

$\dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}$, $-2 < x < 2$

Solution. Quotient rule. The numerator’s derivative is $\dfrac{-1}{\sqrt{1 – x^2/4}} \cdot \dfrac12 = \dfrac{-1}{\sqrt{4-x^2}}$, and the denominator’s is $\dfrac{1}{\sqrt{2x+7}}$:

$$\frac{dy}{dx} = \frac{\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}} – \frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7}$$

Combining the two pieces over $(2x+7)^{3/2}$:

Question 6

$\cot^{-1}\left[\dfrac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} – \sqrt{1-\sin x}}\right]$, $0 < x < \dfrac{\pi}{2}$

Solution. Simplify before differentiating. Writing $1 \pm \sin x = \left(\cos\tfrac{x}{2} \pm \sin\tfrac{x}{2}\right)^2$ and taking positive roots — valid because $0 < \tfrac{x}{2} < \tfrac{\pi}{4}$ makes $\cos\tfrac{x}{2} > \sin\tfrac{x}{2} > 0$:

$$\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}$$

So $y = \cot^{-1}\left(\cot\tfrac{x}{2}\right) = \tfrac{x}{2}$, a straight line.

Question 7

$(\log x)^{\log x}$, $x > 1$

Solution. Logarithms:

$$\log y = \log x\log(\log x)$$

$$\frac1y\frac{dy}{dx} = \frac1x\log(\log x) + \log x \cdot \frac{1}{\log x} \cdot \frac1x = \frac{1}{x} + \frac{\log(\log x)}{x}$$

Question 8

$\cos(a\cos x + b\sin x)$, for some constant $a$ and $b$.

Solution. Chain rule, with the derivative of the inner bracket being $-a\sin x + b\cos x$:

$$\frac{dy}{dx} = -\sin(a\cos x + b\sin x)(-a\sin x + b\cos x)$$

Question 9

$(\sin x – \cos x)^{(\sin x – \cos x)}$, $\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$

Solution. Put $u = \sin x – \cos x$, so $y = u^u$ and $\log y = u\log u$:

$$\frac1y\frac{dy}{dx} = u’\log u + u \cdot \frac{u’}{u} = u'(1 + \log u)$$

with $u’ = \cos x + \sin x$. The stated interval is where $u > 0$, which the logarithm requires.

Question 10

$x^x + x^a + a^x + a^a$, for some fixed $a > 0$ and $x > 0$

Solution. Four terms of four different types — that is the point of the question.

• $x^x$: variable base and exponent, so logarithms give $x^x(1 + \log x)$.
• $x^a$: power rule, giving $ax^{a-1}$.
• $a^x$: exponential with constant base, giving $a^x\log a$.
• $a^a$: a constant, giving $0$.

Question 11

$x^{x^2-3} + (x-3)^{x^2}$, for $x > 3$

Solution. Two terms, each needing logarithms separately.

First: $\log u = \left(x^2-3\right)\log x$, so

$$\frac{u’}{u} = \frac{x^2-3}{x} + 2x\log x$$

Second: $\log v = x^2\log(x-3)$, so

$$\frac{v’}{v} = \frac{x^2}{x-3} + 2x\log(x-3)$$

Question 12

Find $\dfrac{dy}{dx}$, if $y = 12(1 – \cos t)$, $x = 10(t – \sin t)$, $-\dfrac{\pi}{2} < t < \dfrac{\pi}{2}$

Solution. Parametric:

$$\frac{dy}{dt} = 12\sin t, \qquad \frac{dx}{dt} = 10(1 – \cos t)$$

$$\frac{dy}{dx} = \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5} \cdot \frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}$$

Question 13

Find $\dfrac{dy}{dx}$, if $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$, $0 < x < 1$

Solution. Put $x = \sin\theta$ with $\theta \in \left(0, \tfrac{\pi}{2}\right)$. Then $\sqrt{1-x^2} = \cos\theta = \sin\left(\tfrac{\pi}{2} – \theta\right)$, so

$$y = \theta + \left(\frac{\pi}{2} – \theta\right) = \frac{\pi}{2}$$

The function is constant on the whole interval.

Question 14

If $x\sqrt{1+y} + y\sqrt{1+x} = 0$, for $-1 < x < 1$, prove that $\dfrac{dy}{dx} = -\dfrac{1}{(1+x)^2}$.

Solution. Rearrange and square to remove the roots:

$$x\sqrt{1+y} = -y\sqrt{1+x} \quad\Longrightarrow\quad x^2(1+y) = y^2(1+x)$$

$$x^2 – y^2 + x^2y – xy^2 = 0 \quad\Longrightarrow\quad (x-y)(x+y) + xy(x-y) = 0$$

$$(x-y)(x + y + xy) = 0$$

Since $x = y$ would force $x = 0$ in the original equation, the useful factor is $x + y + xy = 0$, giving

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

Differentiating by the quotient rule:

$$\frac{dy}{dx} = \frac{-(1+x) + x}{(1+x)^2} = \frac{-1}{(1+x)^2}$$

Question 15

If $(x-a)^2 + (y-b)^2 = c^2$, for some $c > 0$, prove that $\dfrac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}$ is a constant independent of $a$ and $b$.

Solution. Differentiating implicitly:

$$2(x-a) + 2(y-b)y_1 = 0 \quad\Longrightarrow\quad y_1 = -\frac{x-a}{y-b}$$

Differentiating again by the quotient rule and substituting $y_1$:

$$y_2 = -\frac{(y-b) – (x-a)y_1}{(y-b)^2} = -\frac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\frac{c^2}{(y-b)^3}$$

using the original equation for the numerator. Also

$$1 + y_1^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}$$

so $\left(1+y_1^2\right)^{3/2} = \dfrac{c^3}{|y-b|^3}$. Dividing:

$$\frac{\left(1+y_1^2\right)^{3/2}}{y_2} = -c\,\frac{(y-b)^3}{|y-b|^3}$$

The last fraction is $1$ on the arc above the centre ($y > b$) and $-1$ on the arc below it ($y < b$), so the expression is $-c$ on the upper arc and $c$ on the lower. Each arc is the graph of a function of $x$, which is what $\tfrac{dy}{dx}$ needs, and on either one the value is fixed by the radius alone, not by the centre $(a, b)$. (Geometrically: its size is the radius of curvature, the same at every point of a circle; the sign only records which way the arc bends.)

Question 16

If $\cos y = x\cos(a+y)$, with $\cos a \ne \pm1$, prove that $\dfrac{dy}{dx} = \dfrac{\cos^2(a+y)}{\sin a}$.

Solution. Solve for $x$ and differentiate with respect to $y$ — much easier than the other way round:

$$x = \frac{\cos y}{\cos(a+y)}$$

$$\frac{dx}{dy} = \frac{-\sin y\cos(a+y) + \cos y\sin(a+y)}{\cos^2(a+y)}$$

The numerator is $\sin\big((a+y) – y\big) = \sin a$ by the compound angle formula. So

$$\frac{dx}{dy} = \frac{\sin a}{\cos^2(a+y)} \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}$$

Question 17

If $x = a(\cos t + t\sin t)$ and $y = a(\sin t – t\cos t)$, find $\dfrac{d^2y}{dx^2}$.

Solution. From Exercise 5.6 question 10, the first derivatives simplify to

$$\frac{dx}{dt} = at\cos t, \qquad \frac{dy}{dt} = at\sin t \quad\Longrightarrow\quad \frac{dy}{dx} = \tan t$$

For the second derivative, differentiate $\tfrac{dy}{dx}$ with respect to $t$ and divide by $\tfrac{dx}{dt}$ again — this is the step students most often get wrong:

$$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\tan t)}{\frac{dx}{dt}} = \frac{\sec^2t}{at\cos t}$$

Question 18

If $f(x) = |x|^3$, show that $f”(x)$ exists for all real $x$ and find it.

Solution. Write the function without the modulus:

$$f(x) = \begin{cases}x^3, & x \ge 0 \\ -x^3, & x < 0\end{cases}$$

Then $f'(x) = 3x^2$ for $x > 0$ and $-3x^2$ for $x < 0$; at $x = 0$ both one-sided derivatives are $0$, so $f'(0) = 0$ and $f'(x) = 3x|x|$.

Differentiating again, $f”(x) = 6x$ for $x > 0$ and $-6x$ for $x < 0$; at $x = 0$ both one-sided limits of $\tfrac{f'(h) – f'(0)}{h} = \tfrac{3h|h|}{h} = 3|h|$ tend to $0$, so $f”(0) = 0$.

Question 19

Using the fact that $\sin(\mathrm{A} + \mathrm{B}) = \sin\mathrm{A}\cos\mathrm{B} + \cos\mathrm{A}\sin\mathrm{B}$ and the differentiation, obtain the sum formula for cosines.

Solution. Treat $\mathrm{A} = x$ as the variable and $\mathrm{B}$ as a constant, and differentiate both sides with respect to $x$.

Left side: $\dfrac{d}{dx}\sin(x + \mathrm{B}) = \cos(x + \mathrm{B})$.

Right side: $\dfrac{d}{dx}\left[\sin x\cos\mathrm{B} + \cos x\sin\mathrm{B}\right] = \cos x\cos\mathrm{B} – \sin x\sin\mathrm{B}$.

Equating:

$$\cos(\mathrm{A} + \mathrm{B}) = \cos\mathrm{A}\cos\mathrm{B} – \sin\mathrm{A}\sin\mathrm{B}$$

Question 20

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Solution. Yes. Take

$$f(x) = |x| + |x – 1|$$

Both moduli are continuous everywhere, so their sum is. The only points where differentiability can fail are the corners, at $x = 0$ and $x = 1$, and at each the left and right derivatives differ.

At $x = 0$: to the left $f(x) = -x + 1 – x = 1 – 2x$ with derivative $-2$; to the right $f(x) = x + 1 – x = 1$ with derivative $0$. Since $-2 \ne 0$, $f$ is not differentiable there, and the same argument applies at $x = 1$.

Everywhere else $f$ is locally linear and hence differentiable. So $f$ is continuous on $\mathbf{R}$ and non-differentiable at exactly two points.

Question 21

If $y = \begin{vmatrix}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{vmatrix}$, prove that $\dfrac{dy}{dx} = \begin{vmatrix}f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c\end{vmatrix}$.

Solution. Expand the determinant along the first row:

$$y = f(x)(mc – nb) – g(x)(lc – na) + h(x)(lb – ma)$$

Every bracket is a constant, since $l, m, n, a, b, c$ do not depend on $x$. Differentiating term by term:

$$\frac{dy}{dx} = f'(x)(mc-nb) – g'(x)(lc-na) + h'(x)(lb-ma)$$

which is exactly the expansion of the determinant with the first row differentiated.

Question 22

If $y = e^{a\cos^{-1}x}$, $-1 \le x \le 1$, show that $\left(1-x^2\right)\dfrac{d^2y}{dx^2} – x\dfrac{dy}{dx} – a^2y = 0$.

Solution.

$$y_1 = e^{a\cos^{-1}x} \cdot \frac{-a}{\sqrt{1-x^2}} = \frac{-ay}{\sqrt{1-x^2}}$$

Clear the root before differentiating again — this is what keeps the algebra short:

$$\sqrt{1-x^2}\,y_1 = -ay$$

Differentiating both sides:

$$\sqrt{1-x^2}\,y_2 + y_1 \cdot \frac{-x}{\sqrt{1-x^2}} = -ay_1$$

Multiplying through by $\sqrt{1-x^2}$:

$$\left(1-x^2\right)y_2 – xy_1 = -a\sqrt{1-x^2}\,y_1 = -a(-ay) = a^2y$$

using the boxed relation once more. Rearranging gives the required equation.

Common mistakes

• Differentiating before simplifying. Questions 6 and 13 both have short answers hidden behind a complicated-looking expression.
• Taking the second parametric derivative as $\tfrac{d^2y/dt^2}{d^2x/dt^2}$. Question 17 needs $\tfrac{d}{dt}\left(\tfrac{dy}{dx}\right)$ divided by $\tfrac{dx}{dt}$.
• Using one rule for all four terms of question 10. $x^x$, $x^a$ and $a^x$ each need a different method, and $a^a$ needs none.
• Taking the logarithm of a sum in question 11. Split it into two terms first.
• Assuming $|x|^3$ is not twice differentiable. The cube smooths the corner enough for $f”$ to exist everywhere, unlike $|x|$ itself.
• Choosing the wrong factor in question 14. The factorisation gives $(x-y)(x+y+xy) = 0$, and only the second factor is consistent with the original equation.

Practise next

• Exercise 5.5 — the logarithmic differentiation behind questions 3, 7, 9, 10 and 11.
• Exercise 5.7 — the second-derivative problems that questions 15, 17, 18 and 22 build on.