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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

NCERT – Exercise 5.1

Prove that the function $f(x) = 5x – 3$ is continuous at $x = 0$, at $x = -3$ and at $x = 5$.

SOLUTION

$$f(x) = 5x – 3$$

At $x = 0$:

We have, $f(0) = -3$

$$\mathop{\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 – h\atop\scriptstyle h \to 0} 5(0 – h) – 3 = -3$$

$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 + h\atop\scriptstyle h \to 0} 5(0 + h) – 3 = -3$$

$\therefore \mathop {\lim }\limits_{x \to {0^{}}} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

$\therefore$ $f$ is continuous at $x = 0$.

At $x = -3$:

We have, $f(-3) = 5(-3) – 3 = -18$

$$\mathop {\lim }\limits_{x \to -{3^ – }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 – h\atop\scriptstyle h \to 0} [5(-3 – h) – 3] = -18$$

$$\mathop {\lim }\limits_{x \to -{3^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} [5(-3 + h) – 3] = -18$$

$\therefore \mathop {\lim }\limits_{x \to -{3^ – }} f(x) = \mathop {\lim }\limits_{x \to -{3^ + }} f(x) = f(-3)$

$\therefore$ $f$ is continuous at $x = -3$.

At $x = 5$:

$f(5) = 5(5) – 3 = 22$

$$\mathop {\lim }\limits_{x \to {5^ – }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 5 – h\atop\scriptstyle h \to 0} [5(5 – h) – 3] = 22$$

$$\mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 5 + h\atop\scriptstyle h \to 0} [5(5 + h) – 3] = 22$$

$\therefore \mathop {\lim }\limits_{x \to {5^ – }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} f(x) = f(5)$

$\therefore$ $f$ is continuous at $x = 5$.
Examine the continuity of the function $f(x) = 2x^2 – 1$ at $x = 3$.

SOLUTION

$$f(x) = 2x^2 – 1$$

$$R.H.L. = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} 2{(3 + h)^2} – 1 = 17$$

$$L.H.L. = \mathop {\lim }\limits_{x \to {3^ – }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 – h\atop\scriptstyle h \to 0} 2{(3 – h)^2} – 1 = 17$$

$\therefore R.H.L. = L.H.L.$

Also, $f(3) = 2(3)^2 – 1 = 17$

$\therefore \mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ – }} f(x) = f(3)$

Hence, the given function $f(x) = 2x^2 – 1$ is continuous at $x = 3$.
Examine the following functions for continuity:

(a) $f(x) = x – 5$

(b) $f(x) = \cfrac{1}{{x – 5}}, x \ne 5$

(c) $f(x) = \cfrac{{{x^2} – 25}}{{x + 5}}, x \ne -5$

(d) $f(x) = |x – 5|$

SOLUTION

(a) $f(x) = x – 5$. Let $a$ be a real number.

$$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = a – 5, \quad \mathop {\lim }\limits_{x \to {a^ – }} f(x) = a – 5, \quad f(a) = a – 5$$

Function is continuous.

(b) $f(x) = \cfrac{1}{{x – 5}}$. Let $a \ne 5$ be a real number.

$$\mathop {\lim }\limits_{x \to a} f(x) = \cfrac{1}{a-5} = f(a)$$

Continuous for all $x \in \mathbb{R} \setminus \{5\}$.

(c) $f(x) = \cfrac{{{x^2} – 25}}{{x + 5}} = x – 5$ for $x \ne -5$.

Continuous at every point of its domain.

(d) $f(x) = |x – 5|$.

$\mathop {\lim }\limits_{x \to a} f(x) = |a – 5| = f(a)$. Continuous.
Prove that the function $f(x) = {x^n}$ is continuous at $x = n$, where $n$ is a positive integer.

SOLUTION

Given $f(x) = x^n$. $\mathop {\lim }\limits_{x \to n} f(x) = n^n = f(n)$. Continuous.
Is the function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$ continuous at $x=0$? At $x=1$? At $x=2$?

SOLUTION
- At $x=0$: $LHL = RHL = f(0) = 0$. Continuous.
- At $x=1$: $LHL = 1, RHL = 5$. Discontinuous.
- At $x=2$: $LHL = RHL = f(2) = 5$. Continuous.

Direction: For questions (6 – 12), find all points of discontinuity of function $f(x)$.

$f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x – 3, & \text{if } x > 2 \end{cases}$

SOLUTION

At $x=2$: $LHL = 7, RHL = 1$. $LHL \ne RHL$. Discontinuous at $x=2$.
$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$

SOLUTION

At $x=-3$: $LHL = 6, RHL = 6, f(-3) = 6$. Continuous.

At $x=3$: $LHL = -6, RHL = 20$. Discontinuous at $x=3$.

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$f(x) = \begin{cases} {x^{10}} – 1, & \text{if } x \le 1 \\ {x^2}, & \text{if } x > 1 \end{cases}$

SOLUTION

We observe that $f(x)$ is continuous at real numbers $x < 1$ and $x > 1$ as it is a polynomial function.

Now, checking continuity at $x = 1$:

$$L.H.L. = \mathop {\lim }\limits_{x \to {1^ – }} ({x^{10}} – 1) = 0$$

$$R.H.L. = \mathop {\lim }\limits_{x \to {1^ + }} ({x^2}) = 1$$

Also, $f(1) = 0$. Since $L.H.L. \ne R.H.L.$, $f$ is discontinuous at $x = 1$.

Answer: Point of discontinuity is $x = 1$.
Is the function defined by $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x – 5, & \text{if } x > 1 \end{cases}$ a continuous function?

SOLUTION

Check at $x = 1$:

$$R.H.L. = \mathop {\lim }\limits_{x \to {1^ + }} (x – 5) = -4$$

$$L.H.L. = \mathop {\lim }\limits_{x \to {1^ – }} (x + 5) = 6$$

Since $L.H.L. \ne R.H.L.$, $f(x)$ is not continuous at $x = 1$.
Discuss the continuity of the function $f$, where $f(x) = \begin{cases} 3, & 0 \le x \le 1 \\ 4, & 1 < x < 3 \\ 5, & 3 \le x \le 10 \end{cases}$

SOLUTION

At $x = 1$: $L.H.L. = 3$, $R.H.L. = 4$. Discontinuous.

At $x = 3$: $L.H.L. = 4$, $R.H.L. = 5$. Discontinuous.

Thus, the function is not continuous at $x = 1$ and $x = 3$.
$f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \le x \le 1 \\ 4x, & \text{if } x > 1 \end{cases}$

SOLUTION

At $x = 0$: $L.H.L. = 0, R.H.L. = 0, f(0) = 0$. Continuous.

At $x = 1$: $L.H.L. = 0, R.H.L. = 4$. Discontinuous.
$f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$

SOLUTION

At $x = -1$: $L.H.L. = -2, R.H.L. = -2, f(-1) = -2$. Continuous.

At $x = 1$: $L.H.L. = 2, R.H.L. = 2, f(1) = 2$. Continuous.

Function is continuous everywhere.
Find the relationship between $a$ and $b$ so that $f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases}$ is continuous at $x = 3$.

SOLUTION

For continuity at $x = 3$, $L.H.L. = R.H.L. = f(3)$.

$$3a + 1 = 3b + 3$$

$$3a – 3b = 2 \implies a – b = \cfrac{2}{3}$$
For what value of $\lambda$ is $f(x) = \begin{cases} \lambda ({x^2} – 2x), & x \le 0 \\ 4x + 1, & x > 0 \end{cases}$ continuous at $x = 0$?

SOLUTION

$$L.H.L. = \lambda(0) = 0, \quad R.H.L. = 4(0) + 1 = 1$$

Since $0 \ne 1$, there is no value of $\lambda$ for which the function is continuous at $x = 0$.

At $x = 1$: $f(x) = 4x + 1$, which is continuous for any real value of $\lambda$.
Show that $g(x) = x – [x]$ is discontinuous at all integral points.

SOLUTION

Let $n \in \mathbb{I}$. Then $g(n) = n – n = 0$.

$$L.H.L. = \mathop {\lim }\limits_{x \to {n^ – }} (x – [x]) = n – (n – 1) = 1$$

$$R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} (x – [x]) = n – n = 0$$

Since $L.H.L. \ne R.H.L.$, it is discontinuous at all integers.
Is $f(x) = {x^2} – \sin x + 5$ continuous at $x = \pi$?

SOLUTION

$$f(\pi) = \pi^2 – \sin \pi + 5 = \pi^2 + 5$$

$\mathop {\lim }\limits_{x \to \pi} f(x) = \pi^2 – 0 + 5 = \pi^2 + 5$. Continuous.

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Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

SOLUTION

The cosine function is continuous for all $x \in \mathbb{R}$.

The cosecant function $f(x) = \csc x = \cfrac{1}{\sin x}$ is continuous for all $x \in \mathbb{R}$ except where $\sin x = 0$, i.e., $x = n\pi, n \in \mathbb{I}$.

The secant function $f(x) = \sec x = \cfrac{1}{\cos x}$ is continuous for all $x \in \mathbb{R}$ except where $\cos x = 0$, i.e., $x = (2n+1)\cfrac{\pi}{2}, n \in \mathbb{I}$.

The cotangent function $f(x) = \cot x = \cfrac{\cos x}{\sin x}$ is continuous for all $x \in \mathbb{R}$ except where $\sin x = 0$, i.e., $x = n\pi, n \in \mathbb{I}$.
Find all points of discontinuity of $f(x) = \begin{cases} \cfrac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$

SOLUTION

At $x = 0$:

$$L.H.L. = \mathop {\lim }\limits_{x \to {0^ – }} \cfrac{\sin x}{x} = 1$$

$$R.H.L. = \mathop {\lim }\limits_{x \to {0^ + }} (x + 1) = 1$$

Also, $f(0) = 0 + 1 = 1$. Since $LHL = RHL = f(0)$, the function is continuous at $x = 0$.

Answer: No point of discontinuity.
Determine if $f$ defined by $f(x) = \begin{cases} x^2 \sin\left(\cfrac{1}{x}\right), & \text{if } x \ne 0 \\ 0, & \text{if } x = 0 \end{cases}$ is a continuous function.

SOLUTION

At $x = 0$, we check $\mathop {\lim }\limits_{x \to 0} x^2 \sin\left(\cfrac{1}{x}\right)$.

Since $-1 \le \sin\left(\cfrac{1}{x}\right) \le 1$, multiplying by $x^2$ gives $-x^2 \le x^2 \sin\left(\cfrac{1}{x}\right) \le x^2$.

As $x \to 0$, both $-x^2$ and $x^2 \to 0$. By Squeeze Theorem, $\mathop {\lim }\limits_{x \to 0} f(x) = 0 = f(0)$.

Answer: It is a continuous function.
Examine the continuity of $f$, where $f(x) = \begin{cases} \sin x – \cos x, & \text{if } x \ne 0 \\ -1, & \text{if } x = 0 \end{cases}$

SOLUTION

$$\mathop {\lim }\limits_{x \to 0} (\sin x – \cos x) = \sin 0 – \cos 0 = 0 – 1 = -1$$

Also, $f(0) = -1$. Since limit exists and equals the function value, $f$ is continuous at $x = 0$.
Find the values of $k$ so that the function $f(x) = \begin{cases} \cfrac{k \cos x}{\pi – 2x}, & \text{if } x \ne \cfrac{\pi}{2} \\ 3, & \text{if } x = \cfrac{\pi}{2} \end{cases}$ is continuous at $x = \cfrac{\pi}{2}$.

SOLUTION

For continuity, $\mathop {\lim }\limits_{x \to \pi/2} \cfrac{k \cos x}{\pi – 2x} = 3$.

Let $x = \cfrac{\pi}{2} + h$. As $x \to \cfrac{\pi}{2}, h \to 0$.

$$\mathop {\lim }\limits_{h \to 0} \cfrac{k \cos(\pi/2 + h)}{\pi – 2(\pi/2 + h)} = \mathop {\lim }\limits_{h \to 0} \cfrac{-k \sin h}{-2h} = \cfrac{k}{2} \mathop {\lim }\limits_{h \to 0} \cfrac{\sin h}{h} = \cfrac{k}{2}$$

Setting $\cfrac{k}{2} = 3 \implies k = 6$.
Find $k$ if $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ is continuous at $x = 2$.

SOLUTION

$LHL = f(2) = k(2)^2 = 4k$. $RHL = 3$.

$4k = 3 \implies k = \cfrac{3}{4}$.
Find $k$ if $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ is continuous at $x = \pi$.

SOLUTION

$LHL = k\pi + 1$. $RHL = \cos \pi = -1$.

$k\pi + 1 = -1 \implies k\pi = -2 \implies k = -\cfrac{2}{\pi}$.
Find $k$ if $f(x) = \begin{cases} kx + 1, & \text{if } x \le 5 \\ 3x – 5, & \text{if } x > 5 \end{cases}$ is continuous at $x = 5$.

SOLUTION

$LHL = 5k + 1$. $RHL = 3(5) – 5 = 10$.

$5k + 1 = 10 \implies 5k = 9 \implies k = \cfrac{9}{5}$.
Find the values of $a$ and $b$ such that $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is continuous.

SOLUTION

At $x=2$: $2a + b = 5$.

At $x=10$: $10a + b = 21$.

Subtracting: $8a = 16 \implies a = 2$.

Then $2(2) + b = 5 \implies b = 1$.
Show that the function defined by $f(x) = \cos(x^2)$ is a continuous function.

SOLUTION

Let $g(x) = \cos x$ and $h(x) = x^2$. Both are continuous functions.

The composition $(g \circ h)(x) = g(h(x)) = \cos(x^2)$ is continuous as the composition of two continuous functions is also continuous.
Show that $f(x) = |\cos x|$ is continuous.

SOLUTION

Let $g(x) = |x|$ and $h(x) = \cos x$. Both are continuous.

$(g \circ h)(x) = |\cos x|$ is continuous by composition rule.
Examine that $\sin |x|$ is a continuous function.

SOLUTION

Let $g(x) = \sin x$ and $h(x) = |x|$. Since both are continuous on $\mathbb{R}$, their composition $(g \circ h)(x) = \sin |x|$ is also continuous.
Find all points of discontinuity of $f(x) = |x| – |x + 1|$.

SOLUTION

Since $|x|$ and $|x+1|$ are continuous functions, their difference $f(x)$ is continuous everywhere on $\mathbb{R}$.

Answer: No point of discontinuity.

NCERT – Exercise 5.2

Differentiate the functions with respect to $x$ in Exercises 1 to 8.

$\sin(x^2 + 5)$

SOLUTION

Let $y = \sin(x^2 + 5)$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\sin(x^2 + 5) = \cos(x^2 + 5)\cfrac{d}{dx}(x^2 + 5)$$

$$= \cos(x^2 + 5)(2x + 0) = 2x\cos(x^2 + 5)$$
$\cos(\sin x)$

SOLUTION

Let $y = \cos(\sin x)$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\cos(\sin x) = -\sin(\sin x)\cfrac{d}{dx}\sin x$$

$$= -\sin(\sin x)\cos x$$
$\sin(ax + b)$

SOLUTION

Let $y = \sin(ax + b)$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\sin(ax + b) = \cos(ax + b)\cfrac{d}{dx}(ax + b)$$

$$= \cos(ax + b)(a + 0) = a\cos(ax + b)$$
$\sec(\tan(\sqrt{x}))$

SOLUTION

Let $y = \sec\{\tan(\sqrt{x})\}$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\sec(\tan \sqrt{x}) = \sec(\tan \sqrt{x})\tan(\tan \sqrt{x})\cfrac{d}{dx}\tan \sqrt{x}$$

$$= \sec(\tan \sqrt{x}) \cdot \tan(\tan \sqrt{x}) \cdot \sec^2\sqrt{x}\cfrac{d}{dx}(\sqrt{x})$$

$$= \sec(\tan \sqrt{x}) \cdot \tan(\tan \sqrt{x}) \cdot \sec^2\sqrt{x} \cdot \cfrac{1}{2}x^{\cfrac{1}{2}-1}$$

$$= \sec(\tan \sqrt{x}) \cdot \tan(\tan \sqrt{x}) \cdot \sec^2\sqrt{x} \cdot \cfrac{1}{2\sqrt{x}}$$
$\cfrac{\sin(ax + b)}{\cos(cx + d)}$

SOLUTION

Let $y = \cfrac{\sin(ax + b)}{\cos(cx + d)}$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\left( \cfrac{\sin(ax + b)}{\cos(cx + d)} \right)$$

$$= \cfrac{\cos(cx + d)\cfrac{d}{dx}\sin(ax + b) – \sin(ax + b)\cfrac{d}{dx}\cos(cx + d)}{\cos^2(cx + d)}$$

$$= \cfrac{a\cos(cx + d)\cos(ax + b) + c\sin(ax + b)\sin(cx + d)}{\cos^2(cx + d)}$$

$$= a\cos(ax + b)\sec(cx + d) + c\sin(ax + b)\tan(cx + d) \cdot \sec(cx + d)$$
$\cos x^3 \cdot \sin^2(x^5)$

SOLUTION

Let $y = \cos x^3 \cdot \sin^2(x^5)$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}[\cos x^3 \cdot \sin^2(x^5)]$$

$$= \cos x^3 \cfrac{d}{dx}\sin^2(x^5) + \sin^2(x^5) \cfrac{d}{dx}\cos x^3$$

$$= \cos x^3 \cdot 2\sin(x^5)\cfrac{d}{dx}\sin(x^5) + \sin^2(x^5)(-\sin x^3)\cfrac{d}{dx}(x^3)$$

$$= \cos x^3 \cdot 2\sin(x^5)\cos(x^5)\cfrac{d}{dx}(x^5) + \sin^2(x^5)(-\sin x^3)(3x^2)$$

$$= 10x^4\cos x^3\sin(x^5)\cos(x^5) – 3x^2\sin^2(x^5)\sin x^3$$
$2\sqrt{\cot(x^2)}$

SOLUTION

Let $y = 2\sqrt{\cot(x^2)}$

$$\Rightarrow \cfrac{dy}{dx} = 2\cfrac{d}{dx}\sqrt{\cot(x^2)} = 2 \cdot \cfrac{1}{2}\{\cot(x^2)\}^{\cfrac{-1}{2}} \cdot \cfrac{d}{dx}\cot(x^2)$$

$$= \cfrac{1}{\sqrt{\cot(x^2)}}\{-\csc^2(x^2)\}\cfrac{d}{dx}(x^2)$$

$$= \cfrac{-2x\csc^2(x^2)}{\sqrt{\cot(x^2)}} = \cfrac{-2\sqrt{2}x}{\sin x^2\sqrt{\sin 2x^2}}$$
$\cos(\sqrt{x})$

SOLUTION

Let $y = \cos(\sqrt{x})$

$$\Rightarrow \cfrac{dy}{dx} = \cfrac{d}{dx}\cos(\sqrt{x}) = -\sin \sqrt{x} \cdot \cfrac{d}{dx}(\sqrt{x})$$

$$= -\sin \sqrt{x} \cdot \cfrac{1}{2}x^{\cfrac{-1}{2}} = \cfrac{-\sin \sqrt{x}}{2\sqrt{x}}$$
Prove that the function $f$ given by $f(x) = |x – 1|, x \in \mathbb{R}$ is not differentiable at $x = 1$.

SOLUTION

We have, $f(x) = |x – 1|$ and $f(1) = |1 – 1| = 0$.

$$Rf'(1) = \mathop{\lim}_{h \to 0} \cfrac{f(1 + h) – f(1)}{h} = \mathop{\lim}_{h \to 0} \cfrac{|1 + h – 1| – 0}{h} = \mathop{\lim}_{h \to 0} \cfrac{|h|}{h} = \mathop{\lim}_{h \to 0} \cfrac{h}{h} = 1$$

$$Lf'(1) = \mathop{\lim}_{h \to 0} \cfrac{f(1 – h) – f(1)}{-h} = \mathop{\lim}_{h \to 0} \cfrac{|1 – h – 1| – 0}{-h} = \mathop{\lim}_{h \to 0} \cfrac{|-h|}{-h} = \mathop{\lim}_{h \to 0} \cfrac{h}{-h} = -1$$

Since $Rf'(1) \ne Lf'(1)$, $f(x)$ is not differentiable at $x = 1$.
Prove that the greatest integer function defined by $f(x) = [x], 0 < x < 3$ is not differentiable at $x = 1$ and $x = 2$.

SOLUTION

At $x = 1$:

$$Rf'(1) = \mathop{\lim}_{h \to 0} \cfrac{f(1 + h) – f(1)}{h} = \mathop{\lim}_{h \to 0} \cfrac{[1 + h] – [1]}{h} = \cfrac{1 – 1}{h} = 0$$

$$Lf'(1) = \mathop{\lim}_{h \to 0} \cfrac{f(1 – h) – f(1)}{-h} = \mathop{\lim}_{h \to 0} \cfrac{[1 – h] – [1]}{-h} = \mathop{\lim}_{h \to 0} \cfrac{0 – 1}{-h} = \infty$$

At $x = 2$:

$$Rf'(2) = \mathop{\lim}_{h \to 0} \cfrac{[2 + h] – [2]}{h} = 0$$

$$Lf'(2) = \mathop{\lim}_{h \to 0} \cfrac{[2 – h] – [2]}{-h} = \mathop{\lim}_{h \to 0} \cfrac{1 – 2}{-h} = \infty$$

Since $Rf’ \ne Lf’$ at both points, the function is not differentiable at $x = 1$ and $x = 2$.

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NCERT – Exercise 5.3

Find $\cfrac{dy}{dx}$ in the following:

$2x + 3y = \sin x$

SOLUTION

We are given that, $2x + 3y = \sin x$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$2 + 3\cfrac{dy}{dx} = \cos x \implies \cfrac{dy}{dx} = \cfrac{\cos x – 2}{3}$$
$2x + 3y = \sin y$

SOLUTION

We are given that, $2x + 3y = \sin y$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$2 + 3\cfrac{dy}{dx} = \cos y \cfrac{dy}{dx}$$

$$\implies \cos y \cfrac{dy}{dx} – 3\cfrac{dy}{dx} = 2 \implies \cfrac{dy}{dx} = \cfrac{2}{\cos y – 3}$$
$ax + by^2 = \cos y$

SOLUTION

We are given that, $ax + by^2 = \cos y$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$a + b\left[ 2y\cfrac{dy}{dx} \right] = -\sin y \cfrac{dy}{dx} \implies a + 2by\cfrac{dy}{dx} + \sin y \cfrac{dy}{dx} = 0$$

$$\implies \cfrac{dy}{dx}[2by + \sin y] = -a \implies \cfrac{dy}{dx} = \cfrac{-a}{2by + \sin y}$$
$xy + y^2 = \tan x + y$

SOLUTION

We are given that, $xy + y^2 = \tan x + y$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$x\cfrac{dy}{dx} + y + 2y\cfrac{dy}{dx} = \sec^2 x + \cfrac{dy}{dx}$$

$$\implies x\cfrac{dy}{dx} + 2y\cfrac{dy}{dx} – \cfrac{dy}{dx} = \sec^2 x – y$$

$$\implies \cfrac{dy}{dx}[x + 2y – 1] = \sec^2 x – y \implies \cfrac{dy}{dx} = \cfrac{\sec^2 x – y}{x + 2y – 1}$$
$x^2 + xy + y^2 = 100$

SOLUTION

We are given that, $x^2 + xy + y^2 = 100$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$2x + x\cfrac{dy}{dx} + y + 2y\cfrac{dy}{dx} = 0 \implies \cfrac{dy}{dx}[x + 2y] = -(2x + y)$$

$$\implies \cfrac{dy}{dx} = \cfrac{-(2x + y)}{(x + 2y)}$$
$x^3 + x^2y + xy^2 + y^3 = 81$

SOLUTION

We are given that, $x^3 + x^2y + xy^2 + y^3 = 81$ … (i)

Differentiating (i) on both sides w.r.t. $x$, we get:

$$3x^2 + x^2\cfrac{dy}{dx} + 2xy + y^2 + 2xy\cfrac{dy}{dx} + 3y^2\cfrac{dy}{dx} = 0$$

$$\implies \cfrac{dy}{dx}[x^2 + 2xy + 3y^2] = -(3x^2 + 2xy + y^2)$$

$$\implies \cfrac{dy}{dx} = \cfrac{-(3x^2 + 2xy + y^2)}{x^2 + 2xy + 3y^2}$$
$\sin^2 y + \cos xy = k$

SOLUTION

Differentiating w.r.t. $x$, we get:

$$2\sin y\cos y\cfrac{dy}{dx} – \sin xy \left[ x\cfrac{dy}{dx} + y \right] = 0$$

$$\implies \sin 2y \cfrac{dy}{dx} – x\sin xy\cfrac{dy}{dx} – y\sin xy = 0$$

$$\implies \cfrac{dy}{dx}[\sin 2y – x\sin xy] = y\sin xy \implies \cfrac{dy}{dx} = \cfrac{y\sin xy}{\sin 2y – x\sin xy}$$
$\sin^2 x + \cos^2 y = 1$

SOLUTION

Differentiating w.r.t. $x$, we get:

$$2\sin x\cos x + 2\cos y(-\sin y)\cfrac{dy}{dx} = 0$$

$$\implies \sin 2x – \sin 2y\cfrac{dy}{dx} = 0 \implies \cfrac{dy}{dx} = \cfrac{\sin 2x}{\sin 2y}$$
$y = \sin^{-1} \left( \cfrac{2x}{1 + x^2} \right)$

SOLUTION

Putting $x = \tan \theta$, we get:

$$y = \sin^{-1} \left( \cfrac{2\tan \theta}{1 + \tan^2 \theta} \right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$$

$$\therefore \cfrac{dy}{dx} = \cfrac{2}{1 + x^2}$$
$y = \tan^{-1} \left( \cfrac{3x – x^3}{1 – 3x^2} \right)$

SOLUTION

Putting $x = \tan \theta$, we get:

$$y = \tan^{-1} \left( \cfrac{3\tan \theta – \tan^3 \theta}{1 – 3\tan^2 \theta} \right) = \tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1} x$$

$$\implies \cfrac{dy}{dx} = \cfrac{3}{1 + x^2}$$
$y = \cos^{-1} \left( \cfrac{1 – x^2}{1 + x^2} \right)$

SOLUTION

Putting $x = \tan \theta$, we get:

$$y = \cos^{-1} \left( \cfrac{1 – \tan^2 \theta}{1 + \tan^2 \theta} \right) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1} x$$

$$\implies \cfrac{dy}{dx} = \cfrac{2}{1 + x^2}$$
$y = \sin^{-1} \left( \cfrac{1 – x^2}{1 + x^2} \right)$

SOLUTION

Putting $x = \tan \theta$, we get:

$$y = \sin^{-1} (\cos 2\theta) = \sin^{-1} \left\{ \sin\left( \cfrac{\pi}{2} – 2\theta \right) \right\} = \cfrac{\pi}{2} – 2\theta$$

$$y = \cfrac{\pi}{2} – 2\tan^{-1} x \implies \cfrac{dy}{dx} = -\cfrac{2}{1 + x^2}$$
$y = \cos^{-1} \left( \cfrac{2x}{1 + x^2} \right)$

SOLUTION

Putting $x = \tan \theta$, we get:

$$y = \cos^{-1} (\sin 2\theta) = \cos^{-1} \left\{ \cos\left( \cfrac{\pi}{2} – 2\theta \right) \right\} = \cfrac{\pi}{2} – 2\theta$$

$$y = \cfrac{\pi}{2} – 2\tan^{-1} x \implies \cfrac{dy}{dx} = -\cfrac{2}{1 + x^2}$$
$y = \sin^{-1} \left( 2x\sqrt{1 – x^2} \right)$

SOLUTION

Putting $x = \sin \theta$, we get:

$$y = \sin^{-1}[2\sin \theta \sqrt{1 – \sin^2 \theta}] = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x$$

$$\implies \cfrac{dy}{dx} = \cfrac{2}{\sqrt{1 – x^2}}$$
$y = \sec^{-1} \left( \cfrac{1}{2x^2 – 1} \right)$

SOLUTION

Putting $x = \cos \theta$, we get:

$$y = \sec^{-1} \left( \cfrac{1}{2\cos^2 \theta – 1} \right) = \sec^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1} x$$

$$\implies \cfrac{dy}{dx} = -\cfrac{2}{\sqrt{1 – x^2}}$$

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NCERT – Exercise 5.4

Differentiate the following w.r.t. $x$:

$\cfrac{e^x}{\sin x}$

SOLUTION

Let $y = \cfrac{e^x}{\sin x}$

$$\therefore \cfrac{dy}{dx} = \cfrac{d}{dx}\left( \cfrac{e^x}{\sin x} \right) = \cfrac{\sin x\cfrac{d}{dx}(e^x) – e^x\cfrac{d}{dx}(\sin x)}{\sin^2 x}$$

$$= \cfrac{\sin x \cdot e^x – e^x \cdot \cos x}{\sin^2 x} = \cfrac{e^x(\sin x – \cos x)}{\sin^2 x}, x \ne n\pi, n \in \mathbb{Z}$$
$e^{\sin^{-1} x}$

SOLUTION

Let $y = e^{\sin^{-1} x}$

$$\therefore \cfrac{dy}{dx} = \cfrac{d}{dx}(e^{\sin^{-1} x}) = e^{\sin^{-1} x}\cfrac{d}{dx}(\sin^{-1} x)$$

$$= e^{\sin^{-1} x} \cdot \cfrac{1}{\sqrt{1 – x^2}}, x \in (-1, 1)$$
$e^{x^3}$

SOLUTION

Let $y = e^{x^3}$

$$\therefore \cfrac{dy}{dx} = \cfrac{d}{dx}(e^{x^3}) = e^{x^3}\cfrac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3}$$
$\sin(\tan^{-1} e^{-x})$

SOLUTION

Let $y = \sin(\tan^{-1} e^{-x})$

$$\therefore \cfrac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \cfrac{d}{dx}(\tan^{-1} e^{-x})$$

$$= \cos(\tan^{-1} e^{-x}) \cdot \cfrac{1}{1 + (e^{-x})^2} \cdot \cfrac{d}{dx}(e^{-x})$$

$$= \cos(\tan^{-1} e^{-x}) \cdot \cfrac{-e^{-x}}{1 + e^{-2x}} = \cfrac{-e^{-x}\cos(\tan^{-1} e^{-x})}{1 + e^{-2x}}$$
$\log(\cos e^x)$

SOLUTION

Let $y = \log(\cos e^x)$

$$\therefore \cfrac{dy}{dx} = \cfrac{1}{\cos e^x} \cdot \cfrac{d}{dx}(\cos e^x) = \cfrac{1}{\cos e^x} \cdot (-\sin e^x) \cdot \cfrac{d}{dx}(e^x)$$

$$= -\tan e^x \cdot e^x = -e^x \tan e^x, \text{ where } e^x \ne (2n+1)\cfrac{\pi}{2}$$
$e^x + e^{x^2} + \dots + e^{x^5}$

SOLUTION

Let $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$

$$\therefore \cfrac{dy}{dx} = \cfrac{d}{dx}(e^x) + \cfrac{d}{dx}(e^{x^2}) + \cfrac{d}{dx}(e^{x^3}) + \cfrac{d}{dx}(e^{x^4}) + \cfrac{d}{dx}(e^{x^5})$$

$$= e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}$$
$\sqrt{e^{\sqrt{x}}}, x > 0$

SOLUTION

Let $y = \sqrt{e^{\sqrt{x}}}$

$$\therefore \cfrac{dy}{dx} = \cfrac{1}{2}(e^{\sqrt{x}})^{-1/2} \cdot \cfrac{d}{dx}(e^{\sqrt{x}})$$

$$= \cfrac{1}{2\sqrt{e^{\sqrt{x}}}} \cdot e^{\sqrt{x}} \cdot \cfrac{d}{dx}(\sqrt{x}) = \cfrac{e^{\sqrt{x}}}{2\sqrt{e^{\sqrt{x}}}} \cdot \cfrac{1}{2\sqrt{x}}$$

$$= \cfrac{e^{\sqrt{x}}}{4\sqrt{x \cdot e^{\sqrt{x}}}}, x > 0$$
$\log(\log x), x > 1$

SOLUTION

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

$$\therefore \cfrac{dy}{dx} = \cfrac{1}{\log x} \cdot \cfrac{d}{dx}(\log x) = \cfrac{1}{\log x} \cdot \cfrac{1}{x} = \cfrac{1}{x \log x}, x > 1$$
$\cfrac{\cos x}{\log x}, x > 0$

SOLUTION

Let $y = \cfrac{\cos x}{\log x}$

$$\therefore \cfrac{dy}{dx} = \cfrac{\log x(-\sin x) – \cos x(\cfrac{1}{x})}{(\log x)^2}$$

$$= \cfrac{-(x \sin x \log x + \cos x)}{x(\log x)^2}$$
$\cos(\log x + e^x), x > 0$

SOLUTION

Let $y = \cos(\log x + e^x)$

$$\therefore \cfrac{dy}{dx} = -\sin(\log x + e^x) \cdot \cfrac{d}{dx}(\log x + e^x)$$

$$= -\sin(\log x + e^x) \left[ \cfrac{1}{x} + e^x \right] = -\left( \cfrac{1}{x} + e^x \right) \sin(\log x + e^x)$$

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NCERT – Exercise 5.6 If x and y are connected parametrically by the equations given in questions 1 to 10, without eliminating the parameter, find $\cfrac{{dy}}{{dx}}$ . $x = 2a{t^2}, \quad y = a{t^4}$ SOLUTION Here, $x = 2a{t^2}$ ….(i) and $y = a{t^4}$ …(ii) Differentiating (i) & (ii) w.r.t. t, we get $\cfrac{{dx}}{{dt}} = 2a(2t) = 4at$ and $\cfrac{{dy}}{{dt}} = 4a{t^3}$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{4a{t^3}}}{{4at}} = {t^2}$ $x = a\cos \theta , \quad y = b\cos \theta $ SOLUTION Here, $x = a\cos \theta $ …(i) and $y = b\cos \theta $ …(ii) Differentiating (i) & (ii) w.r.t. $\theta$, we get $\cfrac{{dx}}{{d\theta }} = a( – \sin \theta ) = – a\sin \theta $ and $\cfrac{{dy}}{{d\theta }} = b( – \sin \theta ) = – b\sin \theta $ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{ – b\sin \theta }}{{ – a\sin \theta }} = \cfrac{b}{a}$ $x = \sin t, \quad y = \cos 2t$ SOLUTION Here, $x = \sin t$ …(i) and $y = \cos 2t$ …(ii) Differentiating (i) & (ii) w.r.t. t, we get $\cfrac{{dx}}{{dt}} = \cos t$ and $\cfrac{{dy}}{{dt}} = – \sin 2t \cdot 2 = – 2\sin 2t$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{ – 2\sin 2t}}{{\cos t}} = \cfrac{{ – 2 \cdot 2\sin t\cos t}}{{\cos t}} = – 4\sin t$ $x = 4t, \quad y = \cfrac{4}{t}$ SOLUTION Here, $x = 4t$ …(i) $y = \cfrac{4}{t}$ Differentiating (i) & (ii) w.r.t. t, we get $\cfrac{{dx}}{{dt}} = 4$ and $\cfrac{{dy}}{{dt}} = \cfrac{{ – 4}}{{{t^2}}}$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{ – 4}}{{{t^2}}} \times \cfrac{1}{4} = \cfrac{{ – 1}}{{{t^2}}}$ $x = \cos \theta – \cos 2\theta , \quad y = \sin \theta – \sin 2\theta $ SOLUTION Here, $x = \cos \theta – \cos 2\theta $ ..(i) and $y = \sin \theta – \sin 2\theta $ …(ii) Differentiating (i) & (ii) w.r.t. $\theta $ , we get $\cfrac{{dx}}{{d\theta }} = – \sin \theta – ( – \sin 2\theta ) \cdot 2 = 2\sin 2\theta – \sin \theta$ $\cfrac{{dy}}{{d\theta }} = \cos \theta – \cos 2\theta \cdot 2 = \cos \theta – 2\cos 2\theta $ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{\cos \theta – 2\cos 2\theta }}{{2\sin 2\theta – \sin \theta }}$ $x = a(\theta – \sin \theta ), \quad y = a(1 + \cos \theta )$ SOLUTION Here, $x = a(\theta – \sin \theta )$ and …(i) $y = a(1 + \cos \theta )$ …(ii) Differentiating (i) & (ii) w.r.t. $\theta $, we get $\cfrac{{dx}}{{d\theta }} = a[1 – \cos \theta ]$ and $\cfrac{{dy}}{{d\theta }} = a[ – \sin \theta ] = – a\sin \theta $ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{ – a\sin \theta }}{{a(1 – \cos \theta )}} = \cfrac{{ – \sin \theta }}{{1 – \cos \theta }}$ $ = \cfrac{{ – 2\sin \theta /2\cos \theta /2}}{{2{{\sin }^2}\theta /2}} = – \cot \cfrac{\theta }{2}$ $x = \cfrac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}, \quad y = \cfrac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$ SOLUTION Here $x = \cfrac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}$ …(i) and $y = \cfrac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$ …(ii) Differentiating (i) & (ii) w.r.t. t, we get $\cfrac{{dx}}{{dt}} = \cfrac{{\sqrt {\cos 2t} \cfrac{d}{{dt}}({{\sin }^3}t) – {{\sin }^3}t\cfrac{d}{{dt}}(\sqrt {\cos 2t} )}}{{\cos 2t}}$ $ = \cfrac{{\sqrt {\cos 2t} \cdot 3{{\sin }^2}t\cos t – {{\sin }^3}t \cdot \cfrac{1}{{2\sqrt {\cos 2t} }} \cdot ( – \sin 2t) \cdot 2}}{{\cos 2t}}$ $ = \cfrac{{\sqrt {\cos 2t} \cdot 3{{\sin }^2}t\cos t + \cfrac{{{{\sin }^3}t\sin 2t}}{{\sqrt {\cos 2t} }}}}{{\cos 2t}}$ $ = \cfrac{{3\cos 2t \cdot {{\sin }^2}t\cos t + {{\sin }^3}t\sin 2t}}{{{{(\cos 2t)}^{3/2}}}}$ $\cfrac{{dy}}{{dt}} = \cfrac{{\sqrt {\cos 2t} \cfrac{d}{{dt}}({{\cos }^3}t) – {{\cos }^3}t\cfrac{d}{{dt}}(\sqrt {\cos 2t} )}}{{\cos 2t}}$ $ = \cfrac{{\sqrt {\cos 2t} \cdot 3{{\cos }^2}t( – \sin t) – {{\cos }^3}t \cdot \cfrac{1}{{2\sqrt {\cos 2t} }} \cdot ( – \sin 2t) \cdot 2}}{{\cos 2t}}$ $ = \cfrac{{ – 3{{\cos }^2}t \cdot \sin t \cdot \sqrt {\cos 2t} + \cfrac{{{{\cos }^3}t\sin 2t}}{{\sqrt {\cos 2t} }}}}{{\cos 2t}}$ $ = \cfrac{{{{\cos }^3}t\sin 2t – 3{{\cos }^2}t \cdot \sin t \cdot \cos 2t}}{{{{(\cos 2t)}^{3/2}}}}$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{{{\cos }^3}t\sin 2t – 3{{\cos }^2}t \cdot \sin t\cos 2t}}{{3\cos 2t \cdot {{\sin }^2}t\cos t + {{\sin }^3}t\sin 2t}} = – \cot 3t$ $x = a\left( {\cos t + \log \tan \cfrac{t}{2}} \right), \quad y = a\sin t$ SOLUTION Here, $x = a\left( {\cos t + \log \tan \cfrac{t}{2}} \right)$ …(1) and $y = a\sin t$ …(2) Differentiating (1) & (2) w.r.t. t, we get $\cfrac{{dx}}{{dt}} = a\left[ { – \sin t + \cfrac{1}{{\tan \cfrac{t}{2}}}\cfrac{d}{{dt}}\left( {\tan \cfrac{t}{2}} \right)} \right]$ $ = a\left[ { – \sin t + \cfrac{1}{{\tan \cfrac{t}{2}}}{{\sec }^2}\cfrac{t}{2} \cdot \cfrac{1}{2}} \right] = \cfrac{{a{{\cos }^2}t}}{{\sin t}}$ $\cfrac{{dy}}{{dt}} = a\cos t$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{a\cos t\sin t}}{{a{{\cos }^2}t}} = \tan t$ $x = a\sec \theta , \quad y = b\tan \theta $ SOLUTION Here, $x = a\sec \theta ,$ …(1) and $y = b\tan \theta $ …(2) Differentiating (1) & (2) w.r.t. $\theta $, we get $\cfrac{{dx}}{{d\theta }} = a\sec \theta \tan \theta $ and $\cfrac{{dy}}{{d\theta }} = b{\sec ^2}\theta $ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{b{{\sec }^2}\theta }}{{a\sec \theta \tan \theta }} = \cfrac{b}{a}\cfrac{{\sec \theta }}{{\tan \theta }} = \cfrac{b}{a}\cos ec\theta $ $x = a(\cos \theta + \theta \sin \theta ), \quad y = a(\sin \theta – \theta \sec \theta )$ SOLUTION Here, $x = a(\cos \theta + \theta \sin \theta )$ …..(1) and $y = a(\sin \theta – \theta \sec \theta )$ .…(2) Differentiating (1) & (2) w.r.t. $\theta $, we get $\cfrac{{dx}}{{d\theta }} = a[ – \sin \theta + \theta \cdot \cos \theta + \sin \theta ] = a\theta \cos \theta $ $\cfrac{{dy}}{{d\theta }} = a[\cos \theta – (\theta ( – \sin \theta ) + \cos \theta )]$ $ = a[\cos \theta + \theta \sin \theta – \cos \theta ] = a\theta \sin \theta $ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{a\theta \sin \theta }}{{a\theta \cos \theta }} = \tan \theta $ If $x = \sqrt {{a^{{{\sin }^{ – 1}}t}}} , y = \sqrt {{a^{{{\cos }^{ – 1}}t}}} ,$ show that $\cfrac{{dy}}{{dx}} = – \cfrac{y}{x}.$ SOLUTION Given: $x = \sqrt {{a^{{{\sin }^{ – 1}}t}}} $ and $y = \sqrt {{a^{{{\cos }^{ – 1}}t}}} ,$ Differentiating x and y w.r.t. t, we get $\cfrac{{dx}}{{dt}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\sin }^{ – 1}}t}}} }} \cdot \cfrac{d}{{dt}}{a^{{{\sin }^{ – 1}}t}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\sin }^{ – 1}}t}}} }}{a^{{{\sin }^{ – 1}}t}} \cdot \log a \cdot \cfrac{d}{{dt}}{\sin ^{ – 1}}t$ $ = \cfrac{{\sqrt {{a^{{{\sin }^{ – 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{1}{{\sqrt {1 – {t^2}} }}$ $\cfrac{{dy}}{{dt}} = \cfrac{1}{2}\cfrac{1}{{\sqrt {{a^{{{\cos }^{ – 1}}t}}} }} \cdot \cfrac{d}{{dt}}{a^{{{\cos }^{ – 1}}t}} = \cfrac{1}{2} \cdot \cfrac{1}{{\sqrt {{a^{{{\cos }^{ – 1}}t}}} }} \cdot {a^{{{\cos }^{ – 1}}t}} \cdot \log a \cdot \cfrac{{ – 1}}{{\sqrt {1 – {t^2}} }}$ $ = \cfrac{{\sqrt {{a^{{{\cos }^{ – 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{{ – 1}}{{\sqrt {1 – {t^2}} }}$ $\therefore$ $\cfrac{{dy}}{{dx}} = \cfrac{{\cfrac{{dy}}{{dt}}}}{{\cfrac{{dx}}{{dt}}}} = \cfrac{{\cfrac{{\sqrt {{a^{{{\cos }^{ – 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{{ – 1}}{{\sqrt {1 – {t^2}} }}}}{{\cfrac{{\sqrt {{a^{{{\sin }^{ – 1}}t}}} }}{2} \cdot \log a \cdot \cfrac{1}{{\sqrt {1 – {t^2}} }}}}$ $ = – \cfrac{{\sqrt {{a^{{{\cos }^{ – 1}}t}}} }}{{\sqrt {{a^{{{\sin }^{ – 1}}t}}} }} = \cfrac{{ – y}}{x}$

NCERT – Exercise 5.7 Find the second order derivatives of the functions given in questions 1 to 10. $x^2 + 3x + 2$ SOLUTION Let $y = x^2 + 3x + 2$ $\Rightarrow \cfrac{dy}{dx} = 2x + 3$ $\therefore \cfrac{d^2y}{dx^2} = 2$ $x^{20}$ SOLUTION Let $y = x^{20}$ $\Rightarrow \cfrac{dy}{dx} = 20x^{19}$ $\therefore \cfrac{d^2y}{dx^2} = 20 \cdot 19x^{18} = 380x^{18}$ $x \cdot \cos x$ SOLUTION Let $y = x \cos x$ $\Rightarrow \cfrac{dy}{dx} = x \cdot (-\sin x) + \cos x = -x\sin x + \cos x$ $\Rightarrow \cfrac{d^2y}{dx^2} = -[x\cos x + \sin x] – \sin x = -(x\cos x + 2\sin x)$ $\log x$ SOLUTION Let $y = \log x$ $\Rightarrow \cfrac{dy}{dx} = \cfrac{1}{x}$ $\therefore \cfrac{d^2y}{dx^2} = -\cfrac{1}{x^2}$ $x^3 \log x$ SOLUTION Let $y = x^3 \log x$ $\Rightarrow \cfrac{dy}{dx} = x^3 \cdot \cfrac{1}{x} + \log x \cdot 3x^2 = x^2 + 3x^2 \log x$ $\Rightarrow \cfrac{d^2y}{dx^2} = 2x + 3\left[x^2 \cdot \cfrac{1}{x} + \log x \cdot 2x\right]$ $= 2x + 3[x + 2x\log x] = 2x + 3x + 6x\log x$ $= 5x + 6x\log x = x(5 + 6\log x)$ $e^x \sin 5x$ SOLUTION Let $y = e^x \sin 5x$ $\Rightarrow \cfrac{dy}{dx} = e^x \cdot \cos 5x \cdot 5 + \sin 5x \cdot e^x = e^x[5\cos 5x + \sin 5x]$ $\therefore \cfrac{d^2y}{dx^2} = e^x[5(-\sin 5x) \cdot 5 + \cos 5x \cdot 5] + [5\cos 5x + \sin 5x]e^x$ $= e^x[-25\sin 5x + 5\cos 5x + 5\cos 5x + \sin 5x]$ $= e^x[10\cos 5x – 24\sin 5x] = 2e^x[5\cos 5x – 12\sin 5x]$ $e^{6x} \cos 3x$ SOLUTION Let $y = e^{6x} \cos 3x$ $\Rightarrow \cfrac{dy}{dx} = e^{6x}(-\sin 3x) \cdot 3 + \cos 3x \cdot e^{6x} \cdot 6 = 6e^{6x}\cos 3x – 3e^{6x}\sin 3x$ $\Rightarrow \cfrac{d^2y}{dx^2} = 6[e^{6x}(-\sin 3x) \cdot 3 + \cos 3x \cdot e^{6x} \cdot 6] – 3[e^{6x}\cos 3x \cdot 3 + \sin 3x \cdot e^{6x} \cdot 6]$ $= -18e^{6x}\sin 3x + 36e^{6x}\cos 3x – 9e^{6x}\cos 3x – 18e^{6x}\sin 3x$ $= 27e^{6x}\cos 3x – 36e^{6x}\sin 3x = 9e^{6x}(3\cos 3x – 4\sin 3x)$ $\tan^{-1} x$ SOLUTION Let $y = \tan^{-1} x$ $\Rightarrow \cfrac{dy}{dx} = \cfrac{1}{1 + x^2}$ $\therefore \cfrac{d^2y}{dx^2} = \cfrac{(1+x^2) \cdot 0 – 1 \cdot (2x)}{(1+x^2)^2} = \cfrac{-2x}{(1+x^2)^2}$ $\log(\log x)$ SOLUTION Let $y = \log(\log x)$ $\Rightarrow \cfrac{dy}{dx} = \cfrac{1}{\log x} \cdot \cfrac{1}{x}$ $\therefore \cfrac{d^2y}{dx^2} = \cfrac{1}{\log x}\left(-\cfrac{1}{x^2}\right) + \cfrac{1}{x} \cdot \cfrac{d}{dx}\left(\cfrac{1}{\log x}\right)$ $= -\cfrac{1}{x^2\log x} + \cfrac{1}{x}\left[\cfrac{\log x \cdot 0 – 1 \cdot \frac{1}{x}}{(\log x)^2}\right]$ $= -\cfrac{1}{x^2\log x} + \cfrac{1}{x}\left[-\cfrac{1/x}{(\log x)^2}\right] = -\cfrac{1}{x^2\log x} – \cfrac{1}{x^2(\log x)^2}$ $= -\cfrac{1}{x^2\log x}\left[1 + \cfrac{1}{\log x}\right] = -\cfrac{(1 + \log x)}{(x\log x)^2}$ $\sin(\log x)$ SOLUTION Let $y = \sin(\log x)$ $\Rightarrow \cfrac{dy}{dx} = \cos(\log x) \cdot \cfrac{1}{x}$ $\therefore \cfrac{d^2y}{dx^2} = \cos(\log x) \cdot \left(-\cfrac{1}{x^2}\right) + \cfrac{1}{x} \cdot \{-\sin(\log x)\} \cdot \cfrac{1}{x}$ $= -\cfrac{\cos(\log x)}{x^2} – \cfrac{\sin(\log x)}{x^2} = -\cfrac{1}{x^2}[\cos(\log x) + \sin(\log x)]$ If $y = 5\cos x – 3\sin x$, then prove that $\cfrac{d^2y}{dx^2} + y = 0$. SOLUTION We have, $y = 5\cos x – 3\sin x$ $\Rightarrow \cfrac{dy}{dx} = 5(-\sin x) – 3(\cos x) = -5\sin x – 3\cos x$ $\Rightarrow \cfrac{d^2y}{dx^2} = -5\cos x – 3(-\sin x) = -5\cos x + 3\sin x$ $\Rightarrow \cfrac{d^2y}{dx^2} + y = -5\cos x + 3\sin x + 5\cos x – 3\sin x = 0$ Hence proved. If $y = \cos^{-1} x$, find $\cfrac{d^2y}{dx^2}$ in terms of y alone. SOLUTION $y = \cos^{-1} x \Rightarrow x = \cos y$ …(i) Differentiating (i) w.r.t. x, we get $1 = -\sin y \cfrac{dy}{dx}$ $\Rightarrow \cfrac{dy}{dx} = -\csc y$ …(ii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = \csc y \cot y \cfrac{dy}{dx} = -\csc^2 y \cot y$ If $y = 3\cos(\log x) + 4\sin(\log x)$, then show that $x^2 y_2 + x y_1 + y = 0$. SOLUTION We have, $y = 3\cos(\log x) + 4\sin(\log x)$ …(i) Differentiating (i) w.r.t. x, we get $\cfrac{dy}{dx} = 3[-\sin(\log x)]\cfrac{1}{x} + 4[\cos(\log x)]\cfrac{1}{x}$ …(ii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = 3\left[\left(-\sin(\log x)\right)\left(-\cfrac{1}{x^2}\right) + \cfrac{1}{x}(-\cos(\log x))\cfrac{1}{x}\right]$ $+ 4\left[\cos(\log x)\left(-\cfrac{1}{x^2}\right) + \cfrac{1}{x}\{-\sin(\log x)\}\cfrac{1}{x}\right]$ $= \cfrac{1}{x^2}[3\sin(\log x) – 3\cos(\log x) – 4\cos(\log x) – 4\sin(\log x)]$ $\Rightarrow x^2\cfrac{d^2y}{dx^2} = -x\cfrac{dy}{dx} – y$ $\Rightarrow x^2\cfrac{d^2y}{dx^2} + x\cfrac{dy}{dx} + y = 0$ $\Rightarrow x^2 y_2 + x y_1 + y = 0$ Hence proved. If $y = Ae^{mx} + Be^{nx}$, then show that $\cfrac{d^2y}{dx^2} – (m+n)\cfrac{dy}{dx} + mny = 0$. SOLUTION Let $y = Ae^{mx} + Be^{nx}$ …(i) Differentiating (i) w.r.t. x, we get $\cfrac{dy}{dx} = Ame^{mx} + Bne^{nx}$ …(ii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = Am^2e^{mx} + Bn^2e^{nx}$ …(iii) Now, $\cfrac{d^2y}{dx^2} – (m+n)\cfrac{dy}{dx} + mny$ $= (Am^2e^{mx} + Bn^2e^{nx}) – (m+n)(Ame^{mx} + Bne^{nx}) + mn(Ae^{mx} + Be^{nx})$ $= Am^2e^{mx} + Bn^2e^{nx} – Am^2e^{mx} – Bmne^{nx} – Amne^{mx} – Bn^2e^{nx} + Amne^{mx} + Bmne^{nx} = 0$ Hence proved. If $y = 500e^{7x} + 600e^{-7x}$, then show that $\cfrac{d^2y}{dx^2} = 49y$. SOLUTION Let $y = 500e^{7x} + 600e^{-7x}$ …(i) Differentiating (i) w.r.t. x, we get $\cfrac{dy}{dx} = 500 \cdot e^{7x} \cdot 7 + 600 \cdot e^{-7x} \cdot (-7) = 3500e^{7x} – 4200e^{-7x}$ …(ii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = 3500 \cdot 7 \cdot e^{7x} – 4200 \cdot (-7)e^{-7x} = 24500e^{7x} + 29400e^{-7x}$ $= 49(500e^{7x} + 600e^{-7x}) = 49y$ $\therefore \cfrac{d^2y}{dx^2} = 49y$ If $e^y(x+1) = 1$, then show that $\cfrac{d^2y}{dx^2} = \left(\cfrac{dy}{dx}\right)^2$. SOLUTION $xe^y + e^y = 1$ …(i) Differentiating (i) w.r.t. x, we get $xe^y \cfrac{dy}{dx} + e^y + e^y \cfrac{dy}{dx} = 0$ $\Rightarrow \cfrac{dy}{dx} = -\cfrac{1}{x+1}$ …(ii) From (ii), $\left(\cfrac{dy}{dx}\right)^2 = \left(-\cfrac{1}{x+1}\right)^2 = \cfrac{1}{(x+1)^2}$ …(iii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = \cfrac{1}{(x+1)^2}$ …(iv) From (iii) and (iv), we get $\cfrac{d^2y}{dx^2} = \left(\cfrac{dy}{dx}\right)^2$ Hence proved. If $y = (\tan^{-1} x)^2$, show that $(x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2$. SOLUTION Let $y = (\tan^{-1} x)^2$ …(i) Differentiating (i) w.r.t. x, we get $\cfrac{dy}{dx} = 2\tan^{-1} x \cdot \cfrac{1}{1+x^2}$ …(ii) Differentiating (ii) w.r.t. x, we get $\cfrac{d^2y}{dx^2} = 2\left[\tan^{-1} x \cdot \cfrac{(1+x^2)\cdot 0 – 2x}{(1+x^2)^2} + \cfrac{1}{(1+x^2)} \cdot \cfrac{1}{(1+x^2)}\right]$ $= 2\left[-\cfrac{2x\tan^{-1} x}{(1+x^2)^2} + \cfrac{1}{(1+x^2)^2}\right] = 2\left[\cfrac{1 – 2x\tan^{-1} x}{(1+x^2)^2}\right]$ Now, $(x^2+1)^2 \cfrac{d^2y}{dx^2} + 2x(x^2+1)\cfrac{dy}{dx}$ $= (x^2+1)^2 \cdot 2\left[\cfrac{1 – 2x\tan^{-1} x}{(1+x^2)^2}\right] + 2x(x^2+1) \cdot 2\tan^{-1} x \cdot \cfrac{1}{(1+x^2)}$ $= 2(1 – 2x\tan^{-1} x) + 4x\tan^{-1} x = 2 – 4x\tan^{-1} x + 4x\tan^{-1} x = 2$ Hence proved.

NCERT – Exercise 5.8 Verify Rolle’s theorem for the function $f(x) = x^2 + 2x – 8$, $x \in [-4, 2]$. SOLUTION Consider, $f(x) = x^2 + 2x – 8$ in $[-4, 2]$ which is a polynomial function and so (i) Function $f(x)$ is continuous in $[-4, 2]$ (ii) $f(x)$ is derivable in $(-4, 2)$ and (iii) $f(-4) = 0$ and $f(2) = 0$ $\Rightarrow$ $f(-4) = f(2)$. Hence, conditions of Rolle’s theorem are satisfied. Hence there exists, at least one $c \in (-4, 2)$ such that $f(-4) = f(2)$. Now, $f'(c) = 0 \Rightarrow 2c + 2 = 0 \Rightarrow c = -1 \in (-4, 2)$. Thus, $-1 \in (-4, 2)$ such that $f'(-1) = 0$. Hence, Rolle’s theorem is verified. Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples? (i) $f(x) = [x]$ for $x \in [5, 9]$ (ii) $f(x) = [x]$ for $x \in [-2, 2]$ (iii) $f(x) = x^2 – 1$ for $x \in [1, 2]$. SOLUTION (i) Being greatest integer function, the given function is not differentiable and continuous. Hence, Rolle’s theorem is not applicable. (ii) Being greatest integer function, the given function is not differentiable and continuous. Hence, Rolle’s theorem is not applicable. (iii) $f(x) = x^2 – 1$, $x \in [1, 2]$ is a polynomial function. So, (a) It is continuous in $[1, 2]$ (b) It is derivable in $(1, 2)$ and (c) $f(1) = (1)^2 – 1 = 1 – 1 = 0$ $f(2) = (2)^2 – 1 = 4 – 1 = 3$ As $f(1) \neq f(2)$, hence Rolle’s theorem is not applicable. If $f:[-5,5] \to \mathbb{R}$ is a differentiable function and if $f'(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5)$. SOLUTION For Rolle’s theorem, if (i) $f$ is continuous in $[a, b]$ (ii) $f$ is derivable in $(a, b)$ (iii) $f(a) = f(b)$ then $f'(c) = 0$, $c \in (a, b)$. We are given $f$ is continuous and derivable but $f'(c) \neq 0 \Rightarrow f(a) \neq f(b)$ i.e. $f(-5) \neq f(5)$. Hence proved. Verify Mean Value Theorem, if $f(x) = x^2 – 4x – 3$ in the interval $[a, b]$, where $a = 1$ and $b = 4$. SOLUTION We have, $f(x) = x^2 – 4x – 3$, $x \in [1, 4]$ which is a polynomial function. So (i) $f(x)$ is continuous in $[1, 4]$ (ii) $f(x)$ is derivable in $(1, 4)$ and, hence conditions of mean value theorem are satisfied, so there exists, at least one $c \in (1, 4)$ such that $f'(c) = \cfrac{f(4) – f(1)}{4 – 1} \Rightarrow 2c – 4 = \cfrac{(-3) – (-6)}{3} \Rightarrow 2c – 4 = 1$ $\Rightarrow 2c = 5 \Rightarrow c = \cfrac{5}{2} \in (1, 4)$ Hence, mean value theorem is verified. Verify Mean Value Theorem, if $f(x) = x^3 – 5x^2 – 3x$ in the interval $[a, b]$, where $a = 1$ and $b = 3$. SOLUTION We have, $f(x) = x^3 – 5x^2 – 3x$, $x \in (1, 3)$, which is a polynomial function, so (i) It is continuous in $[1, 3]$. (ii) It is derivable in $(1, 3)$. So, all conditions of mean value theorem are verified. Hence there exists, at least one $c$, such that $f'(c) = \cfrac{f(3) – f(1)}{3 – 1} \Rightarrow 3c^2 – 10c – 3 = \cfrac{-27 – (-7)}{2}$ $\Rightarrow 3c^2 – 10c – 3 = -10 \Rightarrow 3c^2 – 10c + 7 = 0$ $\Rightarrow c = \cfrac{10 \pm \sqrt{100 – 84}}{6} = \cfrac{10 \pm 4}{6}$ $\Rightarrow c = \cfrac{7}{3}, 1$ But, $c = \cfrac{7}{3} \in (1, 3)$ So, mean value theorem is verified. Examine the applicability of Mean Value Theorem for all three functions given in the above question 2. SOLUTION (i) $f(x) = [x]$ for $x \in [5, 9]$ $f(x) = [x]$ in the interval $[5, 9]$ is neither continuous, nor differentiable. $\therefore$ Mean value theorem is not applicable. (ii) $f(x) = [x]$ for $x \in [-2, 2]$ Again, $f(x) = [x]$ in the interval $[-2, 2]$ is neither continuous, nor differentiable. Hence, mean value theorem is not applicable. (iii) $f(x) = x^2 – 1$ for $x \in [1, 2]$ It is a polynomial. Therefore, it is continuous in the interval $[1, 2]$ and differentiable in the interval $(1, 2)$. So all conditions of mean value theorem are satisfied. Therefore, there exists at least one $c \in (1, 2)$ such that $f'(c) = \cfrac{f(2) – f(1)}{2 – 1} \Rightarrow 2c = \cfrac{3 – 0}{2 – 1} = \cfrac{3}{1}$ As $c = \cfrac{3}{2} \in (1, 2)$, so mean value theorem is verified.