# Miscellaneous Exercise Continuity and Differentiability Solution - NCERT Class 12

## Chapter 5 Ex.5.ME Question 1

Differentiate with respect to $$x$$ the function $${\left( {3{x^2} - 9x + 5} \right)^9}$$.

### Solution

Let $$y = {\left( {3{x^2} - 9x + 5} \right)^9}$$

Using chain rule, we get

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}{\left( {3{x^2} - 9x + 5} \right)^9}\\ &= 9{\left( {3{x^2} - 9x + 5} \right)^8}.\frac{d}{{dx}}\left( {3{x^2} - 9x + 5} \right)\\ &= 9{\left( {3{x^2} - 9x + 5} \right)^8}.\left( {6x - 9} \right)\\ &= 9{\left( {3{x^2} - 9x + 5} \right)^8}.3\left( {2x - 3} \right)\\ &= 27{\left( {3{x^2} - 9x + 5} \right)^8}\left( {2x - 3} \right)\end{align}

## Chapter 5 Ex.5.ME Question 2

Differentiate with respect to $$x$$ the function $${\sin ^3}x + {\cos ^6}x$$.

### Solution

Let $$y = {\sin ^3}x + {\cos ^6}x$$

Using chain rule, we get

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{{\sin }^3}\;x} \right) + \frac{d}{{dx}}\left( {{{\cos }^6\;}x} \right)\\& = 3{\;\sin ^2}x.\frac{d}{{dx}}\left( {\sin x} \right) + 6{\;\cos ^5}x.\frac{d}{{dx}}\left( {\cos x} \right)\\& = 3{\;\sin ^2}x.\cos x + 6{\;\cos ^5}x.\left( { - \sin x} \right)\\ &= 3\;\sin x\;\cos x\left( {\sin x - 2{{\;\cos }^4}x} \right)\end{align}

## Chapter 5 Ex.5.ME Question 3

Differentiate with respect to $$x$$ the function $${\left( {5x} \right)^{3\cos 2x}}$$.

### Solution

Let $$y = {\left( {5x} \right)^{3\cos 2x}}$$

Taking logarithm on both the sides, we obtain

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

Differentiating both sides with respect to $$x$$, we get

\begin{align}\frac{1}{y}\frac{{dy}}{{dx}} &= 3\left[ {\log 5x.\frac{d}{{dx}}\left( {\cos 2x} \right) + \cos 2x.\frac{d}{{dx}}\left( {\log 5x} \right)} \right]\\\frac{{dy}}{{dx}} &= 3y\left[ {\log 5x.\left( { - \sin 2x} \right).\frac{d}{{dx}}\left( {2x} \right) + \cos 2x.\frac{1}{{5x}}.\frac{d}{{dx}}\left( {5x} \right)} \right]\\ &= 3y\left[ { - 2\sin 2x.\log 5x + \frac{{\cos 2x}}{x}} \right]\\ &= y\left[ {\frac{{3\cos 2x}}{x} - 6\sin 2x\log 5x} \right]\\ &= {\left( {5x} \right)^{3\cos 2x}}\left[ {\frac{{3\cos 2x}}{x} - 6\sin 2x\log 5x} \right]\end{align}

## Chapter 5 Ex.5.ME Question 4

Differentiate with respect to $$x$$ the function $${\sin ^{ - 1}}\left( {x\sqrt x } \right),\;{\rm{ }}0 \le x \le 1$$.

### Solution

Let $$y = {\sin ^{ - 1}}\left( {x\sqrt x } \right)$$

Using chain rule, we get

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}{\sin ^{ - 1}}\left( {x\sqrt x } \right)\\[5pt]&= \frac{1}{{\sqrt {1 - {{\left( {x\sqrt x } \right)}^2}} }} \times \frac{d}{{dx}}\left( {x\sqrt x } \right)\\[5pt]&= \frac{1}{{\sqrt {1 - {x^3}} }}.\frac{d}{{dx}}\left( {{x^{\frac{3}{2}}}} \right)\\[5pt] &= \frac{1}{{\sqrt {1 - {x^3}} }}.\frac{3}{2}.{x^{\frac{1}{2}}}\\[5pt] &= \frac{{3\sqrt x }}{{2\sqrt {1 - {x^3}} }}\\[5pt]& = \frac{3}{2}\sqrt {\frac{x}{{1 - {x^3}}}} \end{align}

## Chapter 5 Ex.5.ME Question 5

Differentiate with respect to $$x$$ the function $$\frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{\sqrt {2x + 7} }}, \; - 2 < x < 2$$.

### Solution

Let $$y = \frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{\sqrt {2x + 7} }}$$

Using quotient rule, we get

\begin{align}\frac{{dy}}{{dx}}& = \frac{{\sqrt {2x + 7} .\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}\frac{x}{2}} \right) - \left( {{{\cos }^{ - 1}}\frac{x}{2}} \right).\frac{d}{{dx}}\left( {\sqrt {2x + 7} } \right)}}{{{{\left( {\sqrt {2x + 7} } \right)}^2}}}\\[5pt]& = \frac{{\sqrt {2x + 7} \left[ {\frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} }}.\frac{d}{{dx}}\left( {\frac{x}{2}} \right)} \right] - \left( {{{\cos }^{ - 1}}\frac{x}{2}} \right).\frac{1}{{2\sqrt {2x + 7} }}.\frac{d}{{dx}}\left( {2x + 7} \right)}}{{2x + 7}}\\[5pt]& = \frac{{\sqrt {2x + 7} .\frac{{ - 1}}{{\sqrt {4 - {x^2}} }} - \left( {{{\cos }^{ - 1}}\frac{x}{2}} \right).\frac{2}{{2\sqrt {2x + 7} }}}}{{2x + 7}}\\[5pt] &= \frac{{ - \sqrt {2x + 7} }}{{\left( {\sqrt {4 - {x^2}} } \right).\left( {2x + 7} \right)}} - \frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{\left( {\sqrt {2x + 7} } \right)\left( {2x + 7} \right)}}\\[5pt] &= - \left[ {\frac{1}{{\sqrt {4 - {x^2}} \sqrt {2x + 7} }} + \frac{{{{\cos }^{ - 1}}\frac{x}{2}}}{{{{\left( {2x + 7} \right)}^{^{\frac{3}{2}}}}}}} \right]\end{align}

## Chapter 5 Ex.5.ME Question 6

Differentiate with respect to $$x$$ the function \begin{align}{\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right],\;0 < x < \frac{\pi }{2}\end{align}..

### Solution

Let \begin{align}y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Then,

\begin{align}\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} &= \frac{{{{\left( {\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \right)}^2}}}{{\left( {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} } \right)\left( {\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \right)}}\\[10pt] &= \frac{{\left( {1 + \sin x} \right) + \left( {1 - \sin x} \right) + 2\sqrt {\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)} }}{{\left( {1 + \sin x} \right) - \left( {1 - \sin x} \right)}}\\[10pt] &= \frac{{2 + 2\sqrt {1 - {{\sin }^2}x} }}{{2\sin x}} = \frac{{1 + \cos x}}{{\sin x}}\\[10pt]&= \frac{{1 + 2{{\cos }^2}\frac{x}{2} - 1}}{{2\sin \frac{x}{2}\cos \frac{x}{2}}} = \frac{{2{{\cos }^2}\frac{x}{2}}}{{2\sin \frac{x}{2}\cos \frac{x}{2}}}\\[10pt] &= \cot \frac{x}{2}\end{align}

Therefore, equation ($$1$$) becomes,

\begin{align}y &= {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)\\ &\Rightarrow \;y = \frac{x}{2}\end{align}

Thus,

\begin{align} \Rightarrow \;\frac{{dy}}{{dx}} &= \frac{1}{2}\frac{d}{{dx}}\left( x \right)\\&= \frac{1}{2}\end{align}

## Chapter 5 Ex.5.ME Question 7

Differentiate with respect to $$x$$ the function $${\left( {\log x} \right)^{\log x}},\;x > 1$$.

### Solution

Let $$y = {\left( {\log x} \right)^{\log x}}$$

Taking logarithm on both the sides, we obtain

$$\log y = \log x.\log \left( {\log x} \right)$$

Differentiating both sides with respect to $$x$$, we obtain

\begin{align} &\Rightarrow \;\frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\log x.\log \left( {\log x} \right)} \right]\\[5pt]&\Rightarrow \;\frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {\log x} \right).\frac{d}{{dx}}\left( {\log x} \right) + \log x.\frac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right]\\[5pt]& \Rightarrow \;\frac{{dy}}{{dx}} = y\left[ {\log \left( {\log x} \right).\frac{1}{x} + \log x.\frac{1}{{\log x}}.\frac{d}{{dx}}\left( {\log x} \right)} \right]\\[5pt]& \Rightarrow \;\frac{{dy}}{{dx}} = y\left[ {\frac{1}{x}.\log \left( {\log x} \right) + \frac{1}{x}} \right]\\[5pt]&\Rightarrow \;\frac{{dy}}{{dx}} = {\left( {\log x} \right)^{\log x}}\left[ {\frac{1}{x} + \frac{{\log \left( {\log x} \right)}}{x}} \right]\end{align}

## Chapter 5 Ex.5.ME Question 8

Differentiate with respect to $$x$$ the function $$\cos \left( {a\cos x + b\sin x} \right)$$, for some constant $$a$$ and $$b$$.

### Solution

Let $$y = \cos \left( {a\cos x + b\sin x} \right)$$

Using chain rule, we get

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\cos \left( {a\cos x + b\sin x} \right)\\& = - \sin \left( {a\cos x + b\sin x} \right).\frac{d}{{dx}}\left( {a\cos x + b\sin x} \right)\\ &= - \sin \left( {a\cos x + b\sin x} \right).\left[ {a\left( { - \sin x} \right) + b\cos x} \right]\\ &= \left( {a\sin x - b\cos x} \right).\sin \left( {a\cos x + b\sin x} \right)\end{align}

## Chapter 5 Ex.5.ME Question 9

Differentiate with respect to $$x$$ the function $${\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}},\frac{\pi }{4} < x < \frac{{3\pi }}{4}$$

### Solution

Let $$y = {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}$$

Taking log on both the sides, we obtain

\begin{align}\log y &= \log \left[ {{{\left( {\sin x - \cos x} \right)}^{\left( {\sin x - \cos x} \right)}}} \right]\\ &= \left( {\sin x - \cos x} \right)\log \left( {\sin x - \cos x} \right)\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}\frac{1}{y}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\left( {\sin x - \cos x} \right)\log \left( {\sin x - \cos x} \right)} \right]\\[5pt] &\Rightarrow \;\frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {\sin x - \cos x} \right).\frac{d}{{dx}}\left( {\sin x - \cos x} \right) + \left( {\sin x - \cos x} \right).\frac{d}{{dx}}\log \left( {\sin x - \cos x} \right)\\[5pt] &\Rightarrow \;\frac{1}{y}\frac{{dy}}{{dx}} = \log \left( {\sin x - \cos x} \right).\left( {\cos x + \sin x} \right) + \left( {\sin x - \cos x} \right).\frac{1}{{\left( {\sin x - \cos x} \right)}}.\frac{d}{{dx}}\left( {\sin x - \cos x} \right)\\[5pt] &\Rightarrow \;\frac{{dy}}{{dx}} = {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}\left[ {\left( {\cos x + \sin x} \right).\log \left( {\sin x - \cos x} \right) + \left( {\cos x + \sin x} \right)} \right]\\[5pt] &\Rightarrow \;\frac{{dy}}{{dx}} = {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}\left( {\cos x + \sin x} \right)\left[ {1 + \log \left( {\sin x - \cos x} \right)} \right]\end{align}

## Chapter 5 Ex.5.ME Question 10

Differentiate with respect to $$x$$ the function $${x^x} + {x^a} + {a^x} + {a^a}$$, for some fixed $$a > 0$$ and $$x > 0$$.

### Solution

Let $$y = {x^x} + {x^a} + {a^x} + {a^a}$$

Also, let $${x^x} = u,{\;\rm{ }}{x^a} = v,{\;\rm{ }}{a^x} = w$$ and $${a^a} = s$$

Therefore,

\begin{align}& \Rightarrow \;y = u + v + w + s\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}} + \frac{{dw}}{{dx}} + \frac{{ds}}{{dx}} \quad \ldots \left( 1 \right)\end{align}

Now, $$u = {x^x}$$

Taking logarithm on both the sides, we obtain

\begin{align} &\Rightarrow \;\log u = \log {x^x}\\& \Rightarrow \;\log u = x\log x\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}\frac{1}{u}\frac{{du}}{{dx}} &= \log x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log x} \right)\\\frac{{du}}{{dx}} &= u\left[ {\log x.1 + x.\frac{1}{x}} \right]\\& = {x^x}\left[ {\log x + 1} \right] = {x^x}\left( {1 + \log x} \right) \quad \ldots \left( 2 \right)\end{align}

Now, $$v = {x^a}$$

Hence,

\begin{align}\frac{{dv}}{{dx}} &= \frac{d}{{dx}}\left( {{x^a}} \right)\\ &= a{x^{a - 1}}\qquad\ldots \left( 3 \right)\end{align}

Now, $$w = {a^x}$$

Taking logarithm on both the sides, we obtain

\begin{align}& \Rightarrow \;\log w = \log {a^x}\\ &\Rightarrow \;\log w = x\log a\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}\frac{1}{w}\frac{{dw}}{{dx}} &= \log a.\frac{d}{{dx}}\left( x \right)\\\frac{{dw}}{{dx}} &= w\log a\\ &= {a^x}\log a \qquad \ldots \left( 4 \right)\end{align}

Now, $$s = {a^a}$$

Since $$a$$ is constant, $${a^a}$$ is also a constant.

Hence,

\begin{align}\frac{{ds}}{{dx}} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\end{align}

From (1), (2), (3), (4) and (5), we obtain

\begin{align}\frac{{dy}}{{dx}} &= {x^x}\left( {1 + \log x} \right) + a{x^{a - 1}} + {a^x}\log a + 0\\& = {x^x}\left( {1 + \log x} \right) + a{x^{a - 1}} + {a^x}\log a\end{align}

## Chapter 5 Ex.5.ME Question 11

Differentiate with respect to $$x$$ the function $${x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$$, for $$x > 3$$.

### Solution

Let $$y = {x^{{x^2} - 3}} + {\left( {x - 3} \right)^{{x^2}}}$$

Also, let $$u = {x^{{x^2} - 3}}$$ and $$v = {\left( {x - 3} \right)^{{x^2}}}$$

Therefore,

\begin{align}y &= u + v\\ \Rightarrow \;\frac{{dy}}{{dx}} &= \frac{{du}}{{dx}} + \frac{{dv}}{{dx}} \qquad \ldots \left( 1 \right)\end{align}

Now, $$u = {x^{{x^2} - 3}}$$

Taking logarithm on both the sides, we obtain

\begin{align}\log u& = \log \left( {{x^{{x^2} - 3}}} \right)\\ &= \left( {{x^2} - 3} \right)\log x\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}\frac{1}{u}\frac{{du}}{{dx}} &= \log x.\frac{d}{{dx}}\left( {{x^2} - 3} \right) + \left( {{x^2} - 3} \right).\frac{d}{{dx}}\left( {\log x} \right)\\ \Rightarrow \;\;\;\frac{1}{u}\frac{{du}}{{dx}} &= \log x.2x + \left( {{x^2} - 3} \right).\frac{1}{x}\\ \Rightarrow \;\;\;\;\;\;\;\frac{{du}}{{dx}} &= {x^{{x^2} - 3}}\left[ {\frac{{{x^2} - 3}}{x} + 2x\log x} \right] \qquad \ldots \left( 2 \right)\end{align}

Now, $$v = {\left( {x - 3} \right)^{{x^2}}}$$

Taking logarithm on both the sides, we obtain

\begin{align}\log v &= \log {\left( {x - 3} \right)^{{x^2}}}\\& = {x^2}\log \left( {x - 3} \right)\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align} \frac{1}{v}\frac{{dv}}{{dx}} &= \log \left( {x - 3} \right).\frac{d}{{dx}}\left( {{x^2}} \right) + \left( {{x^2}} \right).\frac{d}{{dx}}\left[ {\log \left( {x - 3} \right)} \right]\\ \Rightarrow \;\frac{1}{v}\frac{{dv}}{{dx}} &= \log \left( {x - 3} \right).2x + {x^2}.\frac{1}{{x - 3}}.\frac{d}{{dx}}\left( {x - 3} \right)\\ \Rightarrow \;\;\;\;\frac{{dv}}{{dx}} &= v\left[ {2x\log \left( {x - 3} \right) + \frac{{{x^2}}}{{x - 3}}.1} \right]\\ \Rightarrow \;\;\;\;\frac{{dv}}{{dx}} &= {\left( {x - 3} \right)^{{x^2}}}\left[ {\frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right] \qquad \ldots \left( 3 \right)\end{align}

From (1), (2), and (3), we obtain

\begin{align}\frac{{dy}}{{dx}} = {x^{{x^2} - 3}}\left[ {\frac{{{x^2} - 3}}{x} + 2x\log x} \right] + {\left( {x - 3} \right)^{{x^2}}}\left[ {\frac{{{x^2}}}{{x - 3}} + 2x\log \left( {x - 3} \right)} \right]\end{align}

## Chapter 5 Ex.5.ME Question 12

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

### Solution

The given function is $$y = 12\left( {1 - \cos t} \right),\;x = 10\left( {t - \sin t} \right)$$

Hence,

\begin{align}\frac{{dx}}{{dt}} &= \frac{d}{{dt}}\left[ {10\left( {t - \sin t} \right)} \right]\\[5pt]& = 10.\frac{d}{{dt}}\left( {t - \sin t} \right)\\[5pt]&= 10\left( {1 - \cos t} \right)\\\frac{{dy}}{{dt}} &= \frac{d}{{dt}}\left[ {12\left( {1 - \cos t} \right)} \right]\\[5pt]&= 12.\frac{d}{{dt}}\left( {1 - \cos t} \right)\\[5pt]&= 12.\left[ {0 - \left( { - \sin t} \right)} \right]\\[5pt]&= 12\sin t\end{align}

Therefore,

\begin{align}\frac{{dy}}{{dx}} &= \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{12\sin t}}{{10\left( {1 - \cos t} \right)}}\\[5pt]&= \frac{{12.2\sin \frac{t}{2}.\cos \frac{t}{2}}}{{10.2{{\sin }^2}\frac{t}{2}}}\\& = \frac{6}{5}\cot \frac{t}{2}\end{align}

## Chapter 5 Ex.5.ME Question 13

Find $$\frac{{dy}}{{dx}}$$, if $$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}} ,\;- 1 \le x \le 1$$.

### Solution

The given function is $$y = {\sin ^{ - 1}}x + {\sin ^{ - 1}}\sqrt {1 - {x^2}}$$

Hence,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x + {{\sin }^{ - 1}}\sqrt {1 - {x^2}} } \right]\\[5pt]& = \frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) + \frac{d}{{dx}}\left( {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } \right)\end{align}

\begin{align}\frac{{dy}}{{dx}} &= \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} }}.\frac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right)\\[5pt]&= \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{x}.\frac{1}{{2\sqrt {1 - {x^2}} }}.\frac{d}{{dx}}\left( {1 - {x^2}} \right)\\[5pt]& = \frac{1}{{\sqrt {1 - {x^2}} }} + \frac{1}{{2x\sqrt {1 - {x^2}} }}\left( { - 2x} \right)\\[5pt]& = \frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {x^2}} }}\\[5pt]&= 0\end{align}

## Chapter 5 Ex.5.ME Question 14

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

### Solution

The given function is $$x\sqrt {1 + y}\;+\; y\sqrt {1 + x} = 0$$

$$\Rightarrow \;x\sqrt {1 + y} = - y\sqrt {1 + x}$$

Squaring both sides, we obtain

\begin{align} {x^2}\left( {1 + y} \right) &= {y^2}\left( {1 + x} \right)\\[5pt] \Rightarrow \;{x^2} + {x^2}y &= {y^2} + x{y^2}\\[5pt] \Rightarrow \;{x^2} - {y^2} &= x{y^2} - {x^2}y\\[5pt] \Rightarrow \;{x^2} - {y^2} &= xy\left( {y - x} \right)\\[5pt] \Rightarrow \;\left( {x + y} \right)\left( {x - y} \right) &= xy\left( {y - x} \right)\\[5pt] \Rightarrow \;x + y &= - xy\\[5pt] \Rightarrow \;\left( {1 + x} \right)y &= - x\\[5pt] \Rightarrow \;y &= \frac{{ - x}}{{\left( {1 + x} \right)}}\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}\frac{{dy}}{{dx}} &= - \left[ {\frac{{\left( {1 + x} \right)\frac{d}{{dx}}\left( x \right) - \left( x \right).\frac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}} \right]\\[5pt]&= - \frac{{\left( {1 + x} \right) - x}}{{{{\left( {1 + x} \right)}^2}}}\\[5pt]& = - \frac{1}{{{{\left( {1 + x} \right)}^2}}}\end{align}

Hence proved.

## Chapter 5 Ex.5.ME Question 15

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

### Solution

The given function is $${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$$

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}&\frac{d}{{dx}}\left[ {{{\left( {x - a} \right)}^2}} \right] + \frac{d}{{dx}}\left[ {{{\left( {y - b} \right)}^2}} \right] = \frac{d}{{dx}}\left( {{c^2}} \right)\\ &\Rightarrow \;2\left( {x - a} \right).\frac{d}{{dx}}\left( {x - a} \right) + 2\left( {y - b} \right).\frac{d}{{dx}}\left( {y - b} \right) = 0\\& \Rightarrow \;2\left( {x - a} \right).1 + 2\left( {y - b} \right).\frac{{dy}}{{dx}} = 0\\ &\Rightarrow \;\frac{{dy}}{{dx}} = \frac{{ - \left( {x - a} \right)}}{{y - b}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right]\\ &= - \left[ {\frac{{\left( {y - b} \right).\frac{d}{{dx}}\left( {x - a} \right) - \left( {x - a} \right).\frac{d}{{dx}}\left( {y - b} \right)}}{{{{\left( {y - b} \right)}^2}}}} \right]\\ &= - \left[ {\frac{{\left( {y - b} \right) - \left( {x - a} \right).\frac{{dy}}{{dx}}}}{{{{\left( {y - b} \right)}^2}}}} \right]\\ &= - \left[ {\frac{{\left( {y - b} \right) - \left( {x - a} \right).\left\{ {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right\}}}{{{{\left( {y - b} \right)}^2}}}} \right]\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right)} \right]\\ &= - \left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^3}}}} \right]\end{align}

Hence,

\begin{align}\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}} &= \frac{{{{\left[ {1 + \frac{{{{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^2}}}} \right]}^{\frac{3}{2}}}}}{{ - \left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^3}}}} \right]}}\\ &= \frac{{{{\left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^2}}}} \right]}^{\frac{3}{2}}}}}{{ - \left[ {\frac{{{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}}}{{{{\left( {y - b} \right)}^3}}}} \right]}}\\ &= \frac{{{{\left[ {\frac{{{c^2}}}{{{{\left( {y - b} \right)}^2}}}} \right]}^{\frac{3}{2}}}}}{{ - \frac{{{c^2}}}{{{{\left( {y - b} \right)}^3}}}}}\\ &= \frac{{\frac{{{c^3}}}{{{{\left( {y - b} \right)}^3}}}}}{{ - \frac{{{c^2}}}{{{{\left( {y - b} \right)}^3}}}}}\\ &= - c\end{align}

$$- c$$ is a constant and is independent of $$a$$ and $$b$$.

Hence proved.

## Chapter 5 Ex.5.ME Question 16

If $$\cos y = x\cos \left( {a + y} \right)$$ with $$\cos a \ne \pm 1$$, prove that $$\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}$$.

### Solution

The given function is $$\cos y = x\cos \left( {a + y} \right)$$

Therefore,

\begin{align}& \Rightarrow \;\frac{d}{{dx}}\left[ {\cos y} \right] = \frac{d}{{dx}}\left[ {x\cos \left( {a + y} \right)} \right]\\& \Rightarrow \;- \sin y\frac{{dy}}{{dx}} = \cos \left( {a + y} \right).\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left[ {\cos \left( {a + y} \right)} \right]\\ &\Rightarrow \;- \sin y\frac{{dy}}{{dx}} = \cos \left( {a + y} \right) + x.\left[ { - \sin \left( {a + y} \right)} \right]\frac{{dy}}{{dx}}\\ &\Rightarrow \;\left[ {x\sin \left( {a + y} \right) - \sin y} \right]\frac{{dy}}{{dx}} = \cos \left( {a + y} \right) \qquad \ldots \left( 1 \right)\end{align}

Since, $$\cos y = x\cos \left( {a + y} \right)$$$$\Rightarrow \;x = \frac{{\cos y}}{{\cos \left( {a + y} \right)}}$$

Then, equation (1) becomes,

\begin{align}&\left[ {\frac{{\cos y}}{{\cos \left( {a + y} \right)}}.\sin \left( {a + y} \right) - \sin y} \right]\frac{{dy}}{{dx}} = \cos \left( {a + y} \right)\\& \Rightarrow \;\left[ {\cos y.\sin \left( {a + y} \right) - \sin y.\cos \left( {a + y} \right)} \right].\frac{{dy}}{{dx}} = {\cos ^2}\left( {a + y} \right)\\ &\Rightarrow \;\sin \left( {a + y - y} \right)\frac{{dy}}{{dx}} = {\cos ^2}\left( {a + y} \right)\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}\left( {a + y} \right)}}{{\sin a}}\end{align}

Hence proved.

## Chapter 5 Ex.5.ME Question 17

If $$x = a\left( {\cos t + t\sin t} \right)$$ and $$y = a\left( {\sin t - t\cos t} \right)$$, find $$\frac{{{d^2}y}}{{d{x^2}}}$$.

### Solution

The given function is $$x = a\left( {\cos t + t\sin t} \right)$$ and $$y = a\left( {\sin t - t\cos t} \right)$$

Therefore,

\begin{align}\frac{{dx}}{{dt}} &= a.\frac{d}{{dt}}\left( {\cos t + t\sin t} \right)\\ &= a\left[ { - \sin t + \sin t.\frac{d}{{dx}}\left( t \right) + t.\frac{d}{{dt}}\left( {\sin t} \right)} \right]\\ &= a\left[ { - \sin t + \sin t + t\cos t} \right]\\ &= at\cos t\\\\\frac{{dy}}{{dt}} &= a.\frac{d}{{dt}}\left( {\sin t - t\cos t} \right)\\& = a\left[ {\cos t - \left\{ {\cos t.\frac{d}{{dt}}\left( t \right) + t.\frac{d}{{dt}}\left( {\cos t} \right)} \right\}} \right]\\& = a\left[ {\cos t - \left\{ {\cos t - t\sin t} \right\}} \right]\\ &= at\sin t\\\\\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{at\sin t}}{{at\cos t}} = \tan t\\\frac{{{d^2}y}}{{d{x^2}}}& = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( {\tan t} \right) = {\sec ^2}t.\frac{{dt}}{{dx}}\\ &= {\sec ^2}t.\frac{1}{{at\cos t}} \qquad \left[ {\frac{{dx}}{{dt}} = at\cos t \Rightarrow \;\frac{{dt}}{{dx}} = \frac{1}{{at\cos t}}} \right]\\ &= \frac{{{{\sec }^3}t}}{{at}},0 < t < \frac{\pi }{2}\end{align}

## Chapter 5 Ex.5.ME Question 18

If $$f\left( x \right) = {\left| x \right|^3}$$, show that $$f''\left( x \right)$$ exists for all real $$x$$, and find it.

### Solution

It is known that $$\left| x \right| = \left\{ \begin{array}{l}x,{\rm{ if }}x \ge 0\\ - x,{\rm{ if }}x < 0\end{array} \right.$$

Therefore, when $$x \ge 0,\,\,f\left( x \right) = {\left| x \right|^3} = {x^3}$$

In this case, $$f'\left( x \right) = 3{x^2}$$ and hence, $$f''\left( x \right) = 6x$$

When $$x < 0,\,\,f\left( x \right) = {\left| x \right|^3} = {\left( { - x} \right)^3} = - {x^3}$$

In this case, $$f'\left( x \right) = - 3{x^2}$$ and hence, $$f''\left( x \right) = - 6x$$

Thus, for $$f\left( x \right) = {\left| x \right|^3}$$, $$f''\left( x \right)$$ exists for all real x and is given by,

$$f''\left( x \right) = \left\{ \begin{array}{l}6x,{\rm{ if }}x \ge 0\\ - 6x,{\rm{ if }}x < 0\end{array} \right.$$

## Chapter 5 Ex.5.ME Question 19

Using mathematical induction prove that $$\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$$ for all positive integers $$n$$.

### Solution

To prove: $$P\left( n \right):\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$$ for all positive integers $$n$$.

For $$n = 1$$,

$$P\left( 1 \right):\frac{d}{{dx}}\left( x \right) = 1 = 1 \cdot {x^{1 - 1}}$$

Therefore, $$P\left( n \right)$$ is true for$$n = 1$$.

Let $$P\left( k \right)$$ is true for some positive integer $$k$$.

That is, $$P\left( k \right):\frac{d}{{dx}}\left( {{x^k}} \right) = k{x^{k - 1}}$$

It has to be proved that $$P\left( {k + 1} \right)$$ is also true.

Consider

\begin{align}\frac{d}{{dx}}\left( {{x^{k + 1}}} \right)& = \frac{d}{{dx}}\left( {x \cdot {x^k}} \right)\\& = {x^k}.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {{x^k}} \right) \quad \left[ {{\text{By applying product rule}}} \right]\\ &= {x^k}.1 + x \cdot k \cdot {x^{k - 1}}\end{align}

\begin{align}\frac{d}{{dx}}\left( {{x^{k + 1}}} \right) &= {x^k} + k{x^k}\\ &= \left( {k + 1} \right) \cdot {x^k}\\ &= \left( {k + 1} \right) \cdot {x^{\left( {k + 1} \right) - 1}}\end{align}

Thus, $$P\left( {k + 1} \right)$$ is true whenever $$P\left( k \right)$$ is true.

Therefore, by the principle of mathematical induction, the statement $$P\left( n \right)$$ is true for every positive integer $$n$$.

Hence, proved.

## Chapter 5 Ex.5.ME Question 20

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

### Solution

Given, $$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$$

Differentiating both sides with respect to $$x$$, we obtain

\begin{align}&\frac{d}{{dx}}\left[ {\sin \left( {A + B} \right)} \right] = \frac{d}{{dx}}\left( {\sin A\cos B} \right) + \frac{d}{{dx}}\left( {\cos A\sin B} \right)\\& \Rightarrow \;\cos \left( {A + B} \right).\frac{d}{{dx}}\left( {A + B} \right) = \cos B.\frac{d}{{dx}}\left( {\sin A} \right) \\& \quad+ \sin A.\frac{d}{{dx}}\left( {\cos B} \right) + \sin B.\frac{d}{{dx}}\left( {\cos A} \right) + \cos A.\frac{d}{{dx}}\left( {\sin B} \right)\\ &\Rightarrow \;\cos \left( {A + B} \right).\frac{d}{{dx}}\left( {A + B} \right) = \cos B.\cos A\frac{{dA}}{{dx}} + \sin A\left( { - \sin B} \right)\frac{{dB}}{{dx}} \\& \quad+ \sin B\left( { - \sin A} \right).\frac{{dA}}{{dx}} + \cos A\cos B\frac{{dB}}{{dx}}\\ &\Rightarrow \;\cos \left( {A + B} \right).\left[ {\frac{{dA}}{{dx}} + \frac{{dB}}{{dx}}} \right] = \left( {\cos A\cos B - \sin A\sin B} \right).\left[ {\frac{{dA}}{{dx}} + \frac{{dB}}{{dx}}} \right]\\& \Rightarrow \;\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\end{align}

## Chapter 5 Ex.5.ME Question 21

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

### Solution

Consider, $$y = \left\{ \begin{array}{l}\left| x \right|\;\;\;\;\; - \infty < x \le 1\\2 - x\;\;\;1 \le x \le \infty \end{array} \right.$$

It can be seen from the above graph that the given function is continuous everywhere but not differentiable at exactly two points which are 0 and 1.

## Chapter 5 Ex.5.ME Question 22

If $$y = \left| {\begin{array}{*{20}{c}}{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|$$, prove that $$\frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|$$

### Solution

Given, $$y = \left| {\begin{array}{*{20}{c}}{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|$$

$$\Rightarrow \;y = \left( {mc - nb} \right)f\left( x \right) - \left( {lc - na} \right)g\left( x \right) + \left( {lb - ma} \right)h\left( x \right)$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\left( {mc - nb} \right)f\left( x \right)} \right] - \frac{d}{{dx}}\left[ {\left( {lc - na} \right)g\left( x \right)} \right] + \frac{d}{{dx}}\left[ {\left( {lb - ma} \right)h\left( x \right)} \right]\\[5pt] &= \left( {mc - nb} \right)f'\left( x \right) - \left( {lc - na} \right)g'\left( x \right) + \left( {lb - ma} \right)h'\left( x \right)\\[5pt] &= \left| {\begin{array}{*{20}{c}}{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|\end{align}

Thus, $$\frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}}{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\l&m&n\\a&b&c\end{array}} \right|$$ proved.

## Chapter 5 Ex.5.ME Question 23

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

### Solution

The given function is $$y = {e^{a{{\cos }^{ - 1}}x}}$$

Taking logarithm on both the sides, we obtain

\begin{align} &\Rightarrow \;\log y = a{\cos ^{ - 1}}x\log e\\ &\Rightarrow \;\log y = a{\cos ^{ - 1}}x\end{align}

Differentiating both sides with respect to $$x$$, we obtain

\begin{align} &\Rightarrow \;\frac{1}{y}\frac{{dy}}{{dx}} = a \cdot \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}\\ &\Rightarrow \;\frac{{dy}}{{dx}} = \frac{{ - ay}}{{\sqrt {1 - {x^2}} }}\end{align}

By squaring both the sides, we obtain

\begin{align} &\Rightarrow \;{\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{{{a^2}{y^2}}}{{1 - {x^2}}}\\ &\Rightarrow \;\left( {1 - {x^2}} \right){\left( {\frac{{dy}}{{dx}}} \right)^2} = {a^2}{y^2}\end{align}

Again, differentiating both sides with respect to $$x$$, we obtain

\begin{align} &\Rightarrow \;{\left( {\frac{{dy}}{{dx}}} \right)^2}\frac{d}{{dx}}\left( {1 - {x^2}} \right) + \left( {1 - {x^2}} \right) \times \frac{d}{{dx}}\left[ {{{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right] = {a^2}\frac{d}{{dx}}\left( {{y^2}} \right)\\ &\Rightarrow \;{\left( {\frac{{dy}}{{dx}}} \right)^2}\left( { - 2x} \right) + \left( {1 - {x^2}} \right) \times 2\frac{{dy}}{{dx}}.\frac{{{d^2}y}}{{d{x^2}}} = {a^2}.2y.\frac{{dy}}{{dx}}\\ &\Rightarrow \;- x\frac{{dy}}{{dx}} + \left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} = {a^2}.y \qquad \left[ {\frac{{dy}}{{dx}} \ne 0} \right]\\ &\Rightarrow \;\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} - {a^2}y = 0\end{align}

Hence proved.

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