# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5

Go back toÂ  'Continuity and Differentiability'

## Chapter 5 Ex.5.5 Question 1

Differentiate the function with respect to $$x: \;\; cos x.\cos 2x.\cos 3x$$

### Solution

Let $$y = \cos x.\cos 2x.\cos 3x$$

Taking logarithm on both the sides, we obtain

\begin{align}&\log y = \log \left( {\cos x.\cos 2x.\cos 3x} \right)\\&\Rightarrow \; \log y = \log \left( {\cos x} \right) + \log \left( {\cos 2x} \right) + \log \left( {\cos 3x} \right)\end{align}

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

\begin{align}&\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{\cos x}}.\frac{d}{{dx}}\left( {\cos x} \right) + \frac{1}{{\cos 2x}}.\frac{d}{{dx}}\left( {\cos 2x} \right) + \frac{1}{{\cos 3x}}.\frac{d}{{dx}}\left( {\cos 3x} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = y\left[ { - \frac{{\sin x}}{{\cos x}} - \frac{{\sin 2x}}{{\cos 2x}}.\frac{d}{{dx}}\left( {2x} \right) - \frac{{\sin 3x}}{{\cos 3x}}.\frac{d}{{dx}}\left( {3x} \right)} \right]\\&\therefore\; \frac{{dy}}{{dx}} = - \cos x.\cos 2x.\cos 3x\left[ {\tan x + 2\tan 2x + 3\tan 3x} \right]\end{align}

## Chapter 5 Ex.5.5 Question 2

Differentiate the function with respect to $$x:\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}}$$

### Solution

Let $$y = \sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\log y = \log \sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \\&\Rightarrow \; \log y = \frac{1}{2}\log \left[ {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \right]\\&\Rightarrow \; \log y = \frac{1}{2}\left[ {\log \left\{ {\left( {x - 1} \right)\left( {x - 2} \right)} \right\} - \log \left\{ {\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)} \right\}} \right]\\&\Rightarrow \; \log y = \frac{1}{2}\left[ {\log \left( {x - 1} \right) + \log \left( {x - 2} \right) - \log \left( {x - 3} \right) - \log \left( {x - 4} \right) - \log \left( {x - 5} \right)} \right]\end{align}

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

\begin{align}&\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left[ \begin{array}{l}\frac{1}{{x - 1}}.\frac{d}{{dx}}\left( {x - 1} \right) + \frac{1}{{x - 2}}.\frac{d}{{dx}}\left( {x - 2} \right) - \frac{1}{{x - 3}}.\frac{d}{{dx}}\left( {x - 3} \right)\\ - \frac{1}{{x - 4}}.\frac{d}{{dx}}\left( {x - 4} \right) - \frac{1}{{x - 5}}.\frac{d}{{dx}}\left( {x - 5} \right)\end{array} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{y}{2}\left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]\\&\therefore \; \frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]\end{align}

## Chapter 5 Ex.5.5 Question 3

Differentiate the function with respect to $$x:$$$${\left( {\log x} \right)^{\cos x}}$$

### Solution

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

Taking logarithm on both the sides, we obtain

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

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

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

## Chapter 5 Ex.5.5 Question 4

Differentiate the function with respect to $$x:$$$${x^x} - {2^{\sin x}}$$

### Solution

Let $$y = {x^x} - {2^{\sin x}}$$

Also, let $${x^x} = u$$ and $${2^{\sin x}} = v$$

\begin{align}&\therefore y = u - v\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{du}}{{dx}} - \frac{{dv}}{{dx}}\\&u = {x^x}\end{align}

Taking logarithm on both the sides, we obtain

$$\log u = x\log x$$

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

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

$$v = {2^{\sin x}}$$

Taking logarithm on both the sides, we obtain

$$\log v = \sin x.\log 2$$

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

\begin{align}&\frac{1}{v}.\frac{{dv}}{{dx}} = \log 2.\frac{d}{{dx}}\left( {\sin x} \right)\\&\Rightarrow \; \frac{{dv}}{{dx}} = v\log 2\cos x\\&\Rightarrow \; \frac{{dv}}{{dx}} = {2^{\sin x}}\cos x\log 2\\&\therefore \;\frac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right) - {2^{\sin x}}\cos x\log 2\end{align}

## Chapter 5 Ex.5.5 Question 5

Differentiate the function with respect to $$x:$$$${\left( {x + 3} \right)^2}.{\left( {x + 4} \right)^3}.{\left( {x + 5} \right)^4}$$

### Solution

Let $$y = {\left( {x + 3} \right)^2}.{\left( {x + 4} \right)^3}.{\left( {x + 5} \right)^4}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\log y = \log {\left( {x + 3} \right)^2} + \log {\left( {x + 4} \right)^3} + \log {\left( {x + 5} \right)^4}\\&\Rightarrow \; \log y = 2\log \left( {x + 3} \right) + 3\log \left( {x + 4} \right) + 4\log \left( {x + 5} \right)\end{align}

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

\begin{align}&\frac{1}{y}.\frac{{dy}}{{dx}} = 2.\frac{1}{{x + 3}}.\frac{d}{{dx}}\left( {x + 3} \right) + 3.\frac{1}{{x + 4}}.\frac{d}{{dx}}\left( {x + 4} \right) + 4.\frac{1}{{x + 5}}.\frac{d}{{dx}}\left( {x + 5} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = y\left[ {\frac{2}{{x + 3}} + \frac{3}{{x + 4}} + \frac{4}{{x + 5}}} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = {\left( {x + 3} \right)^2}{\left( {x + 4} \right)^3}{\left( {x + 5} \right)^4}.\left[ {\frac{2}{{x + 3}} + \frac{3}{{x + 4}} + \frac{4}{{x + 5}}} \right]\end{align}

\begin{align}&\Rightarrow \frac{{dy}}{{dx}} = {\left( {x + 3} \right)^2}{\left( {x + 4} \right)^3}{\left( {x + 5} \right)^4}.\left[ {\frac{\begin{array}{l}2\left( {x + 4} \right)\left( {x + 5} \right) + 3\left( {x + 3} \right)\left( {x + 5} \right)\\ + 4\left( {x + 3} \right)\left( {x + 4} \right)\end{array}}{{\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right)}}} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = \left( {x + 3} \right){\left( {x + 4} \right)^2}{\left( {x + 5} \right)^3}.\left[ \begin{array}{l}2\left( {{x^2} + 9x + 20} \right) + 3\left( {{x^2} + 8x + 15} \right)\\ + 4\left( {{x^2} + 7x + 12} \right)\end{array} \right]\\&\therefore \;\frac{{dy}}{{dx}} = \left( {x + 3} \right){\left( {x + 4} \right)^2}{\left( {x + 5} \right)^3}\left( {9{x^2} + 70x + 133} \right)\end{align}

## Chapter 5 Ex.5.5 Question 6

Differentiate the function with respect to $$x:$$$${\left( {x + \frac{1}{x}} \right)^x} + {x^{\left( {1 + \frac{1}{x}} \right)}}$$

### Solution

Let $$y = {\left( {x + \frac{1}{x}} \right)^x} + {x^{\left( {1 + \frac{1}{x}} \right)}}$$

Also, let $$u = {\left( {x + \frac{1}{x}} \right)^x}$$ and $$v = {x^{\left( {1 + \frac{1}{x}} \right)}}$$

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

Then, $$u = {\left( {x + \frac{1}{x}} \right)^x}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow \log u = \log {\left( {x + \frac{1}{x}} \right)^x}{\rm{ }}\\&\Rightarrow \; \log u = x\log \left( {x + \frac{1}{x}} \right)\end{align}

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

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

Now, $$v = {x^{\left( {1 + \frac{1}{x}} \right)}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow \log v = \log \left[ {{x^{\left( {1 + \frac{1}{x}} \right)}}} \right]{\rm{ }}\\&\Rightarrow \; \log v = \left( {1 + \frac{1}{x}} \right)\log x\end{align}

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

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

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$$\frac{{dy}}{{dx}} = {\left( {x + \frac{1}{x}} \right)^x}\left[ {\frac{{{x^2} - 1}}{{{x^2} + 1}} + \log \left( {x + \frac{1}{x}} \right)} \right] + {x^{\left( {1 + \frac{1}{x}} \right)}}\left[ {\frac{{x + 1 - \log x}}{{{x^2}}}} \right]$$

## Chapter 5 Ex.5.5 Question 7

Differentiate the function with respect to $$x:$$$${\left( {\log x} \right)^x} + {x^{\log x}}$$

### Solution

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

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

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

Then, $$u = {\left( {\log x} \right)^x}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow \log u = \log \left[ {{{\left( {\log x} \right)}^x}} \right]\\&\Rightarrow \; \log u = x\log \left( {\log x} \right)\end{align}

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

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

$$v = {x^{\log x}}$$

Taking logarithm on both the sides, we obtain

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

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

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

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$$\frac{{dy}}{{dx}} = {\left( {\log x} \right)^{x - 1}}\left[ {1 + \log x.\log \left( {\log x} \right)} \right] + 2{x^{\log x - 1}}\log x$$

## Chapter 5 Ex.5.5 Question 8

Differentiate the function with respect to $$x:$$$${\left( {\sin x} \right)^x} + {\sin ^{ - 1}}\sqrt x$$

### Solution

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

Also, let $$u = {\left( {\sin x} \right)^x}$$ and $$v = {\sin ^{ - 1}}\sqrt x$$

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

Then, $$u = {\left( {\sin x} \right)^x}$$

Taking logarithm on both the sides, we obtain

\begin{align}& \Rightarrow \;\log u = \log {\left( {\sin x} \right)^x}\\& \Rightarrow \; \log u = x\log \left( {\sin x} \right)\end{align}

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

\begin{align}\frac{1}{u}\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( x \right) \times \log \left( {\sin x} \right) + x\frac{d}{{dx}}\left[ {\log \left( {\sin x} \right)} \right]\\&\Rightarrow \; \frac{{du}}{{dx}} = u\left[ {1 \times \log \left( {\sin x} \right) + x\frac{1}{{\left( {\sin x} \right)}}\frac{d}{{dx}}\left( {\sin x} \right)} \right]\\&\Rightarrow \; \frac{{du}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {\log \left( {\sin x} \right) + \frac{x}{{\left( {\sin x} \right)}}\cos x} \right]\\&\Rightarrow \; \frac{{du}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {x\cot x + \log \sin x} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

$$v = {\sin ^{ - 1}}\sqrt x$$

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

\begin{align}&\frac{{dv}}{{dx}} = \frac{1}{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}\frac{d}{{dx}}\left( {\sqrt x } \right)\\&\Rightarrow \; \frac{{dv}}{{dx}} = \frac{1}{{\sqrt {1 - x} }}\frac{1}{{2\sqrt x }}\\&\Rightarrow \; \frac{{dv}}{{dx}} = \frac{1}{{2\sqrt {x - {x^2}} }} \qquad \ldots \left( 3 \right)\end{align}

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$\frac{{dy}}{{dx}} = {\left( {\sin x} \right)^x}\left[ {x\cot x + \log \sin x} \right] + \frac{1}{{2\sqrt {x - {x^2}} }}$

## Chapter 5 Ex.5.5 Question 9

Differentiate the function with respect to $$x:$$$${x^{\sin x}} + {\left( {\sin x} \right)^{\cos x}}$$

### Solution

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

Also, let $$u = {x^{\sin x}}$$ and $$v = {\left( {\sin x} \right)^{\cos x}}$$

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

Then, $$u = {x^{\sin x}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow\; \log u = \log \left( {{x^{\sin x}}} \right)\\&\Rightarrow \; \log u = \sin x\log x\end{align}

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

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

$$v = {\left( {\sin x} \right)^{\cos x}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow \log v = \log {\left( {\sin x} \right)^{\cos x}}\\&\Rightarrow \; \log v = \cos x\log \left( {\sin x} \right)\end{align}

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

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

\begin{align}&\Rightarrow \frac{{dv}}{{dx}} = {\left( {\sin x} \right)^{\cos x}}\left[ { - \sin x\log \left( {\sin x} \right) + \cot x\cos x} \right]\\&\Rightarrow \; \frac{{dv}}{{dx}} = {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \sin x\log \left( {\sin x} \right)} \right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$$\frac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\cos x\log x + \frac{{\sin x}}{x}} \right] + {\left( {\sin x} \right)^{\cos x}}\left[ {\cot x\cos x - \sin x\log \left( {\sin x} \right)} \right]$$

## Chapter 5 Ex.5.5 Question 10

Differentiate the function with respect to $$x:$$$${x^{x\cos x}} + \frac{{{x^2} + 1}}{{{x^2} - 1}}$$

### Solution

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

Also, let $$u = {x^{x\cos x}}$$ and $$v = \frac{{{x^2} + 1}}{{{x^2} - 1}}$$

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

Then, $$u = {x^{x\cos x}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow\; \log u = \log \left( {{x^{x\cos x}}} \right)\\&\Rightarrow \; \log u = x\cos x\log x\end{align}

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

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

$$v = \frac{{{x^2} + 1}}{{{x^2} - 1}}$$

Taking logarithm on both the sides, we obtain

$$\Rightarrow \log v = \log \left( {{x^2} + 1} \right) - \log \left( {{x^2} - 1} \right)$$

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

\begin{align}&\frac{1}{v}\frac{{dv}}{{dx}} = \frac{{2x}}{{{x^2} + 1}} - \frac{{2x}}{{{x^2} - 1}}\\&\Rightarrow \; \frac{{dv}}{{dx}} = v\left[ {\frac{{2x\left( {{x^2} - 1} \right) - 2x\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}} \right]\\&\Rightarrow \; \frac{{dv}}{{dx}} = \frac{{{x^2} + 1}}{{{x^2} - 1}} \times \left[ {\frac{{ - 4x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}} \right]\\&\Rightarrow \; \frac{{dv}}{{dx}} = \frac{{ - 4x}}{{{{\left( {{x^2} - 1} \right)}^2}}} \qquad \ldots \left( 3 \right)\end{align}

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$$\frac{{dy}}{{dx}} = {x^{x\cos x}}\left[ {\cos x\left( {1 + \log x} \right) - x.\sin x\log x} \right] - \frac{{4x}}{{{{\left( {{x^2} - 1} \right)}^2}}}$$

## Chapter 5 Ex.5.5 Question 11

Differentiate the function with respect to $$x:{\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$$

### Solution

Let $$y = {\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$$

Also, let $$u = {\left( {x\cos x} \right)^x}$$ and $$v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$$

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

Then, $$u = {\left( {x\cos x} \right)^x}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow \; \log u = {\left( {x\cos x} \right)^x}\\&\Rightarrow \; \log u = x\log \left( {x\cos x} \right)\\&\Rightarrow \; \log u = x\left[ {\log x + \log \cos x} \right]\\&\Rightarrow \; \log u = x\log x + x\log \cos x\end{align}

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

\begin{align}&\frac{1}{u}\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {x\log x} \right) + \frac{d}{{dx}}\left( {x\log \cos x} \right)\\&\Rightarrow \; \frac{{du}}{{dx}} = u\left[ {\left\{ {\log x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log x} \right)} \right\} + \left\{ {\log \cos x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log \cos x} \right)} \right\}} \right]\end{align}

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

$$v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow\; \log v = \log {\left( {x\sin x} \right)^{\frac{1}{x}}}\\&\Rightarrow \; \log v = \frac{1}{x}\log \left( {x\sin x} \right)\\&\Rightarrow \; \log v = \frac{1}{x}\left( {\log x + \log \sin x} \right)\\&\Rightarrow \; \log v = \frac{1}{x}\log x + \frac{1}{x}\log \sin x\end{align}

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

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

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

$$\frac{{dy}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] + {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{{x\cot x + 1 - \log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$$

## Chapter 5 Ex.5.5 Question 12

Find $$\frac{{dy}}{{dx}}$$ of the function $${x^y} + {y^x} = 1$$

### Solution

The given function is $${x^y} + {y^x} = 1$$

Let, $${x^y} = u$$ and $${y^x} = v$$

\begin{align}&\therefore \;u + v = 1\\&\Rightarrow \; \frac{{du}}{{dx}} + \frac{{dv}}{{dx}} = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Then, $$u = {x^y}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow\; \log u = \log \left( {{x^y}} \right)\\&\Rightarrow \; \log u = y\log x\end{align}

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

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

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

Taking logarithm on both the sides, we obtain

\begin{align}&\Rightarrow\; \log v = \log \left( {{y^x}} \right)\\&\Rightarrow \; \log v = x\log y\end{align}

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

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

Therefore, from ($$1$$), ($$2$$) and ($$3$$);

\begin{align}&{x^y}\left[ {\log x\frac{{dy}}{{dx}} + \frac{y}{x}} \right] + {y^x}\left[ {\log y + \frac{x}{y}.\frac{{dy}}{{dx}}} \right] = 0\\&\Rightarrow \; \left( {{x^y}\log x + x{y^{x - 1}}} \right)\frac{{dy}}{{dx}} = - \left( {y{x^{y - 1}} + {y^x}\log y} \right)\\&\therefore \; \frac{{dy}}{{dx}} = \frac{{ - \left( {y{x^{y - 1}} + {y^x}\log y} \right)}}{{\left( {{x^y}\log x + x{y^{x - 1}}} \right)}}\end{align}

## Chapter 5 Ex.5.5 Question 13

Find $$\frac{{dy}}{{dx}}$$ of the function $${y^x} = {x^y}$$

### Solution

The given function is $${y^x} = {x^y}$$

Taking logarithm on both the sides, we obtain

$$x\log y = y\log x$$

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

\begin{align}&\log y.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log y} \right) = \log x.\frac{d}{{dx}}\left( y \right) + y.\frac{d}{{dx}}\left( {\log x} \right)\\&\Rightarrow \; \log y.1 + x.\frac{1}{y}.\frac{{dy}}{{dx}} = \log x.\frac{{dy}}{{dx}} + y.\frac{1}{x}\\&\Rightarrow \; \log y + \frac{x}{y}.\frac{{dy}}{{dx}} = \log x.\frac{{dy}}{{dx}} + \frac{y}{x}\\&\Rightarrow \; \left( {\frac{x}{y} - \log x} \right)\frac{{dy}}{{dx}} = \frac{y}{x} - \log y\\&\Rightarrow \; \left( {\frac{{x - y\log x}}{y}} \right)\frac{{dy}}{{dx}} = \frac{{y - x\log y}}{x}\\&\therefore\; \frac{{dy}}{{dx}} = \frac{y}{x}\left( {\frac{{y - x\log y}}{{x - y\log x}}} \right)\end{align}

## Chapter 5 Ex.5.5 Question 14

Find $$\frac{{dy}}{{dx}}$$ of the function $${\left( {\cos x} \right)^y} = {\left( {\cos y} \right)^x}$$

### Solution

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

Taking logarithm on both the sides, we obtain

$$y\log \cos x = x\log \cos y$$

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

\begin{align}&\log \cos x \frac{{dy}}{{dx}} + y.\frac{d}{{dx}}\left( {\log \cos x} \right) = \log \cos y.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log \cos y} \right)\\&\Rightarrow \; \log \cos x.\frac{{dy}}{{dx}} + y.\frac{1}{{\cos x}}.\frac{d}{{dx}}\left( {\cos x} \right) = \log \cos y.1 + x\frac{1}{{\cos y}}.\frac{d}{{dx}}\left( {\cos y} \right)\\&\Rightarrow \; \log \cos x.\frac{{dy}}{{dx}} + \frac{y}{{\cos x}}.\left( { - \sin x} \right) = \log \cos y + \frac{x}{{\cos y}}.\left( { - \sin y} \right).\frac{{dy}}{{dx}}\\&\Rightarrow \; \log \cos x.\frac{{dy}}{{dx}} - y\tan x = \log \cos y - x\tan y\frac{{dy}}{{dx}}\\&\Rightarrow \; \left( {\log \cos x + x\tan y} \right)\frac{{dy}}{{dx}} = y\tan x + \log \cos y\\&\therefore \; \frac{{dy}}{{dx}} = \frac{{y\tan x + \log \cos y}}{{x\tan y + \log \cos x}}\end{align}

## Chapter 5 Ex.5.5 Question 15

Find $$\frac{{dy}}{{dx}}$$ of the function $$xy = {e^{\left( {x - y} \right)}}$$

### Solution

The given function is $$xy = {e^{\left( {x - y} \right)}}$$

Taking logarithm on both the sides, we obtain

\begin{align}&\log \left( {xy} \right) = \log \left( {{e^{x - y}}} \right)\\&\Rightarrow \; \log x + \log y = \left( {x - y} \right)\log e\\&\Rightarrow \; \log x + \log y = \left( {x - y} \right) \times 1\\&\Rightarrow \; \log x + \log y = \left( {x - y} \right)\end{align}

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

\begin{align}&\frac{d}{{dx}}\left( {\log x} \right) + \frac{d}{{dx}}\left( {\log y} \right) = \frac{d}{{dx}}\left( x \right) - \frac{{dy}}{{dx}}\\&\Rightarrow \; \frac{1}{x} + \frac{1}{y}\frac{{dy}}{{dx}} = 1 - \frac{{dy}}{{dx}}\\&\Rightarrow \; \left( {1 + \frac{1}{y}} \right)\frac{{dy}}{{dx}} = 1 - \frac{1}{x}\\&\Rightarrow \; \left( {\frac{{y + 1}}{y}} \right)\frac{{dy}}{{dx}} = \frac{{x - 1}}{x}\\&\therefore \; \frac{{dy}}{{dx}} = \frac{{y\left( {x - 1} \right)}}{{x\left( {y + 1} \right)}}\end{align}

## Chapter 5 Ex.5.5 Question 16

Find the derivative of the function given by $$f\left( x \right) = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right)$$ and hence find $$f'\left( 1 \right)$$.

### Solution

The given function is $$f\left( x \right) = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right)$$

Taking logarithm on both the sides, we obtain

$$\log f\left( x \right) = \log \left( {1 + x} \right) + \log \left( {1 + {x^2}} \right) + \log \left( {1 + {x^4}} \right) + \log \left( {1 + {x^8}} \right)$$

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

\begin{align}&\frac{1}{{f\left( x \right)}}\frac{d}{{dx}}\left[ {f\left( x \right)} \right] = \frac{d}{{dx}}\log \left( {1 + x} \right) + \frac{d}{{dx}}\log \left( {1 + {x^2}} \right)\\ &\qquad + \frac{d}{{dx}}\log \left( {1 + {x^4}} \right) + \frac{d}{{dx}}\log \left( {1 + {x^8}} \right)\\&\Rightarrow \; \frac{1}{{f\left( x \right)}}.f'\left( x \right) = \frac{1}{{1 + x}}.\frac{d}{{dx}}\left( {1 + x} \right) + \frac{1}{{1 + {x^2}}}.\frac{d}{{dx}}\left( {1 + {x^2}} \right)\\&\qquad+ \frac{1}{{1 + {x^4}}}.\frac{d}{{dx}}\left( {1 + {x^4}} \right) + \frac{1}{{1 + {x^8}}}.\frac{d}{{dx}}\left( {1 + {x^8}} \right)\\&\Rightarrow \; f'\left( x \right) = f\left( x \right)\left[ {\frac{1}{{1 + x}} + \frac{1}{{1 + {x^2}}}.2x + \frac{1}{{1 + {x^4}}}.4{x^3} + \frac{1}{{1 + {x^8}}}.8{x^7}} \right]\\&\therefore \;f'\left( x \right) = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right)\left[ {\frac{1}{{1 + x}} + \frac{{2x}}{{1 + {x^2}}} + \frac{{4{x^3}}}{{1 + {x^4}}} + \frac{{8{x^7}}}{{1 + {x^8}}}} \right]\end{align}

Hence,

\begin{align}f'\left( 1 \right) &= \left( {1 + 1} \right)\left( {1 + {1^2}} \right)\left( {1 + {1^4}} \right)\left( {1 + {1^8}} \right)\left[ {\frac{1}{{1 + 1}} + \frac{{2\left( 1 \right)}}{{1 + {1^2}}} + \frac{{4{{\left( 1 \right)}^3}}}{{1 + {1^4}}} + \frac{{8{{\left( 1 \right)}^7}}}{{1 + {1^8}}}} \right]\\&= 2 \times 2 \times 2 \times 2\left[ {\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}} \right]\\&= 16\left( {\frac{{15}}{2}} \right)\\&= 120\end{align}

## Chapter 5 Ex.5.5 Question 17

Differentiate $$\left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)$$ in three ways mentioned below.

By using product rule

By expanding the product to obtain a single polynomial.

By logarithmic differentiation.

Do they all give the same answer?

### Solution

Let $$y = \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)$$

By using product rule

Let $$u = \left( {{x^2} - 5x + 8} \right)$$ and $$v = {x^3} + 7x + 9$$

\begin{align}&\therefore y = uv\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{du}}{{dx}}.v + u.\frac{{dv}}{{dx}} \qquad\left( {{\text{product rule}}} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{x^2} - 5x + 8} \right).\left( {{x^3} + 7x + 9} \right) + \left( {{x^2} - 5x + 8} \right).\frac{d}{{dx}}\left( {{x^3} + 7x + 9} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = \left( {2x - 5} \right)\left( {{x^3} + 7x + 9} \right) + \left( {{x^2} - 5x + 8} \right)\left( {3{x^2} + 7} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = 2x\left( {{x^3} + 7x + 9} \right) - 5\left( {{x^3} + 7x + 9} \right) + {x^2}\left( {3{x^2} + 7} \right)\\&- 5x\left( {3{x^2} + 7} \right) + 8\left( {3{x^2} + 7} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = \left( {2{x^4} + 14{x^2} + 18x} \right) - 5{x^3} - 35x - 45 + \left( {3{x^4} + 7{x^2}} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = - 15{x^3} - 35x + 24{x^2} + 56\\&\therefore \; \frac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11\end{align}

By expanding the product to obtain a single polynomial.

\begin{align}y &= \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)\\&= {x^2}\left( {{x^3} + 7x + 9} \right) - 5x\left( {{x^3} + 7x + 9} \right) + 8\left( {{x^3} + 7x + 9} \right)\\&= {x^5} + 7{x^3} + 9{x^2} - 5{x^4} - 35{x^2} - 45x + 8{x^3} + 56x + 72\\&= {x^5} - 5{x^4} + 15{x^3} - 26{x^2} + 11x + 72\end{align}

Therefore,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^5} - 5{x^4} + 15{x^3} - 26{x^2} + 11x + 72} \right)\\&= \frac{d}{{dx}}\left( {{x^5}} \right) - 5\frac{d}{{dx}}\left( {{x^4}} \right) + 15\frac{d}{{dx}}\left( {{x^3}} \right) - 26\frac{d}{{dx}}\left( {{x^2}} \right) + 11\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {72} \right)\\&= 5{x^4} - 5\left( {4{x^3}} \right) + 15\left( {3{x^2}} \right) - 26\left( {2x} \right) + 11\left( 1 \right) + 0\\&= 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11\end{align}

By logarithmic differentiation.

$$y = \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)$$

Taking logarithm on both the sides, we obtain

$$\log y = \log \left( {{x^2} - 5x + 8} \right) + \log \left( {{x^3} + 7x + 9} \right)$$

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

\begin{align}&\frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\log \left( {{x^2} - 5x + 8} \right) + \frac{d}{{dx}}\log \left( {{x^3} + 7x + 9} \right)\\&\Rightarrow \; \frac{1}{y}.\frac{{dy}}{{dx}} = \frac{1}{{{x^2} - 5x + 8}}.\frac{d}{{dx}}\left( {{x^2} - 5x + 8} \right) + \frac{1}{{{x^3} + 7x + 9}}.\frac{d}{{dx}}\left( {{x^3} + 7x + 9} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = y\left[ {\frac{1}{{{x^2} - 5x + 8}}.\left( {2x - 5} \right) + \frac{1}{{{x^3} + 7x + 9}}.\left( {3{x^2} + 7} \right)} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)\left[ {\frac{{2x - 5}}{{{x^2} - 5x + 8}} + \frac{{3{x^2} + 7}}{{{x^3} + 7x + 9}}} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = \left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)\left[ {\frac{{\left( {2x - 5} \right)\left( {{x^3} + 7x + 9} \right) + \left( {3{x^2} + 7} \right)\left( {{x^2} - 5x + 8} \right)}}{{\left( {{x^2} - 5x + 8} \right)\left( {{x^3} + 7x + 9} \right)}}} \right]\\&\Rightarrow \; \frac{{dy}}{{dx}} = 2x\left( {{x^3} + 7x + 9} \right) - 5\left( {{x^3} + 7x + 9} \right) + 3{x^2}\left( {{x^2} - 5x + 8} \right) + 7\left( {{x^2} - 5x + 8} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = 2{x^4} + 14{x^2} + 18x - 5{x^3} - 35x - 45 + 3{x^5} - 15{x^3} + 24{x^2} + 7{x^2} - 35x + 56\\&\Rightarrow \; \frac{{dy}}{{dx}} = 5{x^4} - 20{x^3} + 45{x^2} - 52x + 11\end{align}

From the above three observations, it can be concluded that all the results of $$\frac{{dy}}{{dx}}$$ are same.

## Chapter 5 Ex.5.5 Question 18

If u, v and w are functions of x, then show that

$$\frac{d}{{dx}}\left( {u.v.w} \right) = \frac{{du}}{{dx}}v.w + u.\frac{{dv}}{{dx}}.w + u.v.\frac{{dw}}{{dx}}$$

in two ways - first by repeated application of product rule, second by logarithmic differentiation.

### Solution

Let $$y = u.v.w = u.\left( {v.w} \right)$$

By applying product rule, we get

\begin{align}&\frac{{dy}}{{dx}} = \frac{{du}}{{dx}}.\left( {v.w} \right) + u.\frac{d}{{dx}}\left( {v.w} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{du}}{{dx}}.\left( {v.w} \right) + u\left[ {\frac{{dv}}{{dx}}.w + v.\frac{{dw}}{{dx}}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Again applying product rule}}} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{du}}{{dx}}.v.w + u.\frac{{dv}}{{dx}}.w + u.v.\frac{{dw}}{{dx}}\end{align}

Taking logarithm on both the sides of the equation $$y = u.v.w$$ , we obtain

$$\log y = \log u + \log v + \log w$$

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

\begin{align}&\frac{1}{y}\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\log u} \right) + \frac{d}{{dx}}\left( {\log v} \right) + \frac{d}{{dx}}\left( {\log w} \right)\\&\Rightarrow \; \frac{1}{y}.\frac{{dy}}{{dx}} = \frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}\\&\Rightarrow \; \frac{{dy}}{{dx}} = y\left( {\frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}} \right)\\&\Rightarrow \; \frac{{dy}}{{dx}} = u.v.w\left( {\frac{1}{u}\frac{{du}}{{dx}} + \frac{1}{v}\frac{{dv}}{{dx}} + \frac{1}{w}\frac{{dw}}{{dx}}} \right)\\&\therefore \; \frac{d}{{dx}}\left( {u.v.w} \right) = \frac{{du}}{{dx}}v.w + u.\frac{{dv}}{{dx}}.w + u.v.\frac{{dw}}{{dx}}\end{align}

Â Â
Related Sections
Related Sections