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# NCERT Solutions for Class 12 Maths Exercise 5.5 of Chapter 5 Continuity and Differentiability

NCERT Solutions for Class 12 Maths Exercise 5.5 of Chapter 5 Continuity and Differentiability are the best study material for clearing your concept on Continuity and Differentiability which is required to solve the questions on the chapter 5 Continuity and Differentiability of Class 12 Maths .These NCERT  Solutions for Class 12 Maths Exercise 5.5 of Chapter 5 Continuity and Differentiability are created by a Maths expert in Future Study Point who has experience of 25 years of Maths teaching .

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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Exercise 5.5-Continuity and Differentiability

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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Differentiate the function with respect to x in exercise 1 to 5

Q1. cos x cos 2x cos 3x

Ans. The given function is cos x cos 2x cos 3x

Let y = cos x cos 2x cos 3x

Taking logarithms both sides

log y = log (cos x cos 2x cos 3x)

log y = log cos x + log cos 2x + log cos 3x

Differentiating both sides

$\fn_cm \frac{1}{y}\frac{dy}{dx}=\frac{1}{cos\: x}\frac{d}{dx}cos\: x+\frac{1}{cos\: 2x}\frac{d}{dx}cos\: 2x+\frac{1}{cos\: 3x}\frac{d}{dx}cos\: 3x$

$\fn_cm \frac{1}{y}\frac{dy}{dx}=\frac{1}{cos\: x}\left ( -sin\: x \right )+\frac{1}{cos\: 2x}\left ( -sin\: 2x \right )\frac{d}{dx}2x+\frac{1}{cos\: 3x}\left ( -sin\: 3x \right )\frac{d}{dx}3x$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=-\frac{sin\: x}{cos\: x}-\frac{sin\: 2x}{cos\: 2x}.2-\frac{sin\: 3x}{cos\: 3x}.3$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=-tan\: x-2tan\: 2x-3tan\: 3x$

$\fn_cm \frac{dy}{dx}=-y(tan\: x+2tan\: 2x+3tan\: 3x)$

$\fn_cm Q2.\sqrt{\frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )}}$

Ans.

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

$\fn_cm \: y=\left ( \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right )^{1/2}$

Taking logarithms on both sides

$\fn_cm log\: y=log\left ( \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right )^{1/2}$

$\fn_cm log\: y=\frac{1}{2}log\left ( \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right )$

$\fn_cm log\: y=\frac{1}{2}\left [ log\left ( x-1 \right )\left ( x-2 \right )-log\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right ) \right ]$

$\fn_cm log\: y=\frac{1}{2}\left [ log\left ( x-1 \right )+log\left ( x-2 \right )-log\left ( x-3 \right )-log\left ( x-4 \right )-log\left ( x-5 \right ) \right ]$

Differentiating the given function with respect to x

$\fn_cm \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left [ \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5} \right ]$

$\fn_cm \frac{dy}{dx}=\frac{y}{2}\left [ \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5} \right ]$

$\fn_cm \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )}}\left [ \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5} \right ]$

$\fn_cm Q3.\left ( logx \right )^{cosx}$

Ans.The given function is      $\fn_cm \left ( logx \right )^{cosx}$

$\fn_cm Let\: y=\left ( logx \right )^{cosx}$

Taking logarithms on both sides

$\fn_cm logy=log\left ( logx \right )^{cosx}$

$\fn_cm logy=cosx\: log\left ( logx \right )$

Differentiating the function with respect to x

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=\frac{d}{dx}\left [ cosx\: log\left ( logx \right )\right ]$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=cosx\frac{d}{dx}log\left ( logx \right )+log\left ( logx \right )\frac{d}{dx}cosx$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=cosx.\frac{1}{logx}\frac{d}{dx}logx+log\left ( logx \right )\left ( -sinx \right )$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=cosx.\frac{1}{logx}.\frac{1}{x}+log\left ( logx \right )\left ( -sinx \right )$

$\fn_cm \frac{dy}{dx}=y\left [\frac{cosx}{xlogx}-sinx.log\left ( logx \right ) \right ]$

$\fn_cm \frac{dy}{dx}=\left ( logx \right )^{cosx}\left [\frac{cosx}{xlogx}-sinx.log\left ( logx \right ) \right ]$

$\fn_cm Q4.x^{x}-2^{sinx}$

Ans.  $\fn_cm Let \: y=x^{x}-2^{sinx}$

Let $\fn_cm u=x^{x}$          $\fn_cm Let \: v=2^{sinx}$

$\fn_cm \frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$

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

Taking logarithms both sides

log u = x log x

Differentiating the function with respect to x

$\fn_cm \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}logx+logx$

$\fn_cm \frac{1}{u}\frac{du}{dx}=x.\frac{1}{x}+logx$

$\fn_cm \frac{du}{dx}=u\left (1+logx \right )$

$\fn_cm \frac{du}{dx}=x^{x}\left (1+logx \right )$

Now, $\fn_cm Let \: v=2^{sinx}$

Taking logarithms both sides

log v = sin x log 2

Differentiating both sides with respect to x

$\fn_cm \frac{1}{v}\frac{dv}{dx}=log2\frac{d}{dx}\left ( sinx \right )$

$\fn_cm \frac{1}{v}\frac{dv}{dx}=log2.cosx$

$\fn_cm \frac{dv}{dx}=v\: log2.cosx$

$\fn_cm \frac{dv}{dx}=2^{sinx}\: log2.cosx$

$\fn_cm \frac{dv}{dx}=cosx\:2^{sinx}log2$

Substituting the value of du/dx and dv/dx in the following equation

$\fn_cm \frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$

$\fn_cm \frac{dy}{dx}=x^{x}\left ( 1+logx \right )-cosx.2^{sinx}.log2$

Q5.(x +3)2(x +4)3(x +5)4

Ans. The given function is (x +3)2(x +4)3(x +5)4

Let y = (x +3)2(x +4)3(x +5)4

Taking logrithms both sides

log y = log [(x +3)2(x +4)3(x +5)4]

log y  = log(x +3)2+log(x +4)3+log(x +5)4

log y  =2 log(x +3)+3log(x +4)+4log(x +5)

Differentiating both sides

$\fn_cm \frac{1}{y}\frac{dy}{dx}=2\frac{d}{dx}log\left ( x+3 \right )+3\frac{d}{dx}log\left ( x+4 \right )+4\frac{d}{dx}log\left ( x+5 \right )$

$\fn_cm \frac{1}{y}\frac{dy}{dx}=2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}$

$\fn_cm \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}$

$\fn_cm \frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )$

$\fn_cm \frac{dy}{dx}=(x +3)^{2}(x +4)^{3}(x +5)^{4}\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )$

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## NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

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### NCERT solutions of class 11 maths

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