ML AGGARWAL CLASS 12 Chapter 5 Continuity and Differentiability Exercise 5.6

 Exercise 5.6

Question 1

(i) $f(x)=\sqrt{4-x^{2}} ;$, x∈(-2,2), find f 'x

Sol :

$f(x)=\sqrt{4-x^{2}}=(4-x^2)^{\frac{1}{2}}$ 

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$f^{\prime}(x)=\frac{1}{2}\left(4-x^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x} ( 4 - x ^ { 2 } )$

$=\frac{1}{2}\left(4-x^{2}\right)^{-\frac{1}{2}}\times (-2 x)$

$=\frac{-x}{\sqrt{4-x^{2}}}$

(ii) $f(x)=\sqrt{1-x^{2}}$ x∈(0,1) ; find f '(x)

Sol :

$f(x)=\sqrt{1-x^{2}}=(1-x^2)^{\frac{1}{2}}$

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$f^{\prime}(x)=\frac{1}{2}\left(1-x^{2}\right)^{-\frac{1}{2}} \frac{d}{d x}\left(1-x^{2}\right)$

$=\frac{1}{2}\left(1-x^{2}\right)^{-1 / 2} \cdot(-2 x)$

$=\frac{-x}{\sqrt{1-x^{2}}}$

Question 2

(i) $\sqrt{2 x-1}$

Sol :

$f(x)=(2 x-1)^{\frac{1}{2}}$

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$f^{\prime}(x)=\frac{1}{2}(2 x-1)^{\frac{1}{2}-1}\cdot \frac{d}{d x}(2 x-1)$

$=\frac{1}{2}(2 x-1)^{-\frac{1}{2}} \cdot 2$

$=\frac{1}{\sqrt{2 x-1}}$

(ii) $\left(3 x^{2}-9 x+5\right)^{9}$
Sol :
$f(x)=\left(3 x^{2}-9 x+5\right)^{9}$

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$f'(x)=9\left(3 x^{2}-9 x+5\right)^{9-1}\cdot \frac{d}{dx}(3x^2-9x+5)$

$f'(x)=9\left(3 x^{2}-9 x+5\right)^{8}\cdot (6x-9)$

$=27(3x^2-9x+5)^8 (2x-3)$

Question 3

(i) $\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^{2}+4}}$

Sol :

$f(x)=(3 x+2)^{\frac{1}{2}}+\frac{1}{(2 x^{2}+4)^{\frac{1}{2}}}$

$f(x)=(3 x+2)^{\frac{1}{2}}+(2 x^{2}+4)^{-\frac{1}{2}}$

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$f'(x)=\frac{1}{2}(3 x+2)^{-\frac{1}{2}} \frac{d}{d x}(3 x+2)+\frac{-1}{2}\left(2 x^{2}+4\right)^{-\frac{3}{2}}\cdot \frac{d}{d x}\left(2 x^{2}+4\right)$

$=\frac{1}{2}(3 x+2)^{-\frac{1}{2}} \cdot 3+ \frac{-1}{2}\left(2 x^{2}+4\right)^{-\frac{3}{2}} (4 x)$

$=\frac{3}{2\sqrt{3 x+2}}-\frac{2x}{(2x^{2}+4)^{\frac{3}{2}}}$

(ii) $\sin ^{3} x+\cos ^{6} x$
Sol :

$f(x)=(\sin x) ^{3}+(\cos x)^{6}$

$f'(x)=3(\sin x) ^{3-1}.\dfrac{d}{dx}(\sin x)+6(\cos x)^{6-1}.\dfrac{d}{dx}(\cos x)$

[Using $\frac{d}{d x} x^{n}=n x^{n-1}$]

$=3 \sin ^{2} x \cos x+\left(-6 \cos ^{5} x \sin x\right)$

$=3\sin x \cos x(\sin x + -2\cos^4 x)$

Question 4

(i) $\sin (x^2)$
Sol :

$f(x)=\sin (x^2)$

$f'(x)=\cos \left(x^{2}\right) .\frac{d}{d x} (x^{2})$

$=2 x \cdot \cos x^{2}$.

(ii) $\cos \sqrt{x}$
Sol :

$f(x)=\cos \sqrt{x}$

$f'(x)=-\sin \sqrt{x}\cdot \frac{d}{d x} \sqrt{x}$

$=-\sin \sqrt{x} \cdot\left(\frac{1}{2} x^{-\frac{1}{2}}\right)$

$=\frac{-1}{2 \sqrt{x}} \cdot \sin \sqrt{x} .$

Question 5

(i) sin(ax+b)

Sol :

f(x)=sin(ax+b)

$f'(x)=\cos (a x+b).\frac{d}{d x}(a x+b)$

=acos(ax+b)

(ii) tan(2x+3)

Sol :

f(x)=tan(2x+3)

$f'(x)=\sec ^{2}(2 x+3).\frac{d}{d x}(2 x+3)$

$=2 \sec ^{2}(2 x+3)$

Question 6

(i) cos (sin x)

Sol :

f(x)=cos (sin x)

$f'(x)=-\sin (\sin x).\frac{d}{dx}(\sin x)$

f '(x)=-[sin (sin x)].cos x

=(-cos x)[sin(sin x)]

(ii) $sin (cos (x^2))$

Sol :

$f(x)=sin (cos (x^2))$

$f'(x)=cos(cos (x^2)).\frac{d}{dx} (\cos x^2).\frac{d}{dx}(x^2)$

$f'(x)=cos(cos (x^2)).(-\sin x^2).(2x)$

$=-2x.\sin x^2. \cos (\cos x^2)$

Question 7

(i) $\cos (\tan \sqrt{x+1})$

Sol :

$f'(x)=-\sin(\tan \sqrt{x+1}).\frac{d}{dx}(\tan \sqrt{x+1}).\frac{d}{dx}(\sqrt{x+1})$

$=-\sin (\tan \sqrt{x+1}).\left(\sec ^{2} \sqrt{x+1}\right) .\frac{1}{2\sqrt{x+1}}.\frac{d}{dx}(x+1)$

$=\frac{-\sin (\tan \sqrt{x+1}).\sec^2 \sqrt{x+1}}{2\sqrt{x+1}}$

(ii) $\sqrt{\tan \sqrt{x}}$

Sol :

[Using $\frac{d}{d x} \sqrt{x}=\frac{1}{2\sqrt{x}}$]

$f'(x)=\frac{1}{2\sqrt{\tan \sqrt{x}}}.\frac{d}{dx}(\tan \sqrt{x}).\frac{d}{dx}(\sqrt{x})$

$f'(x)=\dfrac{1}{2\sqrt{\tan \sqrt{x}}}.(\sec ^2 \sqrt{x}).\bigg(\dfrac{1}{2\sqrt{x}}\bigg)$

$=\frac{\sec ^2 \sqrt{x}}{4\sqrt{x \tan \sqrt{x}}}$

Question 8

$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}+\ldots+x+1$

Show f '(1)=100f '(0)

Sol :

$f(x)=1+ \frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{99}}{99}+\frac{x^{100}}{100}$

$f(x)=1+\sum^{100}_{n=1} \frac{x^{n}}{n}$

$f'(x)=\sum_{n=1}^{100} \frac{nx^{n-1}}{n}$

$f'(x)=\sum_{n=1}^{100} x^{n-1}$

$f^{\prime}(x)=1+x+x^{2}+\cdots+x^{99}$

f '(0)=1+(0+0+....upto 99 terms)

f '(0)=1

f '(1)=1+(1+1+....upto 99 terms)

=1+99=100

=100.(1) or

=100 f '(0)

f '(1)=100 f '(0)

Question 9

$\left|2 x^{2}-3\right|$

Sol :

$f(x)=\left|2 x^{2}-3\right|$

[Using $\frac{d}{d x}|x|=\frac{x}{|x|}$]

$f'(x)=\frac{2x^2-3}{\left|2 x^{2}-3\right|}.\frac{d}{dx}(2x^2-3)$

$=\frac{\left(2x^{2}-3\right)(4 x)}{\left|2 x^{2}-3\right|}$

Question 10

f(x)=|cos x|. Find $f^{\prime}\left(\frac{3 \pi}{4}\right)$

Sol :

$f^{\prime}(x)=\frac{\cos x}{|\cos x|} \frac{d}{d x}(\cos x)$

$=-\frac{\cos x}{\mid \cos x|} \cdot \sin x$

$f^{\prime}\left(\frac{3 \pi}{4}\right)=\frac{-\cos \frac{3 \pi}{4} \sin \frac{3 \pi}{4}}{\mid \cos \left(\frac{3 \pi}{4}\right) \mid}$

$=\frac{-\cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right)}{| \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \mid}$

$=\frac{-\left(-\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}}}{\left|-\frac{1}{\sqrt{2}}\right|}$

$=\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}= \frac{1}{2} \cdot \sqrt{2}=\frac{1}{\sqrt{2}}$

Question 11

(i) $\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}=f(x)$

Sol :

Using $\left[\frac{d}{d x} \frac{u}{v}=\frac{v \frac{d u}{dx}-u \frac{dv}{dx}}{v^{2}}\right]$

$f^{\prime}(x)=\frac{\sqrt{a^{2}+x^{2}} \frac{d}{d x} \sqrt{a^{2}-x^{2}}-\sqrt{a^{2}-x^{2}} \frac{d}{d x} \sqrt{a^{2}+x}}{\left(a^{2}+x^{2}\right)}$

$=\frac{\left[\sqrt{a^{2}+x^{2}}\left[\frac{1}{2}\left(a^{2}-x^{2}\right)^{-1 / 2} \frac{d}{d x}\left(a^{2}-x^{2}\right)\right]\right.{\left.-\sqrt{a^{2}-x^{2}}\left[\frac{1}{2}\left(a^{2}+x^{2}\right)^{\frac{-1}{2}} \frac{d}{d x}\left(a^{2}+x^{2}\right)\right]\right)}}{\left(a^{2}+x^{2}\right)}$

$=\frac{\sqrt{a^{2}+x^{2}} \frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x)-\sqrt{a^{2}-x^{2}}\frac{1}{2 \sqrt{a^{2}+x^{2}}}2 x}{(a^2+x^2)}$

$=\frac{ \frac{-x\sqrt{a^{2}+x^{2}}}{\sqrt{a^{2}-x^{2}}}-\frac{x\sqrt{a^{2}-x^{2}}}{\sqrt{a^{2}+x^{2}}}}{(a^2+x^2)}$

$=\frac{-x\left[a^{2}+x^{2}+a^{2}-x^{2}\right]}{\sqrt{a^{2}-x^{2}} \sqrt{a^{2}+x^{2}}\left(a^{2}+x^{2}\right)}$

$=\frac{-2 a^{2} x}{\sqrt{a^{2}-x^{2}}\left(a^{2}+x^{2}\right)^{3 / 2}}$

(ii) $\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

Sol :
$f(x)=\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

$=\cos \sqrt{x} \times \frac{1}{2\sqrt{x}}+2 \cos \sqrt{x}(-\sin \sqrt{x}) .\frac{d}{d x} \sqrt{x}$

$=\cos \sqrt{x} \frac{1}{2 \sqrt{x}}+(-2 \sin \sqrt{x} \cos \sqrt{x}) \frac{1}{2 \sqrt{x}}$

$=\frac{1}{2 \sqrt{x}}(\cos \sqrt{x}-\sin 2 \sqrt{x})$

Question 12

(i) $\frac{\sin (a x+b)}{\cos (c x+d)}$

Sol :

$f(x)=\frac{\sin (a x+b)}{\cos (c x+d)}$

$f(x)=\sin (a x+b) \cdot(\cos (c x+d))^{-1}$

Using $\left[\frac{d}{d x} u v=u \frac{dv}{d x}+v \frac{d u}{d x}\right]$

$f^{\prime}(x)=\sin (ax+b) .\frac{d}{d x}(\cos (cx+d))^{-1}+(\cos (cx+d))^{-1}.\frac{d}{dx}(\sin x (ax+b))$

$=\sin (a x+b)(-1 \cos (x+a))^{-2}. \frac{d}{d x} \cos (cx+d)+(\cos (c x+d))^{-1} \cdot \cos (a x+b) \cdot \frac{d}{d x}(a x+b)$

$=-\frac{\sin (a x+b)}{\cos ^{2}(c x+d)}\left.(-\sin (cx+d)) .\frac{d}{d x}(cx+d)+\frac{\cos(ax+b)}{\cos(cx+d)}.a\right.$

$=\frac{c \sin (a x+b) \sin (c x+d)+a \cos (a x+6) \cos (cx+d)}{\cos ^{2}(c x+d)}$

(ii) $f(x)=\frac{\sin x+x^{2}}{\cot 2 x}$

Sol :

$f(x)=\left(\sin x+x^{2}\right) \tan 2 x$

$\left[\frac{d}{d x} u v=u \frac{dv}{d x}+v \frac{d u}{d x}\right)$

$f'(x)=(\sin x+x) \frac{d}{d x}(\tan 2 x)+\tan 2 x .\frac{d}{dx}(\sin x+x^2)$

$=\left(\sin x+x^{2}\right)\left(\sec ^{2} 2 x\right) \frac{d}{d x}(2 x)+\tan 2 x(\cos x+2 x$

$=2 x \sec^{2} x\left(\sin x+x^{2}\right)+\tan 2 x(\cos x+2 x)$

Question 13

(i) $\sin ^{m} x \cos ^{x} x$

Sol :

$f(x)=\sin ^{m} x \cos ^{x} x$

Using $\left[\frac{d}{d x} u v=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

$f^{\prime}(x)=\operatorname{sin}^{m} x .\frac{d}{d x} \cos ^{n} x+\cos ^{n} x. \frac{d}{d x} \sin^m x$

$=\sin ^{m} x\left(n \cos ^{n-1} x. \frac{d}{d x} \cos x\right)+\cos ^{n} x\left(m \sin ^{m-1} x. \frac{d}{d x} \sin x\right)$

$=\left(\sin ^{m} x\right)\left(x \cos ^{n-1} x\right)(-\sin x)+\left(\cos ^{n} x\right)\left(m \sin ^{m-1} x\right)(\cos x)$

$=\left(\cos ^{n-1} x\right)\left(\sin ^{m-1} x\right)\left[-n \sin ^{2} x+m \operatorname{cos}^{2} x\right]$

(ii) $\sin ^{n}\left(a x^{2}+b x+c\right)$

Sol :

$f(x)=\sin ^{n}\left(a x^{2}+b x+c\right)$

$=x \sin ^{n-1}\left(a x^{2}+b x+c\right) \cos \left(a x^{2}+b x+c\right).\frac{d}{dx}(ax^2+bx+c)$

$=n \sin ^{n-1}\left(a x^{2}+bx+c\right) \cos \left(ax^{2}+b x+c\right)(2ax+b)$

Question 14

(i) $\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}$

Sol :

$f(x)=\frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \times \frac{\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}$

$=\frac{(\sqrt{a+x}-\sqrt{a-x})^{2}}{(\sqrt{a+x})^{2}-(\sqrt{a-x})^{2}}$

$=\frac{a+x+a-x-2 \sqrt{a^{2}-x^{2}}}{(a+x)-(a-x)}$

$f(x)=\frac{2 a-2 \sqrt{a^{2}-x^{2}}}{2 x}=\frac{a-\sqrt{a^{2}-x^{2}}}{x}$

Using $\left[\frac{d}{d x} \frac{u}{v}=\frac{v \frac{d u}{d x}-u \frac{d v}{dx}}{v^{2}}\right]$

$f'(x)=\frac{x \frac{d}{d x}\left(a-\sqrt{a^{2}- x^{2}}\right)-\left(a-\sqrt{a^{2}-x^{2}}\right) \frac{d}{d x} x}{x^{2}}$

$=\frac{x\left(-\frac{1}{2}\left(a^{2}-x^{2}\right)^{-1 / 2} . \frac{d}{d x}\left(a^{2}-x^{2}\right)\right)-\left(a-\sqrt{a^{2}-x^{2}}\right)}{x^2}$

$\frac{=\frac{-x(-2 x)}{2 \sqrt{a^{2}-x^{2}}}-\left(a-\sqrt{a^{2}-x^{2}}\right)}{x^{2}}$

$= \frac{x^{2}-\left(a \sqrt{a^{2}- x^{2}}\right)+a^{2}-x^{2}}{\sqrt{a^2-x^2}.x^2}$

$=\frac{a^{2}-a \sqrt{a^{2}-x^{2}}}{x \sqrt{a^{2}-x^{2}}}$

(ii) $\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}$

Sol :

$f(x)=\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}} \times \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}$

$=\frac{\left(\sqrt{x^{2}+1}+\sqrt{x^{2}-1}\right)^{2}}{\left(\sqrt{x^{2}+1}\right)^{2}-\left(\sqrt{x^{2}-1}\right)^{2}}$

$=\frac{x^{2} +1+x^{2}-1+2 \sqrt{x^{4}-1}}{\left(x^{2}+1\right)-\left(x^{2}-1\right)}$

$=\frac{2 x^{2}+2 \sqrt{x^{4}-1}}{2}$

$=x^{2}+\sqrt{x^{4}-1}$

$f^{\prime}(x)=2 x+\frac{1}{2}\left(x^{4}-1\right)^{-\frac{1}{2}} .\frac{d}{d x}\left(x^{4}-1\right)$

$=2 x+\frac{1}{2 \sqrt{x^{1-1}}} 4 x^{3}$

$=2 x+\frac{2x^{3}}{\sqrt{x^{1-1}}}$

Question 15

$f(x)=\sqrt{\frac{\sec x-1}{\sec x+1}}$  find $f^{\prime}(x) \text{ and } f^{\prime}\left(\frac{\pi}{3}\right)$

Sol :
$f(x)=\sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}}$

$=\sqrt{\frac{1-\cos x}{1+\cos x}}$

$=\sqrt{\frac{1-(1-2\sin^2 \frac{x}{2})}{1+(2\cos^2 x-1)}}$

$=\sqrt{\frac{\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}}$

$f(x)=\sqrt{\tan ^2 \frac{x}{2}}=\tan \frac{x}{2}$

$f^{\prime}(x)=\sec ^{2} \frac{x}{2} \cdot \frac{d}{d x} \frac{x}{2}$

$=\frac{1}{2} \sec ^{2} \frac{x}{2}$

at $f'\left(\frac{\pi}{3}\right)$

$f^{\prime}(\pi / 3)=\frac{1}{2} \sec ^{2} \frac{\pi}{6}$

$=\frac{1}{2} \cdot\left(\frac{2}{\sqrt{3}}\right)^{2}=\frac{4}{2 \cdot 3}=\frac{2}{3}$

Question 16

(i) $\cos \left(2 x+\frac{\pi}{3}\right)$ ; derivative at $x=\frac{\pi}{3}$

Sol :

$f(x)=\cos \left(2 x+\frac{\pi}{3}\right)$

$f^{\prime}(x)=-\sin \left(2 x+\frac{\pi}{3}\right) \frac{d}{d x}\left(2 x+\frac{\pi}{3}\right)$

$=-2 \sin \left(2 x+\frac{\pi}{3}\right)$

$f^{\prime}\left(\frac{\pi}{3}\right)$

$=-2 \sin \left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)$

=-2sin 𝜋=0

(ii) $\frac{1-\sin x}{1+\cos x}$ ; derivative at $x=\frac{\pi}{2}$

$f(x)=\frac{1-\sin x}{1+\cos x}$

$=(1-\sin x)(1+\cos x)^{-1}$

Using $\left[\frac{d}{d x} u v=u \frac{d v}{d x}+v \frac{d u}{d x}\right\}$

$=(1-\sin x) .\frac{d}{d x}(1+\cos x)^{-1}+(1+\cos x)^{-1} .\frac{d}{d x} (1-\sin x)$

$=(1-\sin n)\left(-(1+\cos x)^{-2} \frac{d}{d x}(1+\cos x)\right)+(1+\cos x)^{-1}(-\cos x)$

$=\frac{(1-\sin x)}{(1+\cos x)^{2}}(\sin x)-\frac{\cos x}{(1+\cos x)}$

$=\frac{(1-\sin x) \sin x-\cos x(1+\operatorname{cos} x)}{(1+\cos x)^{2}}$

$=\frac{\sin x-\operatorname{sin}^{2} x-\operatorname{cos} x-\cos ^{2} x}{(1+\cos x)^{2}}$

$=\frac{\sin x-\cos x-1}{\left(1+\cos x\right)^{2}}$

$f^{\prime}\left(\frac{\pi}{2}\right)=\frac{\sin \frac{\pi}{2}-\cos \frac{\pi}{2}-1}{\left(1+\operatorname{cos} \frac{\pi}{2}\right)^{2}}$

$=\frac{1-0-1}{(1+0)^{2}}=\frac{0}{1^{2}}$

=0

Question 17

$y=\sqrt{x}+\frac{1}{\sqrt{x}}$ ; Prove $2 x \frac{d y}{d x}+y=2 \sqrt{x}$

Sol :

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

$\frac{d y}{d x}=\frac{\sqrt{x} \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x} \sqrt{x}}{x}$

Using $\left[\frac{d}{d x} \frac{u}{v}=\frac{v \frac{d u}{d x}-u \frac{d v}{dx}}{v^{2}}\right]$

$x \frac{d y}{d x}=\sqrt{x}-(x+1) \frac{1}{2} x^{-1 / 2}$

$x \frac{d y}{d x}=\sqrt{x}-\frac{(x+1)}{2 \sqrt{x}}=\frac{2 x-(x+1)}{2 \sqrt{x}}$

$2 x \frac{d y}{d \pi}=\frac{x-1}{\sqrt{x}}$

$2 x \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}$

$2 x \frac{d y}{d x}+\sqrt{x}=2 \sqrt{x}-\frac{1}{\sqrt{x}}$

$2 x \frac{d y}{d x}+\sqrt{x}+\frac{1}{\sqrt{x}}=2 \sqrt{x}$

 $2 x \frac{d y}{d x}+y=2 \sqrt{x}$

Question 18

$y=\sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}$ ; Prove $\frac{d y}{d x}+\sec ^{2}\left(\frac{\pi}{4}-x\right)=0$

Sol :

$y=\sqrt{\frac{1-\cos \left(\frac{\pi}{2}-2 x\right)}{1+\operatorname{sin}\left(\frac{\pi}{2}-2 x\right)}}$

$=\sqrt{\frac{2 \sin ^{2}\left(\frac{\pi}{4}-x\right)}{2 \cos ^{2}\left(\frac{\pi}{4}-x\right)}}$

$y=\sqrt{\tan ^{2}\left(\frac{\pi}{4}-x\right)}=\tan \left(\frac{\pi}{4}-x\right)$

$\frac{d y}{d x}=\sec ^{2}\left(\frac{\pi}{4}-x\right) \frac{d}{d x}\left(\frac{\pi}{4}-x\right)$

$=-\sec ^{2}\left(\frac{\pi}{4}-x\right)$

$\frac{d y}{d x}+\sec ^{2}\left(\frac{\pi}{4}-x\right)=0$

Question 19

|cos x| ; Is this function is differentiable . What about cos |x|

Sol 

f(x)=|cos x| 

Using $\left[\frac{d}{d x}|x|=\frac{x}{|x|}\right]$

$f^{\prime}(x)=\frac{\cos x}{|\cos x|} .\frac{d}{d x}(\cos x)$

$=\frac{\cos x(-\sin x)}{|\cos x \mid}=-\frac{\sin x \cos x}{|\cos x|}$

$f^{\prime}(x)=-\frac{\sin x \cos x}{|\cos x|}$

above function is not differentiable ∀ $x=(2 n+1) \frac{\pi}{2}$ as it forms 0/0 form.

where as cos |x|=cos x which is differentiable for all real values.