Maths Solve the Following Question.(2 Marks)

Let $y=\operatorname{cosec}^{-1}\left(\sec 5^x\right)$
$\therefore y=\operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(\frac{\pi}{2}-5^x\right)\right]$
$\therefore y=\frac{\pi}{2}-5^x$
Differentiating w.r.t. $x$,
$\frac{d y}{d x}=0-5^x \log 5$
$=-5^x \log 5$

$y=\sin ^{-1}(3 x)+\sec ^{-1}\left(\frac{1}{3 x}\right)$
$\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}(3 x)+\frac{d}{d x} \sec ^{-1}\left(\frac{1}{3 x}\right)$
$\frac{d y}{d x}=\frac{3}{\sqrt{1-(3 x)^2}}+\frac{\frac{-1}{3 x^2}}{\frac{1}{3 x} \sqrt{\left(\frac{1}{(3 x)^2}-1\right)}}$
$\frac{d y}{d x}=\frac{3}{\sqrt{1-9 x^2}}-\frac{1}{\times \frac{\sqrt{\left(1-9 x^2\right)}}{3|x|}}$
$=\frac{3}{\sqrt{1-9 x^2}}-\frac{3|x|}{ X \sqrt{1-9 x^2}}$
$=\frac{3}{\sqrt{1-9 x^2}}\left(1-\frac{|X|}{X}\right)$
$=0 \quad X>0$
$=\frac{6}{\sqrt{1-9 x^2}} \quad x<0$

If $y = x \log x$, then $\frac{d^2 y}{d x^2}=$ $\frac{1}{x}$
Explanation:
$y=x \log x$
Differentiating both sides,
$\frac{d y}{d x}=x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)$
$=x \cdot \frac{1}{x}+\log x=1+\log x$
Again differentiating w.r.t.x,
$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(1)+\frac{d}{d x}(\log x)$
$\frac{d^2 y}{d x^2}=0+\frac{1}{x}=\frac{1}{x}$

$y=[\tan (3 \log x)]^2$
differentiate w.r.t. x both side
$\therefore \frac{d y}{d x}=2[\tan (3 \log x)] \times \sec ^2(3 \log x) \cdot \frac{3}{x}$
$\therefore \frac{d y}{d x}=\frac{6}{x} \tan \left(\log x^3\right) \cdot \sec ^2\left(\log x^3\right)$

x sin y + y sin x = 0
Differentiate w.r.t. x both side
$\begin{aligned} & {\left[x \cos y \frac{d y}{d x}+\sin y\right]+\left[y \cos x+\sin x \frac{d y}{d x}\right]=0} \\ & \therefore \sin y+y \cos x=\frac{d y}{d x}(-\sin x-x \cos y) \\ & \therefore \frac{d y}{d x}=-\left(\frac{\sin y+y \cos x}{\sin x+x \cos y}\right)\end{aligned}$

(B) $\frac{\sqrt{3}}{2}$
Let $y=\tan ^3 \theta$, and $x=\sec ^3 \theta$
$\frac{d y}{d \theta}=3 \tan ^2 \theta \cdot \sec ^2 \theta, \frac{d x}{d \theta}=3 \sec ^2 \theta \cdot \sec \theta \tan \theta$
$\frac{d y}{d x}=\sin \theta=\sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$

$y=e^{a x}$
$y=e^{a x} \ldots \ldots \ldots \ldots \ldots .(i)$
$\log y=a x \ldots \ldots \ldots \ldots \ldots . . .(i i)$
$\frac{d y}{d x}=a e^{a x}$
$\frac{d y}{d x}=a y$
$x \frac{d y}{d x}=a x y$
$x \frac{d y}{d x}=y \log y$

$\frac{d y}{d x}=\sin \theta$
$\frac{d x}{d \theta}=-\cos \theta$
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=-\frac{\sin \theta}{\cos \theta}=-\tan \theta$
$\frac{d y}{d x}=-\tan \left(\frac{\pi}{4}\right)$
$\therefore \frac{d y}{d x}=-1$

We have, $y=2 a t$
$\frac{d y}{d t}=2 a \frac{d}{d t}(t)=2 a(1)=2 a$
also $x=a t^2$
$\frac{d x}{d x}=a \frac{d}{d t}\left(t^2\right)=a(2 t)=2 a t$
now $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 a}{2 a t}=\frac{1}{t}$

Let $u=3^x$
$\frac{ du }{d x}=3^x \log 3$
Let $v=\log _3 x=\frac{\log x}{\log 3}$
$\frac{ dv }{d x}=\frac{1}{\log 3} \frac{d}{d x}(\log x)$
$=\frac{1}{\log 3} \frac{1}{x}=\frac{1}{x \log 3}$
$\frac{ du }{ dv }=\frac{\frac{ du }{d x}}{\frac{ dv }{d x}}=\frac{3^x \log 3}{\frac{1}{x \log 3}}=3^x \cdot x(\log 3)^{\wedge} 2$

$y=\sec \sqrt{x}$
$\frac{d y}{d x}=\frac{d}{d x}(\sec \sqrt{x})$
$\frac{d y}{d x}=\sec \sqrt{x} \tan \sqrt{x} \frac{d}{d x}(\sqrt{x})$
$=\sec \sqrt{x} \cdot \tan \sqrt{x} \frac{.1}{2 \sqrt{x}}$
$\frac{d y}{d x}=\frac{\sec \sqrt{x} \tan \sqrt{x}}{2 \sqrt{x}}$