Solve this

Solution:

$y=\sin ^{-1}\left\{6 x \sqrt{1-9 x^{2}}\right\}$

$y=\sin ^{-1}\left\{2 \times 3 x \sqrt{1-(3 x)^{2}}\right\}$

let $3 x=\cos \theta$

$y=\sin ^{-1}\left\{2 \times \sin \theta \sqrt{1-\cos ^{2} \theta}\right\}$

Using $\sin ^{2} \theta+\cos ^{2} \theta=1$

$y=\sin ^{-1}\{2 \times \sin \theta \cos \theta\}$

Using $2 \sin \theta \cos \theta=\sin 2 \theta$

$y=\sin ^{-1}(\sin 2 \theta)$

Considering the limits,

$-\frac{1}{3 \sqrt{2}}

$-\frac{1}{\sqrt{2}}<3 x<\frac{1}{\sqrt{2}}$

$-\frac{1}{\sqrt{2}}<\cos \theta<\frac{1}{\sqrt{2}}$

$-\frac{\pi}{4}<\theta<\frac{\pi}{4}$

$-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}$

For

$0<2 \theta<\frac{\pi}{2}$

Now, $y=\sin ^{-1}(\sin 2 \theta)$

$y=2 \theta$

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

Differentiating w.r.t $\mathrm{x}$, we get

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

For

$-\frac{\pi}{2}<2 \theta<0$

Now, $y=\sin ^{-1}(\sin 2 \theta)$

$y=-2 \theta$

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

Differentiating w.r.t $x$, we get

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