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Question:

If $\mathrm{y}=\sin ^{-1}\left(6 \mathrm{x} \sqrt{1-9 \mathrm{x}^{2}}\right),-\frac{1}{3 \sqrt{2}}<\mathrm{x}<\frac{1}{3 \sqrt{2}}$, then find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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,$y=\sin ^{-1}(\sin 2 \theta)y=2 \thetay=2 \cos ^{-1} x$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,$y=\sin ^{-1}(\sin 2 \theta)y=-2 \thetay=-2 \cos ^{-1} x$Differentiating w.r.t$x$, we get$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\$

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