# arrow_back Show that $\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}$

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

$From \ \frac{d}{dx}(sin^{-1}x) \ we \ ca \ have \ that \ \ y = sin^{-1}x \\ \implies x = siny \qquad \qquad (1)\\ \implies dx = cosy \ dy \qquad \qquad(taking \ the \ derivatives) \\ \implies \frac{dy}{dx} = \frac{1}{cosy} \\ \implies \frac{dy}{dx} = \frac{1}{ \sqrt{1-sin^2y}} \qquad \qquad (using \ the \ trig \ identity \ sin^2 \theta +cos^2 \theta =1) \\ \implies \frac{dy}{dx} = \frac{1}{ \sqrt{1-x^2}} \qquad \qquad (using \ (1))\\ Hence \ shown \$
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