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Question Number 61818 by Kunal12588 last updated on 09/Jun/19

$$\mathrm{Find}\frac{dy}{dx}$$ $$y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$$

Commented byPrithwish sen last updated on 09/Jun/19

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-{\left[\frac{(1-{\mathrm{x}}^2)}{(1+{\mathrm{x}}^2)}\right]}^2}}\frac{-2\mathrm{x}(1+{\mathrm{x}}^2)-2\mathrm{x}(1-{\mathrm{x}}^2)}{{(1+{\mathrm{x}}^2)}^2}$$ $$=\frac{(1+{\mathrm{x}}^2)}{2\mathrm{x}}\frac{(-4\mathrm{x})}{{(1+{\mathrm{x}}^2)}^2}$$ $$=\frac{-2}{(1+{\mathrm{x}}^2)}$$

Commented byKunal12588 last updated on 09/Jun/19

$$seemyway...$$ $$y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)letx=\tan \alpha \Rightarrow \alpha ={\tan }^{-1}x$$ $$y={\sin }^{-1}\left(\cos 2\alpha \right)$$ $$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-{\cos }^{2}2\alpha }}\times (-\sin 2\alpha )\times 2\times \frac{1}{1+x^2}=\frac{-2}{1+x^2}$$

Commented byPrithwish sen last updated on 09/Jun/19

$$\mathrm{Beautiful}$$

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