Question

If $y=\sin \left[2{\tan }^{-1}\left\{\frac{\sqrt{1-x}}{1+x}\right\}\right],\mathrm{find}\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{Here},y=\sin \left[2{\tan }^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right] \\ \mathrm{Put}x=\cos 2\theta \\ \mathrm{We}\mathrm{have},y=\sin \left[2{\tan }^{-1}\left(\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\right)\right] \\ =\sin \left[2{\tan }^{-1}\left(\sqrt{\frac{2{\sin }^2\theta }{2{\cos }^2\theta }}\right)\right] \\ =\sin \left[2{\tan }^{-1}\sqrt{{\tan }^2\theta }\right] \\ =\sin \left[2{\tan }^{-1}\left(\tan \theta \right)\right] \\ =\sin \left(2\theta \right) \\ =\sin \left[2\times \frac{1}{2}{\cos }^{-1}x\right]\left[\mathrm{Since},x=\cos 2\theta \right] \\ =\sin \left({\sin }^{-1}\sqrt{1-x^2}\right) \\ =\sqrt{1-x^2} \end{gathered}$$

Differentiate it with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}\left(1-x^2\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}}\left(-2x\right) \\ \therefore \frac{dy}{dx}=\frac{-x}{\sqrt{1-x^2}} \end{gathered}$$