Question

If y = sin { 2 tan^-1 [1 - x / 1 + x ]^1/2 } show that dy / dx = -x / (1-x²)^1/2

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{y}=\sin \left[2{\tan }^{-1}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right] \\ \mathrm{Let}\mathrm{x}=\mathrm{cos\theta },\mathrm{so} \\ \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}=\sqrt{\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}}=\sqrt{\frac{1-1+2{\sin }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{1+2{\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-1}} \\ =\tan \frac{\mathrm{\theta }}{2}\mathrm{now}\mathrm{putting}\mathrm{this}\mathrm{we}\mathrm{get}, \\ \mathrm{y}=\sin \left[2{\tan }^{-1}\left(\tan \frac{\mathrm{\theta }}{2}\right)\right]=\sin \left(2\times \frac{\mathrm{\theta }}{2}\right)=\mathrm{sin\theta }=\sqrt{1-{\cos }^2\mathrm{\theta }}=\sqrt{1-{\mathrm{x}}^2} \\ \mathrm{Now}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{1-{\mathrm{x}}^2}}\times -2\mathrm{x}=\frac{-\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}} \end{gathered}$$