Question

Find $\frac{dy}{dx}$, if $y=e^{{\sin }^{2}x}\left\{2{\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right\}$.

Answer 1

$y=e^{si{n}^{2}x}\left\{2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}\right\}$

$\frac{dy}{dx}=e^{si{n}^{2}x}.sin2x.2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}+e^{si{n}^{2}x}\frac{d}{dx}\left(2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}\right)$

$\frac{d}{dx}\left(2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}\right)$

Put $x=cos2\theta$

$\therefore 2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}=2ta{n}^{-1}\left(cot\theta \right)=2[\frac{\pi }{2}-\theta ]$

$=\pi -2\times \frac{1}{2}co{s}^{-1}x$

$=\pi -co{s}^{-1}x$

$\frac{dy}{dx}=e^{si{n}^{2}x}.sin2x.2ta{n}^{-1}\sqrt{\frac{1+x}{1-x}}+e^{si{n}^{2}x}.\frac{1}{\sqrt{1-x^2}}$