Question

Differentiate

$\sin \left(\frac{1+x^2}{1-x^2}\right)$

Answer 1

$$\begin{gathered} \mathrm{Let}y=\sin \left(\frac{1+x^2}{1-x^2}\right) \\ \mathrm{Differentiate}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{we}\mathrm{get}, \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left[\sin \left(\frac{1+x^2}{1-x^2}\right)\right] \\ =\cos x\left(\frac{1+x^2}{1-x^2}\right)\frac{d}{dx}\left(\frac{1+x^2}{1-x^2}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\cos x\left(\frac{1+x^2}{1-x^2}\right)\left[\frac{\left(1-x^2\right){\displaystyle \frac{d}{dx}}\left(1+x^2\right)-\left(1+x^2\right){\displaystyle \frac{d}{dx}}\left(1-x^2\right)}{{\left(1-x^2\right)}^2}\right]\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\right] \\ =\cos x\left(\frac{1+x^2}{1-x^2}\right)\left[\frac{\left(1-x^2\right)\left(2x\right)-\left(1+x^2\right)\left(-2x\right)}{{\left(1-x^2\right)}^2}\right] \\ =\cos x\left(\frac{1+x^2}{1-x^2}\right)\left[\frac{2x-2x^3+2x+2x^3}{{\left(1-x^2\right)}^2}\right] \\ =\frac{4x}{{\left(1-x^2\right)}^2}\cos x\left(\frac{1+x^2}{1-x^2}\right) \\ \mathrm{So},\frac{d}{dx}\left\{\sin \left(\frac{1+x^2}{1-x^2}\right)\right\}=\frac{4x}{{\left(1-x^2\right)}^2}\cos x\left(\frac{1+x^2}{1-x^2}\right) \end{gathered}$$