Question

Differentiate the following with respect to x:

$\left(\mathrm{i}\right){\cos }^{-1}\left(\sin x\right)$

(ii) ${\cot }^{-1}\left(\frac{1-x}{1+x}\right)$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let},f\left(x\right)={\cos }^{-1}\left(\sin x\right) \\ \Rightarrow f\left(x\right)={\cos }^{-1}\left[\cos \left(\frac{\mathrm{\pi }}{2}-x\right)\right] \\ \Rightarrow f\left(x\right)=\frac{\mathrm{\pi }}{2}-x \\ \mathrm{Thus},f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}-x\right)=-1 \end{gathered}$$
(ii)
$$\begin{gathered} {\cot }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{d}{dx}\left[{\tan }^{-1}\left(\frac{1+x}{1-x}\right)\right] \\ =\frac{1}{1+{\left({\displaystyle \frac{1+x}{1-x}}\right)}^2}\times \frac{1-x+1+x}{{\left(1-x\right)}^2} \\ =\frac{{\left(1-x\right)}^2}{{\left(1-x\right)}^2+{\left(1+x\right)}^2}\times \frac{2}{{\left(1-x\right)}^2} \\ =\frac{{\left(1-x\right)}^2}{1-2x+x^2+1+2x+x^2} \\ =\frac{{\left(1-x\right)}^2}{2\left(1+x^2\right)}\times \frac{2}{{\left(1-x\right)}^2} \\ =\frac{1}{1+x^2} \end{gathered}$$