Question

Differentiate the following with respect to x:

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

(ii) ${\sin }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

Answer 1

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

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let},y={\sin }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \\ \mathrm{Put}{2}^x=\tan \theta \\ \therefore y={\sin }^{-1}\left(\frac{2^x\times 2}{1+{\left(2^x\right)}^2}\right) \\ \Rightarrow y={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right) \\ \Rightarrow y={\sin }^{-1}\left\{\sin \left(2\theta \right)\right\} \\ \Rightarrow y=2\theta \left[\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\theta \right] \\ \Rightarrow y=2{\tan }^{-1}\left(2^x\right)\left[\mathrm{Since},2^x=\tan \theta \right] \end{gathered}$$

Differentiate with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=\frac{2}{1+{\left(2^x\right)}^2}\frac{d}{dx}\left(2^x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{2\times 2^x\log 2}{1+4^x} \\ \therefore \frac{dy}{dx}=\frac{2^{x+1}}{1+4^x}\log 2 \end{gathered}$$