Question

Differentiate ${\sin }^{-1}\left\{\frac{2^{x+1}·3^x}{1+{\left(36\right)}^x}\right\}$ with respect to x.

Answer 1

We have,
$$\begin{gathered} y={\sin }^{-1}\left\{\frac{2^{x+1}·3^x}{1+{\left(36\right)}^x}\right\} \\ \Rightarrow y={\sin }^{-1}\left\{\frac{2·6^x}{1+6^{2x}}\right\} \\ \mathrm{put}{6}^x=\tan \theta \\ \Rightarrow \theta ={\tan }^{-1}\left(6^x\right) \\ \mathrm{Now}, \\ y={\sin }^{-1}\left\{\frac{2\tan \theta }{1+{\tan }^2\theta }\right\} \\ \Rightarrow y={\sin }^{-1}\left\{\sin 2\theta \right\} \\ \Rightarrow y=2\theta \\ \Rightarrow y=2{\tan }^{-1}\left(6^x\right) \\ \Rightarrow \frac{dy}{dx}=2\times \frac{1}{{\left(6^x\right)}^2}\times 6^x\log 6 \\ \therefore \frac{dy}{dx}=\frac{2\left(\log 6\right)6^x}{{36}^x} \end{gathered}$$