Question

Differentiate

${\tan }^{-1}\left\{\frac{x^{1/3}+a^{1/3}}{1-{\left(ax\right)}^{1/3}}\right\}$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left[\frac{x^{\frac{1}{3}}+a^{\frac{1}{3}}}{1-{\left(ax\right)}^{{\displaystyle \frac{1}{3}}}}\right] \\ \Rightarrow y={\tan }^{-1}\left(x^{\frac{1}{3}}\right)+{\tan }^{-1}\left(a^{\frac{1}{3}}\right)\left[\mathrm{Since},{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right] \end{gathered}$$

Differentiate it with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{1+{\left(x^{{\displaystyle \frac{1}{3}}}\right)}^2}\times \frac{d}{dx}\left(x^{\frac{1}{3}}\right)+0 \\ \Rightarrow \frac{dy}{dx}=\frac{\left({\displaystyle \frac{1}{3}}\times x^{{\displaystyle \frac{1}{3}}-1}\right)}{1+x^{{\displaystyle \frac{2}{3}}}} \\ \therefore \frac{dy}{dx}=\frac{1}{3x^{{\displaystyle \frac{2}{3}}}\left(1+x^{{\displaystyle \frac{2}{3}}}\right)} \end{gathered}$$