Question

Differentiate

${\tan }^{-1}\left(\frac{x}{1+6x^2}\right)$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left(\frac{x}{1+6x^2}\right) \\ \Rightarrow y={\tan }^{-1}\left(\frac{3x-2x}{1+\left(3x\right)\left(2x\right)}\right) \\ \Rightarrow y={\tan }^{-1}3x-{\tan }^{-1}2x\left[\mathrm{Since},{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)\right] \end{gathered}$$

Differentiate it with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=\frac{1}{1+{\left(3x\right)}^2}\frac{d}{dx}\left(3x\right)-\frac{1}{1+{\left(2x\right)}^2}\frac{d}{dx}\left(2x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+9x^2}\left(3\right)-\frac{1}{1+4x^2}\left(2\right) \\ \therefore \frac{dy}{dx}=\frac{3}{1+9x^2}-\frac{2}{1+4x^2} \end{gathered}$$