Question

If $y={\tan }^{-1}\left(2x\right)$, then $\frac{dy}{dx}$ is:

Answer 1

Explanation:

$\frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}\left(2x\right)\right)$

Now using chain rule, we get

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left({\tan }^{-1}\left(2x\right)\right)=\frac{1}{1+{\left(2x\right)}^2}\times \frac{d}{dx}\left(2x\right)\mathbf{}\mathbf{[}\mathbf{D}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{]} \\ \Rightarrow \frac{d}{dx}\left({\tan }^{-1}\left(2x\right)\right)=\frac{2}{1+4x^2} \end{gathered}$$

Therefore, $\frac{dy}{dx}=\frac{2}{1+4x^2}$.