# How do you differentiate y=tan^-1(2x^4)?

Dec 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 {x}^{3}}{1 + 4 {x}^{8}}$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 x\right) = \frac{1}{1 + {x}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}} \to \left(A\right)$

$\text{let } u = 2 {x}^{4} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 8 {x}^{3}$

$\text{and } y = {\tan}^{-} 1 \left(u\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{1 + {u}^{2}}$

Substitute into (A) and change u back to x

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {u}^{2}} \times 8 {x}^{3} = \frac{8 {x}^{3}}{1 + 4 {x}^{8}}$