How do you differentiate y=tan^-1(5x)?

Oct 8, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{1 + 25 {x}^{2}}$

Explanation:

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

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

let $u = 5 x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 5$

and $y = {\tan}^{-} 1 u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{1 + {u}^{2}}$

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

substitute results into (A) and change u back into terms of x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {u}^{2}} \times 5 = \frac{5}{1 + 25 {x}^{2}}$