# What's the derivative of arctan (x/2)?

Jul 30, 2016

$\frac{2}{4 + {x}^{2}}$

#### Explanation:

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

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

Note that $\frac{x}{2} = \frac{1}{2} x$

let $\textcolor{b l u e}{u = \frac{1}{2} x} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2}$

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

and y=arctancolor(blue)(u)rArr(dy)/(du)=1/(1+color(blue)(u)^2

Substitute these values into (A) and convert u back into terms of x.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\left(\frac{x}{2}\right)}^{2}} \times \frac{1}{2} = \frac{\frac{1}{2}}{\left(1 + {x}^{2} / 4\right)} = \frac{\frac{1}{2}}{\frac{1}{4} \left(4 + {x}^{2}\right)}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{4 + {x}^{2}}$