# How do you find the derivative of the function: arctan (cos x)?

Sep 1, 2016

$\frac{- \sin x}{1 + {\cos}^{2} x}$

#### Explanation:

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

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

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

let $u = \cos x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = - \sin x$

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

substitute these values into (A) changing u back to x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {u}^{2}} \times \left(- \sin x\right) = \frac{- \sin x}{1 + {\cos}^{2} x}$