# How do you find the derivative of f(x) = arctan(cos(3x))?

May 29, 2017

$= - \frac{3 \sin \left(3 x\right)}{1 + {\cos}^{2} \left(3 x\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(\cos \left(3 x\right)\right)\right)$

Step 1. Use the chain rule
$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(\cos \left(3 x\right)\right)\right) = \frac{d \left({\tan}^{-} 1 \left(u\right)\right)}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$,

where $u = \cos \left(3 x\right)$ and $\frac{d}{\mathrm{du}} \left({\tan}^{-} 1 \left(u\right)\right) = \frac{1}{1 + {u}^{2}}$

$= \frac{\frac{d}{\mathrm{dx}} \left(\cos \left(3 x\right)\right)}{1 + {\cos}^{2} \left(3 x\right)}$

Step 2. Using the chain rule again,

$\frac{d}{\mathrm{dx}} \left(\cos \left(3 x\right)\right) = \frac{d \left(\cos \left(u\right)\right)}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$,

where $u = 3 x$ and $\frac{d}{\mathrm{du}} \left(\cos \left(u\right)\right) - \sin \left(u\right)$, gives

$= \frac{- \frac{d}{\mathrm{dx}} \left(3 x\right) \sin \left(3 x\right)}{1 + {\cos}^{2} \left(3 x\right)}$

Step 3. Factor out constants

$= - 3 \frac{\sin \left(3 x\right)}{1 + {\cos}^{2} \left(3 x\right)} \frac{d}{\mathrm{dx}} \left(x\right)$

ANSWER: $= - \frac{3 \sin \left(3 x\right)}{1 + {\cos}^{2} \left(3 x\right)}$