# How do you find the derivative of y= sin{cos^2(tanx)}?

Nov 8, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {\sec}^{2} \left(x\right) \sin \left(\tan x\right) \cos \left(\tan x\right) \cos \left({\cos}^{2} \left(\tan \left(x\right)\right)\right)$

#### Explanation:

$y = \sin \left({\cos}^{2} \left(\tan x\right)\right) = \sin \left(f \left(x\right)\right)$

$y = \sin \left(f \left(x\right)\right) \implies \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) \cos \left(f \left(x\right)\right)$

$f \left(x\right) = {\cos}^{2} \left(g \left(x\right)\right) = {\left(\cos \left(g \left(x\right)\right)\right)}^{2}$
$f ' \left(x\right) = 2 \cdot \frac{d}{\mathrm{dx}} \left[\cos \left(g \left(x\right)\right)\right] \cdot \cos \left(g \left(x\right)\right)$
$f ' \left(x\right) = 2 \cdot - g ' \left(x\right) \sin \left(g \left(x\right)\right) \cdot \cos \left(g \left(x\right)\right)$
$f ' \left(x\right) = - 2 g ' \left(x\right) \sin \left(g \left(x\right)\right) \cos \left(g \left(x\right)\right)$

$g \left(x\right) = \tan x$
$g ' \left(x\right) = {\sec}^{2} x$

Therefore, $f ' \left(x\right) = - 2 {\sec}^{2} \left(x\right) \sin \left(\tan x\right) \cos \left(\tan x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {\sec}^{2} \left(x\right) \sin \left(\tan x\right) \cos \left(\tan x\right) \cos \left({\cos}^{2} \left(\tan \left(x\right)\right)\right)$