# How do you differentiate f(x) =cosx*sec^2(x) ?

Mar 7, 2016

$\frac{d}{d x} f \left(x\right) = {\sec}^{2} x$

#### Explanation:

$\frac{d}{d x} f \left(x\right) = - \sin x \cdot {\sec}^{2} x + 2 \sec x \cdot \sec x \cdot \tan x \cdot \cos x$
$\frac{d}{d x} f \left(x\right) = - \sin x \cdot {\sec}^{2} x + 2 {\sec}^{2} x \cdot \tan x \cdot \cos x$
$\frac{d}{d x} f \left(x\right) = {\sec}^{2} \left(- \sin x + \tan x \cdot \cos x\right) \text{ , } \tan x = \frac{\sin x}{\cos x}$
$\frac{d}{d x} f \left(x\right) = {\sec}^{2} x \left(- \sin x + \frac{\sin x}{\cancel{\cos x}} \cdot \cancel{\cos x}\right)$
$\frac{d}{d x} f \left(x\right) = {\sec}^{2} x \left(\cancel{- \sin x} + \cancel{\sin x}\right)$
$\frac{d}{d x} f \left(x\right) = {\sec}^{2} x$

Mar 7, 2016

$f ' \left(x\right) = \sec x \tan x$. Here are two ways to get this answer.

#### Explanation:

Rewrite first

$f \left(x\right) = \cos x {\sec}^{2} x$

$= \cos x \frac{1}{\cos x} ^ 2$

$= \frac{1}{\cos} x$

$= \sec x$.

So,

$f ' \left(x\right) = \sec x \tan x$.

Use product rule first
$f \left(x\right) = \cos x {\sec}^{2} x$

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

$= - \sin x {\sec}^{2} x + 2 {\sec}^{2} x \left[\tan x \cos x\right]$

$= - \sin x {\sec}^{2} x + 2 {\sec}^{2} x \left[\sin x\right]$

$= {\sec}^{2} x \sin x$

$= \sec x \tan x$