# How do you differentiate g(x) =e^(1-x)tanx using the product rule?

Apr 8, 2016

$g ' \left(x\right) = {e}^{1 - x} {\sec}^{2} x - {e}^{1 - x} \tan x$

#### Explanation:

let $f = {e}^{1 - x}$ and $g = \tan x$

$f ' = {e}^{1 - x} \cdot - 1 , g ' = {\sec}^{2} x$

$g ' \left(x\right) = f g ' + g f '$

$g ' \left(x\right) = {e}^{1 - x} {\sec}^{2} x - {e}^{1 - x} \tan x$