# How do you differentiate f(x)=(-e^x+secx)(3x^3-x) using the product rule?

Jan 20, 2016

#### Answer:

$f ' \left(x\right) = \left(\sec x \tan x - {e}^{x}\right) \left(3 {x}^{3} - x\right) + \left(\sec x - {e}^{x}\right) \left(9 {x}^{2} - 1\right)$

#### Explanation:

Product rule states that the derivative of a function $f \left(x\right) = g \left(x\right) h \left(x\right)$ is

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

In this particular case, we have

$g \left(x\right) = - {e}^{x} + \sec x$
$h \left(x\right) = 3 {x}^{3} - x$

$g ' \left(x\right) = - {e}^{x} + \sec x \tan x$
$h ' \left(x\right) = 9 {x}^{2} - 1$

Applying this, we see that

$f ' \left(x\right) = \left(\sec x \tan x - {e}^{x}\right) \left(3 {x}^{3} - x\right) + \left(\sec x - {e}^{x}\right) \left(9 {x}^{2} - 1\right)$