# How do you differentiate f(x)=(e^x-3lnx)(tanx+2x) using the product rule?

Jan 5, 2016

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

#### Explanation:

The product rule states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$,

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

Thus

$f ' \left(x\right) = \left(\tan x + 2 x\right) \frac{d}{\mathrm{dx}} \left({e}^{x} - 3 \ln x\right) + \left({e}^{x} - 3 \ln x\right) \frac{d}{\mathrm{dx}} \left(\tan x + 2 x\right)$

Find the derivative of each part separately.

$\frac{d}{\mathrm{dx}} \left({e}^{x} - 3 \ln x\right) = {e}^{x} - \frac{3}{x}$

$\frac{d}{\mathrm{dx}} \left(\tan x + 2 x\right) = {\sec}^{2} x + 2$

Hence

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