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

Feb 20, 2018

d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x), which can be simplified as necessary

Explanation:

Given:
$f \left(x\right) = \left(\ln x + x\right) \left(\tan x + {e}^{x}\right)$
Let
$y = f \left(x\right)$

if $u = \ln x + x$, then
$\frac{\mathrm{du}}{\mathrm{dx}} = \tan x + {e}^{x}$

if $v = \tan x + {e}^{x}$, then
$\frac{\mathrm{dv}}{\mathrm{dx}} = {\sec}^{2} x + {e}^{x}$
$\frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x} + {\sec}^{2} x$

$f \left(x\right) = u v$
$y = u v$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(y\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(u v\right)$

Thus,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(u v\right)$

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$
Substituting
d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x)