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

Dec 31, 2015

The product rule $\left(u v\right) ' = u v ' + v u '$ where $u$ and $v$ are functions of $x$ Our answer would be $f ' \left(x\right) = \left(x - 3 \ln \left(x\right)\right) \left({\sec}^{2} \left(x\right) + 2\right) + \left(\tan \left(x\right) + 2 x\right) \left(1 - \frac{3}{x}\right)$
$f \left(x\right) = \left(x - 3 \ln \left(x\right)\right) \left(\tan \left(x\right) + 2 x\right)$
$f ' \left(x\right) = \left(x - 3 \ln \left(x\right)\right) \frac{d}{\mathrm{dx}} \left(\tan \left(x\right) + 2 x\right) + \left(\tan \left(x\right) + 2 x\right) \frac{d}{\mathrm{dx}} \left(x - 3 \ln \left(x\right)\right)$
f'(x) = (x-3ln(x)){d/dx(tan(x) + d/dx(2x)} + (tan(x)+2x){d/dx(x) - d/dx(3ln(x))}
$f ' \left(x\right) = \left(x - 3 \ln \left(x\right)\right) \left({\sec}^{2} \left(x\right) + 2\right) + \left(\tan \left(x\right) + 2 x\right) \left(1 - \frac{3}{x}\right)$ Answer