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

Mar 17, 2018

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

or

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

Explanation:

Start:

$f \left(x\right) = \left({e}^{x} + x\right) \left({x}^{2} - 3\right)$

Product rule applied:

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

Rearrange if you wish:

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

$= \left({x}^{2} + 2 x - 3\right) {e}^{x} + 3 \left({x}^{2} - 1\right)$