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

May 18, 2017

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

Explanation:

We have: $f \left(x\right) = \left({x}^{2} + \sin \left(x\right)\right) \left({e}^{x} - \cos \left(x\right)\right)$

$R i g h t a r r o w f ' \left(x\right) = \left({x}^{2} + \sin \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{x} - \cos \left(x\right)\right) + \left({e}^{x} - \cos \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + \sin \left(x\right)\right)$

$R i g h t a r r o w f ' \left(x\right) = \left({x}^{2} + \sin \left(x\right)\right) \left({e}^{x} + \sin \left(x\right)\right) + \left({e}^{x} - \cos \left(x\right)\right) \left(2 x + \cos \left(x\right)\right)$

$R i g h t a r r o w f ' \left(x\right) = {x}^{2} {e}^{x} + {x}^{2} \sin \left(x\right) + \sin \left(x\right) {e}^{x} + {\sin}^{2} \left(x\right) + 2 x {e}^{x} + \cos \left(x\right) {e}^{x} - 2 x \cos \left(x\right) - {\cos}^{2} \left(x\right)$