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

Apr 22, 2018

If $g \left(y\right)$: $g ' \left(y\right) = 0$

If $g \left(x\right)$: $g ' \left(x\right) = \left(2 {x}^{3} + 7 x + 1\right) {e}^{x}$

Explanation:

$g \left(y\right) = \left(2 {x}^{3} + x\right) {e}^{x}$

$g ' \left(y\right) = 0$

If you meant $g \left(x\right) = \left(2 {x}^{3} + x\right) {e}^{x}$, then

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\left(2 {x}^{3} + x\right)\right] {e}^{x} + \left(2 {x}^{3} + x\right) \frac{d}{\mathrm{dx}} \left[{e}^{x}\right]$

$g ' \left(x\right) = \left(6 x + 1\right) {e}^{x} + \left(2 {x}^{3} + x\right) {e}^{x}$

$g ' \left(x\right) = \left(6 x + 1 + 2 {x}^{3} + x\right) {e}^{x}$

$g ' \left(x\right) = \left(2 {x}^{3} + 7 x + 1\right) {e}^{x}$