# How do you differentiate g(y) =e^x(x^3-1)  using the product rule?

Dec 27, 2015

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

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

#### Explanation:

The product rule states that

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f \left(x\right) \cdot g ' \left(x\right) + g \left(x\right) \cdot f ' \left(x\right)$.

Application of this rule in this particular function yields

Case 1 - g(y) as written in the question

The function only contains x terms and so the derivative $g ' \left(y\right) = 0$.

Case 2 - Assuming g(x) as then the product rule becomes applicable

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

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