# How do you find the derivative of y=2xe^x-2e^x?

Jun 4, 2016

I found: $y ' = 2 x {e}^{x}$

#### Explanation:

We can use the Product Rule to tackle $2 x {e}^{x}$ and remember that the derivative of ${e}^{x}$ is ${e}^{x}$ itself.
We get:
$y ' = \cancel{2 {e}^{x}} + 2 x {e}^{x} \cancel{- 2 {e}^{x}} = 2 x {e}^{x}$

Jun 4, 2016

$\frac{d y}{d x} = 2 {e}^{x} - 2 {e}^{x} l n \left(e\right) + 2 x {e}^{x} l n \left(e\right)$

#### Explanation:

(d y)/(d x)=2 x e^x-2 e^x=?

$\frac{d y}{d x} = 2 \cdot {e}^{x} + 2 x \cdot {e}^{x} \cdot l n \left(e\right) - 2 {e}^{x} \cdot l n \left(e\right)$

$\frac{d y}{d x} = 2 {e}^{x} - 2 {e}^{x} l n \left(e\right) + 2 x {e}^{x} l n \left(e\right)$