# How do you differentiate e^(2x^2+x)  using the chain rule?

Jun 17, 2017

$\frac{d}{\mathrm{dx}} = {e}^{2 {x}^{2} + x} \left(4 x + 1\right)$

#### Explanation:

The derivative of ${e}^{x}$ is itself.
When you differentiate ${e}^{x}$ you take the derivative of $x$ basically you take the derivative of whatever $x$ may be.

$\frac{d}{\mathrm{dx}} = {e}^{2 {x}^{2} + x} = {e}^{2 {x}^{2} + x}$

Then take the derivative of what's inside:

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 1\right) = 4 x + 1$

Now you multiply them:

$\frac{d}{\mathrm{dx}} = {e}^{2 {x}^{2} + x} \left(4 x + 1\right)$

$\frac{d}{\mathrm{dx}} = 4 {e}^{2 {x}^{2} + x} x + {e}^{2 {x}^{2} + x}$

Put you can simply leave it as:

$\frac{d}{\mathrm{dx}} = {e}^{2 {x}^{2} + x} \left(4 x + 1\right)$