# How do you find the derivative of 2e^(sqrtx)?

Jun 26, 2016

$= {e}^{\sqrt{x}} / \left(\sqrt{x}\right)$

#### Explanation:

using the chain rule you know that $\left(\alpha {e}^{f \left(x\right)}\right) ' = \alpha f ' \left(x\right) {e}^{f \left(x\right)}$

or you can see for yourself by noting that if

$y = \alpha {e}^{f \left(x\right)}$

then

$\frac{y}{\alpha} = {e}^{f \left(x\right)}$

and

$\ln \left(\frac{y}{\alpha}\right) = f \left(x\right)$

so $\left(\ln \left(\frac{y}{\alpha}\right)\right) ' = f ' \left(x\right)$

ie $\frac{1}{\frac{y}{\alpha}} \left(\frac{1}{\alpha}\right) y ' = f ' \left(x\right)$

so $y ' = \left(y f ' \left(x\right)\right) = \alpha f ' \left(x\right) {e}^{f \left(x\right)}$

here that means that

$\left(2 {e}^{\sqrt{x}}\right) ' = 2 {e}^{\sqrt{x}} \cdot \left(\sqrt{x}\right) '$

$= 2 {e}^{\sqrt{x}} \cdot \frac{1}{2} \frac{1}{\sqrt{x}}$

$= {e}^{\sqrt{x}} / \left(\sqrt{x}\right)$