# What is the second derivative of f(x)= e^sqrt(3x-7)?

Jan 12, 2016

$f ' ' \left(x\right) = \frac{\left(\left(2 \sqrt{3 x - 7}\right) \cdot \frac{9}{2} {e}^{\sqrt{3 x - 7}} {\left(3 x - 7\right)}^{- \frac{1}{2}}\right) - 3 {e}^{\sqrt{3 x - 7}} \cdot 3 {\left(3 x - 7\right)}^{- \frac{1}{2}}}{4 \left(3 x - 7\right)}$

#### Explanation:

By use of the chain rule, power rule, and quotient rule, we get :

$f ' \left(x\right) = {e}^{\sqrt{3 x - 7}} \cdot \frac{1}{2} {\left(3 x - 7\right)}^{- \frac{1}{2}} \cdot \left(3\right)$

$= \frac{3 {e}^{\sqrt{3 x - 7}}}{2 \sqrt{3 x - 7}}$.

$\therefore f ' ' \left(x\right) = \frac{\left(\left(2 \sqrt{3 x - 7}\right) \cdot \frac{9}{2} {e}^{\sqrt{3 x - 7}} {\left(3 x - 7\right)}^{- \frac{1}{2}}\right) - 3 {e}^{\sqrt{3 x - 7}} \cdot 3 {\left(3 x - 7\right)}^{- \frac{1}{2}}}{4 \left(3 x - 7\right)}$