# What is the second derivative of f(x)= ln sqrt(xe^x)?

Jan 6, 2016

You'll need product rule and chain rule, as well as acknowledging that the derivative of $\ln x = \frac{1}{x}$.

#### Explanation:

• Product rule: $\left(a b\right) ' = a ' b + a b '$
• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

renaming $u = x {e}^{x}$ and $v = \sqrt{u}$, we can proceed:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{v} \frac{1}{2 \sqrt{u}} \left({e}^{x} + x {e}^{x}\right) = \frac{{e}^{x} \left(1 + x\right)}{2 v \sqrt{u}}$

Substituting $v$, then $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} \left(1 + x\right)}{2 \sqrt{u} \sqrt{u}} = \frac{{e}^{x} \left(1 + x\right)}{2 u} = \frac{\cancel{{e}^{x}} \left(1 + x\right)}{2 x \cancel{{e}^{x}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + x}{2 x}$

The second derivative will demand quotient rule: $\left(\frac{a}{b}\right) ' = \frac{a ' b - a b '}{b} ^ 2$

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