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

Mar 4, 2017

$f ' ' \left(x\right) = - \frac{1}{2 {x}^{2}}$

#### Explanation:

We have:

$f \left(x\right) = \ln \left(\sqrt{\frac{x}{e} ^ x}\right)$

which, using the rule of logs, we can write as:

$f \left(x\right) = \ln \left({\left(\frac{x}{e} ^ x\right)}^{\frac{1}{2}}\right)$
$\text{ } = \frac{1}{2} \ln \left(\frac{x}{e} ^ x\right)$
$\text{ } = \frac{1}{2} \left\{\ln \left(x\right) - \ln \left({e}^{x}\right)\right\}$
$\text{ } = \frac{1}{2} \left\{\ln \left(x\right) - x\right\}$

Differentiating wrt $x$ gives:

$f ' \left(x\right) = \frac{1}{2} \left(\frac{1}{x} - 1\right)$

Differentiating a second time wrt $x$ gives:

$f ' ' \left(x\right) = \frac{1}{2} \left(- \frac{1}{x} ^ 2\right)$
$\text{ } = - \frac{1}{2 {x}^{2}}$