# How do you differentiate  f(x)=1/sqrt(ln(xe^x)) using the chain rule.?

Jul 10, 2016

$f ' = \left(- \frac{1}{2}\right) \left(1 + x\right) {e}^{x} {f}^{3} , x \ge 0.5672$, nearly.

#### Explanation:

The function is differentiable for $x {e}^{x} > 1$ to make $\sqrt{\ln} \left(x {e}^{x}\right)$

value real. Numerical, method(s) give the corresponding value for x

to be $\ge 0.5672$, nearly..

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

$= \left(- \frac{1}{2}\right) \left(1 + x\right) {e}^{x} {f}^{3}$