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

Nov 15, 2016

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

#### Explanation:

${e}^{x} > 0$. So, for ln to be real, $x {e}^{x} > 0 \to x > 0$

Also, for sqrt to be real, ln > 0, .

So, $x {e}^{x} > 1$.

By a numerical method,

x < 0.567, nearly.

Thus, f(x) is defined for x in ( 0, 0.567 )

Use ln(1/(xe^x))=-ln(xe^x)=-lnx - ln(e^x=-ln x - x#

For this x,

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

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

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

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