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

Mar 15, 2016

Just chain rule over and over again.

$f ' \left(x\right) = {e}^{x} \frac{1 + x}{4} \sqrt{\frac{x {e}^{x}}{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}}$

#### Explanation:

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

Okay, this is gonna be hard:

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

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

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

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

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

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

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

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

$= \frac{\sqrt{x {e}^{x}}}{4 \sqrt{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right)}} \frac{1}{\sqrt{{\left(x {e}^{x}\right)}^{3}}} \left(x {e}^{x}\right) ' =$

$= \frac{\sqrt{x {e}^{x}}}{4 \sqrt{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}} \left(x {e}^{x}\right) ' =$

$= \frac{1}{4} \sqrt{\frac{x {e}^{x}}{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}} \left(x {e}^{x}\right) ' =$

$= \frac{1}{4} \sqrt{\frac{x {e}^{x}}{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}} \left[\left(x\right) ' {e}^{x} + x \left({e}^{x}\right) '\right] =$

$= \frac{1}{4} \sqrt{\frac{x {e}^{x}}{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}} \left({e}^{x} + x {e}^{x}\right) =$

$= {e}^{x} \frac{1 + x}{4} \sqrt{\frac{x {e}^{x}}{\ln \left(\frac{1}{\sqrt{x {e}^{x}}}\right) {\left(x {e}^{x}\right)}^{3}}}$

P.S. These exercises should be illegal.