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

Dec 23, 2015

Using two power rules, we can simplify the expression and then proceed to derivate it more simply.

#### Explanation:

Two power rules to be remembered:

• ${a}^{- n} = \frac{1}{{a}^{n}}$
• ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$

Rewriting the expression: $f \left(x\right) = {\left(x {e}^{x} - x\right)}^{- \frac{1}{2}}$, then.

Using the chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, let's rename $u = x {e}^{x} - x$ and derivate it:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = - \left(\frac{1}{2}\right) {u}^{- \frac{3}{2}} \left({e}^{x} + x {e}^{x} - 1\right) = - \frac{{e}^{x} + x {e}^{x} - 1}{2 {\left(x {e}^{x} - x\right)}^{\frac{3}{2}}}$

Just organizing:

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