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

Sep 18, 2016

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

#### Explanation:

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

$f ' \left(x\right) = \frac{1}{\sqrt{{e}^{x} - x}} \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{{e}^{x} - x}\right)$ (Standard differential and chain rule)

$= \frac{1}{\sqrt{{e}^{x} - x}} \cdot \frac{1}{2} {\left({e}^{x} - x\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left({e}^{x} - x\right)$ (Power rule and chain rule)

$= \frac{1}{\sqrt{{e}^{x} - x}} \cdot \frac{1}{2 \left(\sqrt{{e}^{x} - x}\right)} \cdot \left({e}^{x} - 1\right)$ (Standard differential and power rule)

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