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

Mar 19, 2016

Chain rule states that if $f \left(g \left(h \left(x\right)\right)\right)$,

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

Hence, here for $f \left(x\right) = \ln \left(\frac{1}{\sqrt{{e}^{x} - x}}\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} =$

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

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

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