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

May 26, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = - {e}^{\left(\frac{\sqrt{- \ln x}}{\sqrt{2}}\right)} / \left(2 \sqrt{2} x \sqrt{- \ln x}\right)$

#### Explanation:

Note that, using the properties of logarithms:

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

so:

e^sqrt(ln(1/sqrtx)) = e^sqrt(-1/2lnx) = e^((sqrt(-lnx)/sqrt2)

Pose:

$y \left(x\right) = \frac{\sqrt{- \ln x}}{\sqrt{2}}$

We have:

$f \left(y \left(x\right)\right) = {e}^{y \left(x\right)}$

so:

$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\left(\frac{\sqrt{- \ln x}}{\sqrt{2}}\right)} \frac{d}{\mathrm{dx}} \left(\frac{\sqrt{- \ln x}}{\sqrt{2}}\right)$

Similarly with $y \left(x\right) = - \ln \left(x\right)$:

$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{\left(\frac{\sqrt{- \ln x}}{\sqrt{2}}\right)} / \left(2 \sqrt{2} \sqrt{- \ln x}\right) \frac{d}{\mathrm{dx}} \left(- \ln x\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = - {e}^{\left(\frac{\sqrt{- \ln x}}{\sqrt{2}}\right)} / \left(2 \sqrt{2} x \sqrt{- \ln x}\right)$