# How do you differentiate f(x)=1/(x^sqrt(x-3)) using the chain rule?

Nov 19, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{- \sqrt{x - 3}} \left(\frac{\sqrt{x - 3}}{x} + \ln \frac{x}{2 \sqrt{x - 3}}\right)$

#### Explanation:

So we have

$y = {x}^{- \sqrt{x - 3}}$

Taking the log of both sides we have

$\ln \left(y\right) = - \sqrt{x - 3} \cdot \ln \left(x\right)$

Derivating we have

$\frac{d}{\mathrm{dx}} \ln \left(y\right) = - \sqrt{x - 3} \frac{d}{\mathrm{dx}} \ln \left(x\right) - \ln \left(x\right) \frac{d}{\mathrm{dx}} \sqrt{x - 3}$

Using the chain rule on the left hand side we have

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{x - 3} \frac{d}{\mathrm{dx}} \ln \left(x\right) - \ln \left(x\right) \frac{d}{\mathrm{dx}} \sqrt{x - 3}$

Isolating $\frac{\mathrm{dy}}{\mathrm{dx}}$ and solving the other, simple derivatives, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{x - 3} \cdot \frac{1}{x} \cdot {x}^{- \sqrt{x - 3}} - \ln \left(x\right) \cdot \left(\frac{1}{2 \sqrt{x - 3}}\right) \cdot {x}^{- \sqrt{x - 3}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{- \sqrt{x - 3}} \left(\frac{\sqrt{x - 3}}{x} + \ln \frac{x}{2 \sqrt{x - 3}}\right)$

It's important to notice we can't simply say $- \sqrt{x - 3} = u$ and do

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} {x}^{u} \frac{\mathrm{du}}{\mathrm{dx}} = u {x}^{u - 1} \frac{\mathrm{du}}{\mathrm{dx}}$

Because that only works for constant values on the exponent! Nor can we say

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} {x}^{u} \frac{\mathrm{du}}{\mathrm{dx}} = \ln \left(x\right) {x}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$

Because that only works when the base is a constant! If both base and exponent change, you need to use implicit differentiation with logs.