# How do you use the chain rule to differentiate 1/ln(4x)?

Mar 28, 2017

Let $u = 4 x$, $v = \ln \left(u\right) = \ln \left(4 x\right)$.

Differentiating the above equation yields

$\frac{\mathrm{du}}{\mathrm{dx}} = 4$
$\frac{\mathrm{dv}}{\mathrm{du}} = \frac{1}{u}$

Then,

$\frac{1}{\ln} \left(4 x\right) = \frac{1}{v}$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{\ln} \left(4 x\right)\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{v}\right)$

$= \frac{d}{\mathrm{dv}} \left(\frac{1}{v}\right) \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$= \left(- \frac{1}{v} ^ 2\right) \cdot \left(\frac{1}{u}\right) \cdot \left(4\right)$

$= \left(- \frac{1}{\ln \left(4 x\right)} ^ 2\right) \cdot \left(\frac{1}{4 x}\right) \cdot \left(4\right)$

$= - \frac{1}{x {\left(\ln \left(4 x\right)\right)}^{2}}$