# What is the derivative of Ln(x)/(1+ln(2x))?

Nov 8, 2016

$\frac{d}{\mathrm{dx}} \ln \frac{x}{1 + \ln \left(2 x\right)} = \frac{1 + \ln \left(2 x\right) - \ln x}{x {\left(1 + \ln \left(2 x\right)\right)}^{2}}$

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{\ln x}{1 + \ln \left(2 x\right)}$ Then

$\left\{\begin{matrix}\text{Let "u=lnx & => & (du)/dx=1/xx \\ "And } v = 1 + \ln \left(2 x\right) & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + \ln \left(2 x\right)\right) \left(\frac{1}{x}\right) - \left(\ln x\right) \left(\frac{1}{x}\right)}{1 + \ln \left(2 x\right)} ^ 2$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{x}\right) \frac{1 + \ln \left(2 x\right) - \ln x}{1 + \ln \left(2 x\right)} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + \ln \left(2 x\right) - \ln x}{x {\left(1 + \ln \left(2 x\right)\right)}^{2}}$