# Differentiate [ln(u)]/[1+ln(2u)]

1. Jul 29, 2008

### illjazz

1. The problem statement, all variables and given/known data
Differentiate the function

$$f(u)=\frac{\ln u}{1+\ln(2u)}$$

2. Relevant equations
- Quotient rule?
- Logarithmic differentiation?

3. The attempt at a solution
I just learned about logarithmic differentiation so I think the idea here was to use logarithmic differentiation.. however, with ln's on the top and bottom, I was unsure of how to go about that and used the quotient rule instead, which got nasty.

$$f(u)=\frac{\ln u}{1+\ln(2u)}$$

$$f'(u)=\frac{(1+\ln(2u))\frac{d}{du}\ln u-\ln u\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^2}$$

$$\frac{(1+\ln(2u))\frac{1}{u}-\ln u(\frac{2}{2u})}{(1+2\ln(2u)+(\ln(2u))^2}$$

$$\frac{\frac{1+\ln(2u)}{u}-\frac{\ln u}{u}}{(1+2\ln(2u)+(\ln(2u))^2}$$

$$\frac{1+\ln(2u)-\ln u}{u(1+2\ln(2u)+(\ln(2u))^2)}$$

The book gives

$$f'(u)=\frac{1+\ln2}{u[1+\ln(2u)]^2}$$

Oh my god that was a **** to type up :(

2. Jul 29, 2008

### Integral

Staff Emeritus
note that ln(2u) = ln2 + lnu

3. Jul 29, 2008

### illjazz

Thanks! I actually did not know that.. logs aren't my strength :/

That helps with the numerator, but not with the denumerator. Continuing from the last step:

$$\frac{1+\ln 2 + \ln u-\ln u}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)}$$

$$\frac{1+\ln 2}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)}$$

4. Jul 29, 2008

### Integral

Staff Emeritus
They did not expand the square, look at your first step.

5. Jul 29, 2008

### illjazz

Ah! Brilliant! Thank you! So I did get it :). Much appreciated.