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

[illjazz]

Homework Statement

Differentiate the function

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

Homework Equations

- Quotient rule?
- Logarithmic differentiation?

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 :(

 

[Integral]

note that ln(2u) = ln2 + lnu

 

[illjazz]

note that ln(2u) = ln2 + lnu

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)}

 

[Integral]

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

 

[illjazz]

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

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