# Math Help - Derivative involving ln

1. ## Derivative involving ln

I need help with this problem.

find the derivative of f(u) = ln(u)/1 + ln(2u)

I started by finding ln of ln(u) and 1 + ln(2u) . . .

I got ln(u) = 1/u and 1 + ln(2u) = 1. Then I just used the quotient rule, but I can't get the right answer. The answer is supposed to be 1 + ln2/ u[1 + ln(2u)]^2.

2. Originally Posted by zachb
I need help with this problem.

find the derivative of f(u) = ln(u)/1 + ln(2u)

I started by finding ln of ln(u) and 1 + ln(2u) . . .

I got ln(u) = 1/u and 1 + ln(2u) = 1. Then I just used the quotient rule, but I can't get the right answer. The answer is supposed to be 1 + ln2/ u[1 + ln(2u)]^2.

Using the quotient rule, which is

d/dx f(x)/g(x) = [g(x)f '(x) - f(x)g'(x)]/g(x)^2

f(u) = ln(u)/1 + ln(2u)
=> f '(u) = {[1 + ln(2u)](1/u) - ln(u)[1/u]}/(1 + ln(2u))^2
=> f '(u) = {(1 + ln(2u) - ln(u))/u}/(1 + ln(u))^2
=> f '(u) = [1 + ln(2u) - ln(u)]/u(1 + ln(u))^2
=> f '(u) = [1 + ln2]/u(1 + ln(u))^2 ........................we got to this step by combining the ln's. remember, lnx - lny = ln(x/y)

where did you make the mistake? first off, the derivative of ln(2u) is 1/u by the chain rule. if this was not your fatal mistake, perhaps you used the quotient rule incorrectly

3. firstly you write down the derivatives of ln(u) and 1 + ln(2u);

d/du ln(u) = 1/u
d/du 1+ ln(2u) = d/du 1+ ln(u) + ln(2) ;by using the rule log[ab]=log[a]+log[b]
= 1/u

then quotient rule you get;
[1/u[1+ln(2u)] - (ln(u))(1/u) ]/(1+ln2u)^2

factorise;
[1/u(1+ln(2u)-ln(u) )]/(1+ln2u)^2

again use a log rule; ln[a]-ln[b] = ln[a/b] and put the 1/u as a u down the bottom;
[1+ln(2u/u)]/[u(1+ln2u)^2]
=[1+ln(2)]/[u(1+ln2u)^2]

which is your result.