calculus problem?

Question

differentiate the function:
f(u)= ln u / (1+ln(2u))

The answer is f'(u)= (1+ln2) / u(1+ln(2u))^2
but I have no idea how to get it!! Thanks!

Answer 1

Differentiate the function:
f(u) = (ln u) / (1 + ln(2u))
________

Use the rule for differentiating quotients.

f(u) = g(u) / h(u)
f'(u) = [g'(u)h(u) - g(u)h'(u)] / h²(u)

For the problem at hand we have:

f(u) = (ln u) / (1 + ln(2u))

f'(u) = [(1/u)(1 + ln(2u)) - (ln u)(2/(2u))] / (1 + ln(2u))²
f'(u) = [(1/u)(1 + ln(2u)) - (ln u)(1/u)] / (1 + ln(2u))²
f'(u) = [(1 + ln(2u)) - (ln u)] / [u(1 + ln(2u))²]
f'(u) = [(1 + ln(2u/u)] / [u(1 + ln(2u))²]
f'(u) = [(1 + ln(2)] / [u(1 + ln(2u))²]

Answer 2

Let's use the Quotient Rule?

f'(u) = (bottom d top - top d bottom)/bottom^2

= ((1 + ln(2u))*(1/u) - (ln u * 2/2u)))/(1 + ln(2u))^2
=((1 + ln(2u)) - (lnu))/u(1 + ln(2u))^2 by multiplying throughout top and bottom by u
=(1+ln(2u/u))/u(1 + ln(2u))^2 by laws of logs
=(1 + ln 2)/(u(1+ln(2u))^2