Question

Differentiate f and find the domain of f.
f(x) = x/[1-ln(x-1)]

Answer 1

use quotient rule.

Answer 2

yeah. this i know.

Answer 3

i got [2-ln(x-1)]/[(1-ln(x-1))^2] but that's wrong

Answer 4

I think it looks more like this

u = x du/dx = 1

v = 1-ln(x -1) dv/dx = -1/(x-1)

\[dy/dx = (1 - \ln(x -1) + x/(x-1))/(1-\ln(x -1))^2\]

Answer 5

The answer in the texbtook is f'(x) = [2x - 1 - (x - 1) ln(x - 1)] / [(x - 1)[1 - ln(x - 1)]^2]; domain of f = (1, 1+e) U (1 + e, infinity)

Answer 6

I need this explained to me.

Answer 7

still.

Answer 8

Optionally, use the reverse multiplication rule. Say:
f(x) = x*(1-ln(x-1))^(-1)
Quotient rule tend to be quite messy. But can you show what you did so far? I may be able to guide you through it. :-)

Answer 9

\[y=\frac{x}{1-\ln(x-1)}\]
You could do log diff.
\[\ln(y)=\ln(\frac{x}{1-\ln(x-1)})\]
\[\ln(y)=\ln(x)-\ln(1-\ln(x-1))\]
Now diff both sides
\[\frac{y'}{y}=\frac{1}{x}-\frac{0-\frac{1}{x-1}}{1-\ln(x-1)}\]
So Simplifying right hand side first we have
\[\frac{y'}{y}=\frac{1}{x}-\frac{-1}{(x-1)(1-\ln(x-1))}\]
Now multiply y on both sides
\[y'=y(\frac{1}{x}+\frac{1}{(x-1)(1-\ln(x-1))})\]
And replace y with
\[\frac{x}{1-\ln(x-1)}\]
So this is another way :)

Answer 10

http://www.wolframalpha.com/input/?i=differentiate+x%2F%5B1-ln%28x-1%29%5D

good old classic way
click on <show steps> to take a peek at the steps.