## calyne Group Title Differentiate f and find the domain of f. f(x) = x/[1-ln(x-1)] 2 years ago 2 years ago

1. experimentX Group Title

use quotient rule.

2. calyne Group Title

yeah. this i know.

3. calyne Group Title

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

4. campbell_st Group Title

I think it looks more like this u = x du/dx = 1 v = 1-ln(x -1) dv/dx = -1/(x-1) $dy/dx = (1 - \ln(x -1) + x/(x-1))/(1-\ln(x -1))^2$

5. calyne Group Title

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)

6. calyne Group Title

I need this explained to me.

7. calyne Group Title

still.

8. bmp Group Title

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. :-)

9. myininaya Group Title

$y=\frac{x}{1-\ln(x-1)}$ 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 :)

10. experimentX Group Title

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.