# Thread: differentiate and find the domain of f

1. ## differentiate and find the domain of f

Hello, I was able to do part one but have problem finding the domain

f(x) = x / [1-ln(x-1)]

The solution says:

x<=1 and ln(x-1) = 1

x<=1 and ln(x-1) = ln e

x<=1 and x-1 = ln e
--> Here I do not understand how to get from ln(x-1) to x-1

x<=1 and x = (e + 1) --> I do not get this step too

Can someone explain that to me. Thanks

2. Originally Posted by DBA
Hello, I was able to do part one but have problem finding the domain

f(x) = x / [1-ln(x-1)]

The solution says:

x<=1 and ln(x-1) = 1

x<=1 and ln(x-1) = ln e

x<=1 and x-1 = ln e
--> Here I do not understand how to get from ln(x-1) to x-1

x<=1 and x = (e + 1) --> I do not get this step too

Can someone explain that to me. Thanks
$\displaystyle f'(x)=\frac{1\cdot[1-\ln(x-1)]-x\cdot\frac{-1}{x-1}}{[1-\ln(x-1)]^2}=\frac{(x-1)[1-\ln(x-1)]+x}{(x-1)[1-\ln(x-1)]^2}$

Note that

$\displaystyle \ln(x-1)$ implies $\displaystyle x>1$

And the denominator cannot be equal to zero, which mens that

$\displaystyle 1-\ln(x-1)\neq0$

But

$\displaystyle 1-\ln(x-1)=0$ occurs when

$\displaystyle x-1=e$

$\displaystyle x=e+1$

therfore, the domain of the f prime is

All $\displaystyle x>1\neq{e+1}$