1. ## Find the derivative

f(x)=ln(ln(1-x))

Find f '(x)

2. Hi

The derivative of $\ln(u(x))$ is $\frac{u'(x)}{u(x)}$

Here $u(x) = \ln(1-x)$

Can you get it from here ?

3. well the thing is i already have to correct answer. its -1/(1-x)ln(1-x). but depending on the methods i have been using, i get both that and -1/(1-x)^2ln(1-x)

4. i get the -1/(x-1)^2ln(x-1) when i use the chain method

5. The derivative of $\ln(u(x))$ is $\frac{u'(x)}{u(x)}$

Here $u(x) = \ln(1-x) = \ln(v(x))$ where $v(x) = 1-x$

$u'(x) = \frac{v'(x)}{v(x)} = \frac{-1}{1-x}$

Therefore $f'(x) = \frac{\frac{-1}{1-x}}{\ln(1-x)} = -\frac{1}{(1-x)\ln(1-x)}$

6. Originally Posted by emory
i get the -1/(x-1)^2ln(x-1) when i use the chain method
Could you explain how you get this result ?

7. take the derivative of the outside, which is inside over derivative of inside. (-1/(1-x))/ln(1-x). Then take the derivative of the inside which is -1/(x-1) then derivative of that which is -1

8. Originally Posted by emory
take the derivative of the outside, which is inside over derivative of inside. (-1/(1-x))/ln(1-x).
This first part is actually 1/ln(1-x).

(fog)' = f'og . g'

f(x) = ln(x) and g(x) = ln(1-x)
f'(x) = 1/x and g'(x) = -1/(1-x)

f'og(x) = f'(g(x)) = 1/g(x) = 1/ln(1-x)

(fog)'(x) = f'og(x) . g'(x) = 1/ln(1-x) . (-1)/(1-x) = -1/[(1-x)ln(1-x)]

9. oh ok gotcha thanks a bunch