If f(x) = ln(ln(1-x)), then f'(x) =
a) -1/ln(1-x)
b) 1/[(1-x)ln(1-x)]
c) 1/[(1-x)^2]
d) -1/[(1-x)ln(1-x)]
e) -1/[ln(1-x)^2]
$\displaystyle \ln \ln x $ ?
you use the chain rule.
$\displaystyle u = \ln x$
$\displaystyle \frac{du}{dx} = \frac{1}{x}$
$\displaystyle \ln \ln x \rightarrow \ln u$
$\displaystyle \frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{du}{dx} $
$\displaystyle \frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{1}{x} $
$\displaystyle \frac {d}{dx} ( \ln \ln x ) = \frac{1}{x \ln x} $
Ya others already said it, but if you are new to this,solve any such problem like this :
When you see ln (>something here<), at first step dont worry about >something here< just write 1/>something here< on paper. Now look at that >somethin here< it could be another function, say sin(>more something<), again forget what ever is inside parentheses just write derivative of sin (that is cos....) Keep on proceeding until you run out of functions.
Example:
say you are asked to differentiate $\displaystyle (x+\ln(\sin(x^2+1)^3)))^4$
Yuck! Ugly function but don't worry...
Start Identifying functions from the >outermost< parts.
What I do?
***I tell myself, hey looks like some Y^4 ,so I write down: $\displaystyle 4(Y)^3(...........)$. Now let's fill "............"
***Now I see what's inside, hey looks like some $\displaystyle x+ \ln Y$(This is not the old Y, I am using Y as a variable!). I say to myself, I know it's derivative, it's $\displaystyle 1 + \frac1{Y}$, so I write down $\displaystyle 4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(............))$.
***Proceeding like this I notice next up is sin Y and its derivative is cos Y. So now I can happily write $\displaystyle 4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(\cos(x^2+1)^3)(.........))$
Hope you get the idea
I know it looks complicated but it is just a matter of practice. So can you complete my answer ??
Hope you do well
Instead takin' the derivative quickly, do the make-up:
$\displaystyle f(x) = \ln (\ln (1 - x))\iff e^{f(x)} = \ln (1 - x).$
Now you can easily contemplate its derivative
$\displaystyle f'(x)e^{f(x)} = - \frac{1}
{{1 - x}}\,\therefore \,f'(x) = - \frac{1}
{{(1 - x)\ln (1 - x)}}.$
And we're done.