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]

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- Dec 26th 2007, 10:10 AMDINOCALC09derivative involving "ln"
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] - Dec 26th 2007, 10:24 AMcolby2152
- Dec 26th 2007, 10:38 AMDINOCALC09
how does the double ln work [ln(ln(...]

- Dec 26th 2007, 10:49 AMcolby2152
- Dec 26th 2007, 10:51 AMTKHunny
All 'ln's work the same. It's just an application of the chain rule.

[f(g(h(x)))]' = f'(g(h(x)))*g'(h(x))*h'(x)

If f, g, and h are all the same, it just looks funny, it doesn't work any differently. - Dec 26th 2007, 10:54 AMbobak
$\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} $ - Dec 26th 2007, 10:59 AMJhevon
$\displaystyle \mbox{derivative of} \ln (\mbox{ANYTHING}) = \frac 1{\mbox{ANYTHING}} \times \mbox{derivative of ANYTHING}$

Just a matter of applying the chain rule as others said. and of course, we assume 1/(anything) is defined and what not... - Dec 26th 2007, 11:23 AMIsomorphism
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 :D

I know it looks complicated but it is just a matter of practice. So can you complete my answer ??

Hope you do well :) - Dec 26th 2007, 11:49 AMcolby2152
Good explanation of the chain rule guys! I always liked using the word "junk" when teaching the chain rule. Junk can contain a carrot, and you would have to take the derivative of carrot and multiply that by junk.

- Dec 26th 2007, 12:22 PMIsomorphism
- Dec 26th 2007, 12:55 PMKrizalid
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.