1. ## derivative 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]

2. Originally Posted by DINOCALC09
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]
You should go over the chain rule and derivatives of logarithms.

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

3. how does the double ln work [ln(ln(...]

4. Originally Posted by DINOCALC09
how does the double ln work [ln(ln(...]
If x = ln(ln(y)), then x is basically the power you put on e^e (not e squared, but e^e) to get y.

5. 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.

6. $\ln \ln x$ ?

you use the chain rule.

$u = \ln x$
$\frac{du}{dx} = \frac{1}{x}$

$\ln \ln x \rightarrow \ln u$

$\frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{du}{dx}$

$\frac {d}{dx} ( \ln u ) = \frac{1}{u} \cdot \frac{1}{x}$
$\frac {d}{dx} ( \ln \ln x ) = \frac{1}{x \ln x}$

7. $\mbox{derivative of} \ln (\mbox{ANYTHING}) = \frac 1{\mbox{ANYTHING}} \times \mbox{derivative of ANYTHING}$

Just a matter of applying the chain rule as others said. and of course, we assume 1/(anything) is defined and what not...

8. Originally Posted by DINOCALC09
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]
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 $(x+\ln(\sin(x^2+1)^3)))^4$

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: $4(Y)^3(...........)$. Now let's fill "............"
***Now I see what's inside, hey looks like some $x+ \ln Y$(This is not the old Y, I am using Y as a variable!). I say to myself, I know it's derivative, it's $1 + \frac1{Y}$, so I write down $4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(............))$.
***Proceeding like this I notice next up is sin Y and its derivative is cos Y. So now I can happily write $4(x+\ln(\sin(x^2+1)^3)))^3 (1 + \frac{1}{\sin(x^2+1)^3}(\cos(x^2+1)^3)(.........))$

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

9. 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.

10. Originally Posted by colby2152
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.

In fact, I used to explain it with the same vegetable!!!. That's really a strange co-incidence OR it's a common mathematical parlance

11. Originally Posted by DINOCALC09
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]
Instead takin' the derivative quickly, do the make-up:

$f(x) = \ln (\ln (1 - x))\iff e^{f(x)} = \ln (1 - x).$

Now you can easily contemplate its derivative

$f'(x)e^{f(x)} = - \frac{1}
{{1 - x}}\,\therefore \,f'(x) = - \frac{1}
{{(1 - x)\ln (1 - x)}}.$

And we're done.