# Two functions involving derivatives

1. Oct 7, 2008

### abbykeck

1. The problem statement, all variables and given/known data let f be a ffunction for which f(2)=-3 and f'(x)=square root of (x^2+5) if g(x)= x^2*f(x/(x-1)), what is g'(2)?

2. Relevant equations

3. The attempt at a solutioni am almost positive that i need to find the opposite of derivative (antiderivative?) for f'(x), so i can find f(x) and then substitute that into g(x) to find g'(x). however. i'm not sure how to do antiderivatives. i think it is something like square root of "?" +5x, but i don't know what the antideravitive of x^2 is. i would have guessed x^3, but that would yeild 3x^2 instead of x^2.

2. Oct 7, 2008

### dirk_mec1

You can integrate to find f then use f(2) =-3 to find the integration constant. Then just substitute 2 in the derative of g.

3. Oct 7, 2008

### abbykeck

Umm... if you don't mid, could you try to explaint to me what the intergration method is. i don't think i've heard of it. either that or i have forgotten it. is it dealing with algebra or calculus?

4. Oct 7, 2008

### CompuChip

Or, if you don't like integrating (or don't know what it is), you can just use the chain rule to write down the derivative of g and plug in x = -2.
Since you only do one derivation you will only get stuff with f(-2) and f'(-2), both of which you know.

dirk's method also works but I don't see the need to find f(x) first.

5. Oct 7, 2008

### abbykeck

but in order to find g(x), don't i have to find f(x)? since f(x) is in the function of g(x)? maybe that's what you were saying at the end of your explination, but i don't really understand it.

6. Oct 7, 2008

### gabbagabbahey

Suppose you were given h(x)=2f(x) taking the derivative of that would give you h'(x)=2f'(x) would it not? Similarly, if you were given h(x)=3f(3x^2) you would get h'(x)=3f'(3x^2)*(6x). So as long as you know f', you don't need to know f in order to determine g'.

7. Oct 7, 2008

### abbykeck

ok, so i think i know what your'e saying. so in my problem, would i end up with g'(x) = 2x*f'(-1/(x-1)^2)? then plug in 2 for all x's to get g'(2)!?!?!

8. Oct 7, 2008

### Hootenanny

Staff Emeritus

9. Oct 7, 2008

### ashleyrc

Uh oh... secret's out, yeah, i didn't realize i couldn't have multiple accounts. so "abbykeck" is technically still online...

10. Oct 7, 2008

### CompuChip

Yes, something like that. I mean: the approach is all right, but I don't really think you did the derivation well. You will need both the product rule and the chain rule!

11. Oct 7, 2008

### ashleyrc

wha am i using the chain rule for? the f(x/(x-1))? the product rule would be for x^2 and whatever that chain rule produces, correct? i'll try it out...

12. Oct 7, 2008

### CompuChip

g(x) is of the form a(x)*b(y(x))
so
g'(x) = a'(x) b(y(x)) + a(x) b'(y(x))
where you need the chain rule for b'(y(x)) = d/dx b(y(x))

13. Oct 7, 2008

### ashleyrc

g'(x)=2x(f(x/(x-1))+x^2(f(-1/(x-1)^2))?

14. Oct 7, 2008

### gabbagabbahey

Your first term is right, but you've incorrectly taken the derivative of f(x/(x-1)). Remember, the chain rule gives:

$$\frac{d}{dx} f(u)=f'(u) \frac{du}{dx}$$

In your case, $u=\frac{x}{x-1}$