Calculus

[masterhickle]

This problem was from my daily test today in calculus. Even after seeing it worked out, I still don't quite understand it >.>

Here's the problem:
f(x) = (g(x))^5
g(2) = -1
f '(2) = 5
g '(2 = ?

Hopefully someone on here can help me to understand it =/

[thyraxicmole]

Re: Calculus

i forgot the rules i learned in calculus otherwise that problem doesnt really look so hard, i could have probabli done it in a flash if i were still in my old class.

[masterhickle]

Re: Calculus

Im probably forgetting something extremely simple to this problem, but I just can't understand how to work it out >.>

50 credits says I feel retarded for not knowing how to do the problem

[Doug31]

Re: Calculus

Well, f'(x) = 5(g(x))^4*g'(x), right? So, 5(-1)^4*5=25

Edit: I think I answered the wrong question.

g'(2)=5

Edit2: What have I done? I don't even remember how I came up with g'(2)=5 in that first part.

Edit3: Okay, I think I fixed it now. f'(x) = 5(g(x))^4*g'(x) so 5=5(g'(2))^4, so 1=(g'(2))^4, so g'(2)=1

[GuidoHunter]

Re: Calculus

Thanks for your astounding help, thyrax.

First, differentiate the given expression (make note of the chain rule)

f'(x) = [5(g(x))^4]g'(x)

Then just substitute in your values for x = 2

f'(2) = [5(g(2))^4]g'(2)

5 = [5(-1)^4]g'(2)

5 = 5(1)g'(2)

g'(2) = 1

--Guido

http://andy.mikee385.com

[Doug31]

Re: Calculus

Guido's way also works.

[masterhickle]

Re: Calculus

Haha, wow, I realized I was forgetting something extremely simple >.> I failed to realize all I had to do, after the chain rule, was simply substitute in a 2 for each x value.

Thanks Guido for the work, and Doug31 for at least trying =D

[Doug31]

Re: Calculus

I fixed it so that my way works now, before I saw guido's answer.

[masterhickle]

Re: Calculus

Actually Doug, your way allows for 1 and -1 to be answers =P

[Doug31]

Re: Calculus

I know, I often get answers that are negative that I know should be positive, but then I just get rid of the negative sign.