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Calculus
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 =/ |
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.
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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 |
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 |
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 |
Re: Calculus
Guido's way also works.
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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 |
Re: Calculus
I fixed it so that my way works now, before I saw guido's answer.
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Re: Calculus
Actually Doug, your way allows for 1 and -1 to be answers =P
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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.:)
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