
#1
Oct112, 11:44 AM

P: 459

1. The problem statement, all variables and given/known data
f'(u) = 1 / (1 + u^3) g(x) = f(x^2) Find g'(x) and g'(2) 2. Relevant equations 3. The attempt at a solution So the derivative of function f at u is: 1 / (1 + u^3) That means g'(x) would be f'(x^2), but to find the general derivative of f at u is 1 / (1 + u^3) so can I just plug in x^2 for u so I get: 1 / (1 + x^6) ? 



#2
Oct112, 12:33 PM

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Your subject has nothing to do with this problem. There are no antiderivatives here. You need to use the chain rule to compute g'(x). It is NOT true that g'(x) = f'(x^2).




#3
Oct112, 12:51 PM

P: 459

Don't you need to find f(u) which would be going backwards from 1 / (1 + u^3), which is an antiderivative?
And why is g'(x) != f'(x^2)? I just took the derivative of both functions on each side with respect to x 



#4
Oct112, 01:24 PM

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Finding antiderivative without integration 



#5
Oct112, 02:04 PM

P: 459

We learned the Chain Rule but not sure how to apply it here. My best guess would be:
g'(x) = f'(x) * 2x 



#6
Oct112, 02:10 PM

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#7
Oct112, 03:56 PM

P: 459

Sorry, that's what I meant. I write it down on paper but I'm not very good at typing
So for g'(x) I have f'(x^2) * 2x Do I have to somehow use f'(u) = 1 / (1 + u^3) in order to find the values for 'x' for f'(x^2) * 2x ? Do I need to use the Chain Rule again? If so, how would I apply it in this situation? 



#8
Oct112, 03:58 PM

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#10
Oct112, 08:07 PM

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