Finding antiderivative without integration


by PhizKid
Tags: antiderivative, integration
PhizKid
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#1
Oct1-12, 11:44 AM
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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) ?
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jbunniii
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Oct1-12, 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).
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Oct1-12, 12:51 PM
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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

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Oct1-12, 01:24 PM
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Finding antiderivative without integration


Quote Quote by PhizKid View Post
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
No, you don't need to find f(u). Regarding why g'(x) != f'(x^2), please apply the chain rule to differentiate both sides of g(x) = f(x^2).
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Oct1-12, 02:04 PM
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We learned the Chain Rule but not sure how to apply it here. My best guess would be:

g'(x) = f'(x) * 2x
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Oct1-12, 02:10 PM
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Quote Quote by PhizKid View Post
We learned the Chain Rule but not sure how to apply it here. My best guess would be:

g'(x) = f'(x) * 2x
Since g(x) = f(x2), then g'(x) = f'(x2) * 2x
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Oct1-12, 03:56 PM
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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?
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Oct1-12, 03:58 PM
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Quote Quote by PhizKid View Post
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?
You don't need to use the chain rule again. If f'(u) = 1/(1 + u^3) then what is f'(x^2)? Just substitute u = x^2. You are simply evaluating the function defined by f'(u) = 1/(1 + u^3) at the particular value u = x^2.
PhizKid
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Oct1-12, 07:52 PM
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How do you know that u = x^2 ?
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Oct1-12, 08:07 PM
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Quote Quote by PhizKid View Post
How do you know that u = x^2 ?
What does f'(x^2) mean? It means "evaluate the function f' at x^2." We have a formula for f'(u), i.e. a formula to "evaluate the function f' at u." If I want to evaluate f'(5), then I replace u with 5. If I want to evaluate f'(x^2), then I replace u with x^2.


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