Finding antiderivative without integration

PhizKid

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) ?

jbunniii

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).

PhizKid

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

jbunniii

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).

PhizKid

We learned the Chain Rule but not sure how to apply it here. My best guess would be:

g'(x) = f'(x) * 2x

Mark44

Since g(x) = f(x²), then g'(x) = f'(x²) * 2x

PhizKid

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?

jbunniii

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

How do you know that u = x^2 ?

jbunniii

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.