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    Hi, I have this question: Integrate x f'(x^2) dx. My answer should be in terms of f. What does this mean? Because I know f' of x^2 is 2x but then I'd be left with x times 2x dx which is 2x^2 dx which is (2x^3)/3? But that seems way too easy so I'm stuck...
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    Clue: think about what the differential of f(x^2) is and don't forget to use the chain rule.

    Substitute in x^2 =y if you need to then remember df/dx =df/dy * dy/dx

    Random note f is any arbitrary function
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    You need to tighten up your definitions a bit or mistakes will follow. f' is an abbreviation for the derivative of a single-variable function f(.) with respect to that variable. Saying "f' of x^2 is 2x" is therefore sloppy and meaningless.

    Correct statements: "the derivative (with respect to x) of x^2 is 2x", "if f(x) = x^2 then the derivative of f (with respect to x) is 2x", "if f(x) = x^2 then f' = 2x"

    Anyway back to your problem, what is (f(x^2))'? or d/dx [f(x^2)] if you prefer (it's the same thing).
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    Sorry but I'm still confused. I know dy/dx is f'(x) but... just don't get it :/
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    The problem is that you don't understand the chain rule. Telling you the solution won't address that.

    Try to understand what the derivative of f(g(x)) is in general (the chain rule tells you!) then apply it to this case.
 
 
 
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