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Math Help - f'(5x^4) of f(x)=x^5

  1. #1
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    f'(5x^4) of f(x)=x^5

    Hi, I was wondering how to do these types of derivatives... I have several to do, but I would like to see only one of them solved so that I can do the others. The course is about several variable calculus...

    The problem says:
    "If f(x)=x^5, find

    f'(5x^4), f(x^2), d/dx(f(x^2)), f'(x^2)"
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  2. #2
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    Re: f'(5x^4) of f(x)=x^5

    Quote Originally Posted by juanma101285 View Post
    The problem says:
    "If f(x)=x^5, find
    f'(5x^4), f(x^2), d/dx(f(x^2)), f'(x^2)"
    If f(x)=x^5 then f'(x)=5x^4.

    Thus f'(5x^4)=5(5x^4)^4=5^5x^{16}.
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  3. #3
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    Re: f'(5x^4) of f(x)=x^5

    This is just composition of functions. If I asked you to find f'(x), you'd probably have no problem doing that. Here's a different example:

    f(x)=x^2+5

    Find f'(x+3), f(x^5), d/dx(f(x^3)), f'(x^3)


    First, let's calculate f'(x).
    f'(x)=2x

    Now, remember how to do composition of functions?
    f'(x+3)=2(x+3)=2x+6

    The second function is even easier, we just compose f with x^5:
    f(x^5)=(x^5)^2+2 = x^10+2

    Now we're back to derivatives, and some tricky notation. The book is trying to trip you up into thinking that, since d/dx(f(x))=f'(x), then d/dx(f(x^3))=f'(x^3). This is not the case. Remember the chain rule:

    d/dx(f(g(x)) = f'(g(x))g'(x)

    So, with g(x)=x^3, g'(x)=3x^2 and therefore
    d/dx(f(x^3))=[f'(x^3)]g'(x)=[2(x^3)](3x^2)=6x^5

    Finally, the last one is just there to possibly alert those people who tripped up on the third one that they should reconsider their answer. It's just simple composition, like before:

    f'(x^2)=2(x^2)=2x^2
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