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Calculus I problem I can't solve

  1. Apr 5, 2008 #1
    Here's the problem:

    [tex]d/dx(f(3x^5)) = 8x^2 [/tex]

    Find [tex]f'(x)[/tex]

    After applying the chain rule:

    [tex]f'(3x^5)(15x^4) = 8x^2[/tex]

    [tex]f'(3x^5) = 8/(15x^2) [/tex]

    It's not apparent to me how I proceed from here to find [tex]f'(x)[/tex]. I tried dividing the expression on the right by three and taking the fifth root but that does not seem to be right. Any help would be appreciated!
     
  2. jcsd
  3. Apr 5, 2008 #2

    HallsofIvy

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    Since you want to know f(x) without all that "stuff" inside the parentheses, let u= 3x5. Then f(3x5)= f(u) and df/dx= (df/du)(du/dx)= (df/du)(15x)= 8x2. Now you have df/du= (8/15)x. Since u= 3x5, x5= u/3 and x= (u/3)1/5. df/du= (8/15)(u1/5)/31/5. Now just replace u by x to get f'(x)
     
  4. Apr 5, 2008 #3
    Since you want to know f(x) without all that "stuff" inside the parentheses, let u= 3x5. Then f(3x5)= f(u) and df/dx= (df/du)(du/dx)= (df/du)(15x)= 8x2. Now you have df/du= (8/15)x. Since u= 3x5, x5= u/3 and x= (u/3)1/5. df/du= (8/15)(u1/5)/31/5. Now just replace u by x to get f'(x)

    I see what you're saying, but isn't du/dx equal to [tex]15x^4[/tex], not [tex]15x[/tex]? That would make df/du equal to [tex](8/15x^2)[/tex]. Therefore, substituting x= (u/3)1/5 gives a result of (8/15)(3^(2/5)/u^(2/5)).

    Substituting u by x to get f'(x) would get a final solution of [tex]8/15x^2[/tex], which is what I started with above for [tex]f'(3x^5)[/tex], which I don't think is the correct answer.
     
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