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Homework Help: Derivative of function

  1. Jan 23, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]y'=((3x^2+2x+5)^{8x^3+2x^2 +4})'=?[/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex]((3x^2+2x+5)^{8x^3+2x^2 +4})'=(8x^3+2x^2+4)(3x^2+2x+5)^{8x^3+2x^2 +4-1}(24x^2+4x)(6x+2)[/tex]
  2. jcsd
  3. Jan 23, 2008 #2
    The power rule only holds when the exponent is a constant (not a function of x).
    Last edited: Jan 23, 2008
  4. Jan 23, 2008 #3
    The function [tex]f(x)=g(x)^{h(x)}[/tex] can be written

    [tex]f(x)=e^{\ln g(x)^{h(x)}}=e^{h(x)\,\ln g(x)}[/tex]

    Now you can take the derivative, i.e.

    [tex]f'(x)=e^{h(x)\,\ln g(x)}\left(h(x)\,\ln g(x)\right)'\Rightarrow f'(x)=f(x)\left(h(x)\,\ln g(x)\right)'[/tex]
  5. Jan 23, 2008 #4
    [tex]((3x^2+2x+5)^{8x^3+2x^2 +4})'=(3x^2+2x+5)^{8x^3+2x^2 +4}((24x^2+4x)\ln(3x^2+2x+5)+(8x^3+2x^2 +4)\frac{6x+2}{3x^2+2x+5})[/tex]
  6. Jan 23, 2008 #5
    You missed a parethensis after [tex](3x^2+2x+5)^{8x^3+2x^2 +4}[/tex], but you are correct :smile:
  7. Jan 23, 2008 #6


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    Or, much the same thing, write ln(f(x))= h(x)ln(g(x)) and use the product and chain rules: (1/f)f '= h'(x) ln(g(x))+ (h(x)/g(x)) g'(x) so f '= [h'(x) ln(g(x)+(h(x)/g(x))g'(x)]f(x).
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