1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook