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F'(x) of a fraction

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the definition of a derivitive to find f'(x).
    f(x)=1/(1-4X)

    2. Relevant equations
    Lim as h approaches 0 [f(x+h)-f(x)]/h


    3. The attempt at a solution
    I know that the answer is supposed to be -1/(1-4X)2 but I keep getting 4/(1-4X)2. This is what I have done so far (I hope this isn't too hard to understand):

    f'(x)= (1/(1-4(x+h))-(1/(1-4x)))/h
    = ((1-4x-(1-4(x+h)))/((1-4(x+h))(1-4x)))/h
    = ((1-4x-1+4x+4h)/((1-4(x+h))(1-4x)))/h
    *cancel the numerator values*
    =(4h)/((1-4(x+h))(1-4x))/h
    *divide by 1/h and let h=0*
    =4/(1-4(x-0))(1-4x)
    =4/(1-4X)2

    What am I doing wrong?
     
  2. jcsd
  3. Feb 7, 2012 #2

    Mark44

    Staff: Mentor

    Nothing - that's the right answer.

    BTW, you don't take "f'(x) of a fraction" as you have in the title. You can take the derivative with respect to x of a fraction (in symbols, d/dx(f(x)) ), but f'(x) already represents the derivative of some function f.
     
  4. Feb 7, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The derivative of 1/u, with respect to u, is -1/u^2. But that "4" in the numerator is from the chain rule. If u is a function of x, the derivative of 1/u, with respect to x is (-1/u^2) du/dx. Here, u= 1- 4x so du/dx= -4. The derivative of 1/(1- 4x), with respect to x, is -1/(1- 4x)^2(-4)= 4/(1- 4x)^2
     
  5. Feb 7, 2012 #4
    oh, ok so it's right. I did that problem six times because I thought I was doing it wrong haha thank you! and also thank you for correcting me on the terms! :)
     
  6. Feb 7, 2012 #5
    oh yeah the chain rule!! i definitely need to remember that! thanks!
     
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