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: 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).

    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*
    *divide by 1/h and let h=0*

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


    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


    User Avatar
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook