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Differential calculus, derivatives

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Find f'(x)= 16x - x-2 using first principles.

    2. Relevant equations
    http://img153.imageshack.us/img153/8403/597137697c1f605c7a43d34qz4.png [Broken]

    3. The attempt at a solution
    I used dy/dx and got 2x-3 + 16 but I get something different when I use the formular I attempted several times and I cant get the same answer as the dy/dx.
    Please need help!!!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 20, 2009 #2


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    Science Advisor

    Do you mean "find f'(x) if f(x)= 16x- x-1"? That's quite different from what you say!

    If f(x)= 16x- x-2, then f(x+h)= 16(x+h)- (x+ h)-2

    f(x+h)- f(x)= 16(x+ h)- 16x - (x+h)-2+ x-2

    The first part of that is just 16x+ 16h- 16x= 16h.

    The second part is
    [tex]-\frac{1}{(x+h)^2}+ \frac{1}{x^2}[/tex]
    [tex]= \frac{-x^2}{x^2(x+h)^2}+ \frac{(x+h)^2}{x^2(x+h)^2}[/tex]
    [tex]= \frac{-x^2+ x^2+ 2hx+h^2}{x^2(x+h)^2}= \frac{2hx+ h^2}{x^2(x+h)^2}[/tex]
    [tex]f(x+h)- f(x)= 16h+ \frac{2hx+h^2}{x^2(x+h)^2}[/tex]
    Now, what is (f(x+h)- f(x))/h and what is the limit of that as h goes to 0.
  4. Feb 21, 2009 #3
    Last edited by a moderator: May 4, 2017
  5. Feb 21, 2009 #4


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    Homework Helper

    you've lost a square from the denominator for no reason during your calc, they should work
  6. Feb 21, 2009 #5
    Thanks sorry about that it works perfectly!
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