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Homework Help: Derivative help

  1. Jan 29, 2007 #1
    1. The problem statement, all variables and given/known data
    This problem is not hard at all its just that this stupid online homework program is a problem. Anyways,

    f(x)=1/(5x+7)

    The quotient:
    f(7+h)+f(7)\h

    This can be simplified to:
    1\(A+Bh)

    What is A & B
    What is f'(7)?

    2. Relevant equations

    [tex]lim_{h\rightarrow0}\frac{\f(7+h)-f(7)}{h}[/tex]


    3. The attempt at a solution

    The derivative is easy to get.
    The final form is [tex]\frac{-5}{210h+1764}[/tex]
    How do I get it to:

    [tex]\frac{1}{A+Bh}[\tex]
     
    Last edited: Jan 29, 2007
  2. jcsd
  3. Jan 29, 2007 #2

    cristo

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    Why not multiply top and bottom by -1/5?
     
  4. Jan 29, 2007 #3
    but when I input B=210 and A=1764 it tells me I'm incorrect
     
  5. Jan 29, 2007 #4

    cristo

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    Well, you haven't multiplied them by -1/5. In order to get [tex]\frac{-5}{210h+1764}[/tex] into the required form, you need to multiply top and bottom by -1/5 (to get unity in the numerator).
     
  6. Jan 29, 2007 #5
    Ah..of coarse.
    There is another problem that requires the form Ah^2 + B h + C.
    f(x) = 2x^2 + 9 x + 4, find f'(2).

    Now I got :
    2h+17, A is 0, B is 2 but
    when i input 17 for c it is incorrect. Why?
     
    Last edited: Jan 29, 2007
  7. Jan 29, 2007 #6

    Gib Z

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    Homework Helper

    Umm do you have to do it by first principles? It seems the quotient rule would work fine here.

    For your 2nd problem, f'(2)=17, I dont see what the problem is.
     
  8. Jan 29, 2007 #7
    Well the question wants the form Ah^2+Bh+C
    I know that
    A=0
    B=2
    Shouldnt C=17? When i input that it is wrong.
     
  9. Jan 29, 2007 #8
    If not c=17, I don't understand what else it could be!
     
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