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CAlculus derivative help please?

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculus Derivative help?
    when x = 0, f=2, f '=1, g=5, g '= -4
    when x = 1, f=3, f '=2, g=3, g '= -3
    when x = 2, f=5, f '=3, g=1, g '= -2
    when x = 3, f=10, f '=4, g=0, g '= -1
    based on the above table, if a = f +2g, then a'(3) =?
    please show steps.thanks


    2. Relevant equations


    if a = f +2g, then a'(3) =?
    3. The attempt at a solution
    a=f'+0(g')
    a'(3)=f'3+0
    a'(3)=4????
     
  2. jcsd
  3. Dec 21, 2011 #2
    I don't understand what the problem is =(
     
  4. Dec 21, 2011 #3

    NascentOxygen

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    Does the textbook give you the correct answer?
     
  5. Dec 21, 2011 #4

    vela

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    You didn't differentiate this correctly. You may want to review the properties of differentiation.
     
  6. Dec 22, 2011 #5
    1. The problem statement, all variables and given/known data

    Calculus Derivative help?
    when x = 0, f=2, f '=1, g=5, g '= -4
    when x = 1, f=3, f '=2, g=3, g '= -3
    when x = 2, f=5, f '=3, g=1, g '= -2
    when x = 3, f=10, f '=4, g=0, g '= -1
    based on the above table, if a = f +2g, then a'(3) =?
    please show steps.thanks


    2. Relevant equations


    if a = f +2g, then a'(3) =?

    3. The attempt at a solution
    a=f'+0(g')
    a'(3)=f'3+0
    a'(3)=4????


    can anyone just please give me the steps on how to do it and just be direct
     
  7. Dec 22, 2011 #6
    Think about it.

    a = f + 2g really means that a at 3 will equal f at 3 added to twice the value of g at 3. The latter information is given you in your table; just do the addition.

    But how to calculate the derivative of a? That's where the point of the whole problem comes into play. The goal of the problem is to get you to think about one of the rules of derivatives. What rule have you learned that allows you to calculate the derivative of a sum of functions of which you already know the derivatives?
     
  8. Dec 22, 2011 #7

    HallsofIvy

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    A very relevant equation is "(af(x)+ bg(x))'= af'(x)+ bg'(x)"


    Where did this "0" come from? a'(3)= f'(3)+ 2g'(3)

     
  9. Dec 22, 2011 #8

    Mark44

    Staff: Mentor

    That's not how it works here at Physics Forums. We provide help so that you can do a problem, but we don't do the work for you.
     
  10. Dec 22, 2011 #9

    eumyang

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    To the OP: also, stop posting the same question in multiple threads. I believe that's also against the forum rules.
     
  11. Dec 22, 2011 #10

    berkeman

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    (Two threads merged)
     
  12. Dec 25, 2011 #11

    NascentOxygen

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    Indeed it can be solved that way. But it may not help the OP's understanding of how to solve further exercises of this nature.

    You do no one any favors by handing them the complete solution before they have had the opportunity to be guided to achieving it for themselves.
     
  13. Dec 25, 2011 #12
    you are right about that
    thanks
     
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