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Find the derivative of the function(chain rule)

  1. Aug 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Various values of the functions f(x) and g(x) and their derivatives are given in the table below. Find the derivative of f(x+g(x)) at x=0.

    at x=0 f(x)=5 f'(x)=2 g(x)=1 g'(x)=3
    at x=1 f(x)=7 f'(x)=3 g(x)=-2 g'(x)=-5
    2. Relevant equations
    Chain rule



    3. The attempt at a solution

    So f'(x)(x+g(x))[(1+g'(x))(f(x)]
    2(0+1)(1+3)(5)=40

    Is this correct or did I overlook something chain rule always has a way of making me feel I missed something


     
  2. jcsd
  3. Aug 4, 2015 #2

    Lok

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    Could you detail the chain rule used? Just the base formula.
     
  4. Aug 4, 2015 #3
    Derivative of the outside multiplied by the derivative of the inside
    (f*g)'=(f'*g)+(g'*f)

    Where f=f(x)
    g=(x+g(x))

    I'm getting 40 but not sure if I attempted this correctly
     
  5. Aug 4, 2015 #4

    Lok

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    I asked as this did not seem the right chain rule use. If i understood the problem correctly.
    One issue is that f(x+g(x)) is not equal to f*(x+g(x)).
    f(u) is a function and in your case u=x+g(x).
    What chain rule do you need to apply in this case?
     
  6. Aug 4, 2015 #5
    This has been written wrong .

    d( f(x+g(x)) )/dx = f'(x+g(x))*(1+g'(x)) .
     
  7. Aug 4, 2015 #6

    Lok

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    Ahhhh. Much better.
    And now, just solve :P
     
  8. Aug 4, 2015 #7
    f'(x+g(x))
    How do I solve this with the given information? Would it just be the derivative f a constant which makes the whole thing 0?
     
  9. Aug 4, 2015 #8

    Lok

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    Well you don't know anything about the function f, but solving for x=0 does just mean f'(0+g(0)).
     
  10. Aug 4, 2015 #9
    I don't follow you .
     
  11. Aug 4, 2015 #10
    I would plug in the value for x and g(0) getting derivative of 1 which is 0. Making the final answer 0.

    I think thats right I really confused myself on the chain rule
     
  12. Aug 4, 2015 #11

    Lok

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    but f'(1) is not zero by the conditions above.
     
  13. Aug 4, 2015 #12

    Lok

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    Sorry about the mixed responses to both of you. I have been switching through screens all morning.
     
  14. Aug 4, 2015 #13
    You don't find derivative by putting the value before differentiating .
     
  15. Aug 4, 2015 #14
    oh wow that x=1 did come in handy I got it finally 12 was correct

    thanks for the patience really helped with my confusion
     
  16. Aug 4, 2015 #15
    My suggestion to you is that you revise differentiation once again .
     
    Last edited: Aug 4, 2015
  17. Aug 4, 2015 #16

    Lok

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    Anytime there is any time :P.
     
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