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(f of g)' (0) Problem

  1. Mar 30, 2008 #1
    Find the value of (f of g)' at the given value of x.
    f(u)= (6u)/(u^2+5)

    2. The attempt at a solution
    f'(u) = (6u) x -1(u^2+5)^(-2) x 2u + (u^2+5)^(-1) x 6
    '= -12u^2 x (u^2+5)^(-2) + 6(u^2+5)^(-1)

    g'(x) = 8x+5
    g'(0) = 5

    (f of g)'(0) = -12(5)^2 x (5^2 + 5)^(-2) + 6(5^2+5)^(-1)
    = (-12x25)/900 + 6/30
    = (-300/900) + 180/900 = -2/15

    Correct answer: 10/3

    Thank you ahead of time, I have been working on this simple calculus question for at least 2 hours and just can not get the right answer. I am guessing it is a simple mathematical error in the differentiating somewhere so I have posted it here so hopefully someone could point it out.

    Thank you, b0mberman

    Attached Files:

  2. jcsd
  3. Mar 30, 2008 #2
    Apply the chain rule

    f(g(x))' = f'(g(x)) * g(x)
  4. Mar 30, 2008 #3
    I have if you did not notice I applied chain rule to f than to g then did (f of g), unless you mean something else
  5. Mar 30, 2008 #4
    [tex]f'(u) = \frac{-6(u^-5)}{(x^2+5)^2[/tex] -- I think you might've applied the quotient rule incorrectly :-/

    [tex]g(0) = 1[/tex]

    [tex]g'(x) = 8x+5[/tex]

    It's f'(g(0)) so,

    (f o g)' = f'(1)*g'(0), right?
    Last edited: Mar 30, 2008
  6. Mar 30, 2008 #5
    Yes, but where did f'(1) come from? Sorry...
  7. Mar 30, 2008 #6
    Well it's


    and g(0) = 1 so it's f'(1)*g'(0)
  8. Mar 30, 2008 #7
    OK I see now, I was just thinking of that after your first post(where the 1 came from), OK I understand now. I did not do (f of g)' right. Thank you very much!

    The funny thing is I used the same method for the problem before(the wrong method) and it came out right! f(u) = u^7 + 8
    u=g(x)=4x^2 +5x+1, x=0
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