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(f of g)' (0) Problem

  • Thread starter b0mberman
  • Start date
  • #1
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1.Question
Find the value of (f of g)' at the given value of x.
f(u)= (6u)/(u^2+5)
u=g(x)=4x^2+5x+1
x=0

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
 

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Answers and Replies

  • #2
1,341
3
Apply the chain rule

f(g(x))' = f'(g(x)) * g(x)
 
  • #3
4
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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
 
  • #4
1,341
3
[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:
  • #5
4
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Yes, but where did f'(1) come from? Sorry...
 
  • #6
1,341
3
Well it's

f'(g(x))*g'(x)

and g(0) = 1 so it's f'(1)*g'(0)
 
  • #7
4
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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!

P.S.
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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