# Derivative of composite function

• Oct 30th 2010, 04:58 PM
Vamz
Derivative of composite function
http://img52.imageshack.us/img52/841...01030at653.png

so we have f(g(x))

U=g(x)
f'(U)*U*U'
when, x=1, U=3 & U'=2
f'(3)*3*2
-4*3*2=-12

-12 is totally wrong apparently. Can someone please explain how its done?
Thanks!
• Oct 30th 2010, 05:17 PM
Plato
If this is graded homework, then it is our policy not to give an answer.
• Oct 30th 2010, 05:26 PM
Vamz
its not graded, I have unlimited attempts.
• Oct 31st 2010, 04:49 AM
HallsofIvy
Do you know the "chain rule"? That is what is being tested here. Using your notation, with U= g(x), f(g(x))= f(U) and the derivative is f'(U)U', NOT the "f'(U)UU'" that you have.
• Oct 31st 2010, 08:33 AM
Soroban
Hello, Vamz!

Quote:

$\displaystyle \text{Given the following table of values, find }(f\circ g)'(1)$

. . $\displaystyle \begin{array}{|c||c|c|c|c|} x & f(x) & g(x) & f'(x) & g'(x) \\ \hline \hline 1 & 2 & 3 & 2 & 4 \\ \hline 2 & 3 & 1 & -1 & -2 \\ \hline 3 & 1 & 2 & 4 & 2 \\ \hline\end{array}$

We have: .$\displaystyle f\circ g \;=\;f(g(x))$

Then: .$\displaystyle (f\circ g)' \;=\;f'(g(x))\cdot g'(x)$

And: .$\displaystyle (f\circ g)'(1) \;=\;f'(g(1))\cdot g'(1)$
. . . . . . . . . . . . . . . . . $\displaystyle \uparrow\qquad\;\; \uparrow$
. - - - . . . . . . . . . . . . .$\displaystyle 3 \qquad\;\;4$

. . . . . . . . . . . . $\displaystyle =\;\;\;f'(3)\cdot 4$

. . . . . . . . . . . . $\displaystyle =\quad\;\;4\cdot 4$

. . . . . . . . . . . . $\displaystyle =\qquad 16$