Derivative of composite function

**Vamz** · October 30th 2010, 04:58 PM

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!

**Plato** · October 30th 2010, 05:17 PM

If this is graded homework, then it is our policy not to give an answer.

**Vamz** · October 30th 2010, 05:26 PM

its not graded, I have unlimited attempts.

**HallsofIvy** · October 31st 2010, 04:49 AM

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.

**Soroban** · October 31st 2010, 08:33 AM

Hello, Vamz!

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

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

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

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

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

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

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

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

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