# Thread: Derivative of composite function

1. ## Derivative of composite function

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!

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

3. its not graded, I have unlimited attempts.

4. 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.

5. Hello, Vamz!

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

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

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

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

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

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

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

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