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!
Hello, Vamz!
$\displaystyle \text{Given the following table of values, find }(f\circ g)'(1)$
. . $\displaystyle \begin{array}{|c||c|c|c|c|}
x & f(x) & g(x) & f'(x) & g'(x) \\ \hline \hline
1 & 2 & 3 & 2 & 4 \\ \hline
2 & 3 & 1 & -1 & -2 \\ \hline
3 & 1 & 2 & 4 & 2 \\ \hline\end{array}$
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 $