$\begin{array} { | l | l | l | l | l | }
\hline \mathrm { X } & \mathrm { g } ( \mathrm { x } ) & \mathrm { h } ( \mathrm { x } ) & \mathrm { g } ^ { \prime } ( \mathrm { x } ) & \mathrm { h } ^ { \prime } ( \mathrm { x } ) \
\hline - 3 & 0 & 3 & 1 & 0 \
\hline - 2 & 1 & 2 & 2 & - 3 \
\hline - 1 & 3 & 0 & - 1 & - 2 \
\hline 0 & 2 & 3 & - 2 & 3 \
\hline 1 & 0 & - 1 & - 2 & - 2 \
\hline 2 & - 2 & - 2 & - 1 & 0 \
\hline 3 & - 3 & 0 & 0 & 1 \
\hline
\end{array}$
-Use the table above to find $f ^ { \prime } ( - 1 )$ if $f ( x ) = h \left( x ^ { 2 } g ( x ) \right)$
A) 1
B) - 1
C) 7
D) - 7

Question

$\begin{array} { | l | l | l | l | l | } 
\hline \mathrm { X } & \mathrm { g } ( \mathrm { x } ) & \mathrm { h } ( \mathrm { x } ) & \mathrm { g } ^ { \prime } ( \mathrm { x } ) & \mathrm { h } ^ { \prime } ( \mathrm { x } ) \
\hline - 3 & 0 & 3 & 1 & 0 \
\hline - 2 & 1 & 2 & 2 & - 3 \
\hline - 1 & 3 & 0 & - 1 & - 2 \
\hline 0 & 2 & 3 & - 2 & 3 \
\hline 1 & 0 & - 1 & - 2 & - 2 \
\hline 2 & - 2 & - 2 & - 1 & 0 \
\hline 3 & - 3 & 0 & 0 & 1 \
\hline
\end{array}$
-Use the table above to find $f ^ { \prime } ( - 1 )$ if $f ( x ) = h \left( x ^ { 2 } g ( x ) \right)$
A) 1
B) - 1
C) 7
D) - 7

Quizplus · Accepted Answer

To find $f'(x)$ for $f(x) = h(x^2g(x))$, we use the chain rule. Let $u = x^2g(x)$, so $f(x) = h(u)$. First, find $du/dx = 2xg(x) + x^2g'(x)$ and then $df/du = h'(u)$. At $x = -1$, $u = (-1)^2 \cdot 3 = 3$, so $h'(u) = h'(3) = 1$. For $du/dx$ at $x = -1$, $2(-1) \cdot 3 + (-1)^2 \cdot (-1) = -6 - 1 = -7$. Thus, $f'(-1) = h'(u) \cdot du/dx = 1 \cdot (-7) = -7$.

-Use the Table Above to Find f′(−1)f ^ { \prime } ( - 1 )f′(−1)

-Use the Table Above to Find $f ^ { \prime } ( - 1 )$