$\begin{array} { | r | c | c | c | c | }
\hline x & f ( x ) & f ^ { \prime } ( x ) & g ( x ) & g ^ { \prime } ( x ) \
\hline - 6 & 7 & - 8 & - 6 & 7 \
- 4 & 1 & - 5 & 0 & 5 \
- 2 & 3 & - 2 & 4 & 3 \
0 & 5 & 0 & 6 & 1 \
2 & - 5 & 1 & 6 & - 1 \
4 & - 3 & 3 & 4 & - 3 \
6 & 1 & 5 & 0 & - 5 \
\hline
\end{array}$
-Find $\frac { d } { d x } \left( \frac { 2 \cdot g ( x ) } { x ^ { 2 } + 1 } \right) \text { when } x = 0$
A) $- 10$
B) $- 6$
C)2
D)10
E)14
F)16
G)20
H)24

Question

$\begin{array} { | r | c | c | c | c | } 
\hline x & f ( x ) & f ^ { \prime } ( x ) & g ( x ) & g ^ { \prime } ( x ) \
\hline - 6 & 7 & - 8 & - 6 & 7 \
- 4 & 1 & - 5 & 0 & 5 \
- 2 & 3 & - 2 & 4 & 3 \
0 & 5 & 0 & 6 & 1 \
2 & - 5 & 1 & 6 & - 1 \
4 & - 3 & 3 & 4 & - 3 \
6 & 1 & 5 & 0 & - 5 \
\hline
\end{array}$
-Find $\frac { d } { d x } \left( \frac { 2 \cdot g ( x ) } { x ^ { 2 } + 1 } \right) \text { when } x = 0$
A) $- 10$ 
B) $- 6$
C)2
D)10
E)14
F)16
G)20
H)24

Quizplus · Accepted Answer

To find the derivative of $\frac{2 \cdot g(x)}{x^2 + 1}$ and evaluate it at $x = 0$, we use the quotient rule and the chain rule. The derivative of $2 \cdot g(x)$ is $2g'(x)$, and the derivative of $x^2 + 1$ is $2x$. Applying the quotient rule:$$\frac{d}{dx} \left( \frac{2 \cdot g(x)}{x^2 + 1} \right) = \frac{(2g'(x))(x^2 + 1) - (2 \cdot g(x))(2x)}{(x^2 + 1)^2}$$Substituting $x = 0$, $g(0) = 6$, and $g'(0) = 1$:$$\frac{d}{dx} \left( \frac{2 \cdot g(x)}{x^2 + 1} \right) \bigg|_{x=0} = \frac{(2 \cdot 1)(0^2 + 1) - (2 \cdot 6)(2 \cdot 0)}{(0^2 + 1)^2} = \frac{2}{1} = 2$$Therefore, the derivative evaluated at $x = 0$ is 2.

-Find ddx(2⋅g(x)x2+1) when x=0\frac { d } { d x } \left( \frac { 2 \cdot g ( x ) } { x ^ { 2 } + 1 } \right) \text { when } x = 0dxd​(x2+12⋅g(x)​) when x=0

-Find $\frac { d } { d x } \left( \frac { 2 \cdot g ( x ) } { x ^ { 2 } + 1 } \right) \text { when } x = 0$