Given $g(1) = 6$, $g'(1) = -1$, find $d/dx(2 g(x)/(x^2 + 1))$ when $x = 1$ Why is the answer $-7$? I plugged $1$ into the equation and I ended up with $12/2$ and got $6$. Can someone explain to me what I did wrong?

Applying Quotient Rule, we find that:
\begin{align*}
\frac{d}{dx}\left[\frac{2g(x)}{x^2+1} \right]
&= \frac{(x^2+1)\frac{d}{dx}[2g(x)] - 2g(x)\frac{d}{dx}[x^2+1]}{(x^2+1)^2} \\
&= \frac{(x^2+1)[2g'(x)] - 2g(x)[2x]}{(x^2+1)^2} \\
&= \frac{2(x^2+1)g'(x) - 4xg(x)}{(x^2+1)^2} \\
\end{align*}
Hence, substituting $x = 1$, we obtain:
$$
\frac{2(1^2+1)g'(1) - 4(1)g(1)}{(1^2+1)^2}
= \frac{4(-1) - 4(6)}{4}
= -1 - 6 = -7
$$
