Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 132

Use the graphs of the functions below to decide which of the following seems to be
the correct derivative of the function $\displaystyle f(x) = \frac{4x}{x^2 + 1}$


$
\begin{equation}
\begin{aligned}
y_1 &= \frac{2}{x}\\
\\
y_2 &= \frac{4-4x}{x^2 + 1}\\
\\
y_3 &= \frac{4 - 4x^2}{(x^2 + 1)^2}\\
\\
y_4 &= \frac{4x^2 - 4}{(x^2 + 1)^2}
\end{aligned}
\end{equation}
$





Based from the graph of $f$ the points at which the tangent line to $f$ is horizontal (slope = 0) are
at $ x = -1$ and $x = 1$













Notice that the functions $y_3$ and $y_4$ crossed the $x$-axis or has a zero value points $x = -1$ and $x = 1$.
However, you can see in the interval of $f$ for $-1 < x < 1$ that $f$ is increasing, meaning that the derivative of that
interval is positive. Therefore, the answer is $\displaystyle y_3 = \frac{4 - 4x^2}{(x^2 + 1)^2}$