Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 96

Determine the derivative of the function $f(x) = x\sqrt{4 - x^2}$ analytically. Then use a calculator to check your results.

By using Product Rule and Chain Rule, we get


$
\begin{equation}
\begin{aligned}
f'(x) &= x \cdot \frac{d}{dx} (4 - x^2)^{\frac{1}{2}} + (4 - x^2)^{\frac{1}{2}} \cdot \frac{d}{dx} (x)\\
\\
&= x \cdot \frac{1}{2} (4 - x^2)^{\frac{1}{2} - 1} \cdot \frac{d}{dx} (4 - x^2) + (4 - x^2)^{\frac{1}{2}} (1)\\
\\
&= x \cdot \frac{1}{2} (4 - x^2)^{\frac{1}{2} - 1} (-2x) + (4 - x^2)^{\frac{1}{2}}\\
\\
&= -x^2( 4- x^2)^{-\frac{1}{2}} + (4 -x ^2)^{\frac{1}{2}}\\
\\
&= \frac{-x^2}{(4 - x^2)^{\frac{1}{2}}} + (4 - x^2)^{\frac{1}{2}}\\
\\
&= \frac{-x^2 + 4 - x^2}{(4 - x^2)^{\frac{1}{2}}}\\
\\
&= \frac{-2x^2 + 4}{(4 - x^2)^{\frac{1}{2}}}
\end{aligned}
\end{equation}
$


Thus, the graph of the function and its derivative is



Based from the graph, we can see that the function has a positive slope or positive derivative when it is increasing.
On the other hand, the function has a negative slope or negative derivative when the function is decreasing.
Also, the function has a zero slope at the minimum and maximum point of the graph.

So we can say that both functions agree.