Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 98

Determine the derivative of the function $f(x) = \left( \sqrt{2x - 1} + x^3 \right)^5$ analytically. Then use a calculator to check your results.

By using Product Rule and Chain Rule, we get


$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{d}{dx} \left[ (\sqrt{2x - 1} + x^3)^5 \right]\\
\\
&= 5 \left( \sqrt{2x - 1} + x^3 \right)^{5 - 1} \cdot \frac{d}{dx} \left( \sqrt{2x - 1} + x^3 \right)\\
\\
&= 5 \left( \sqrt{2x - 1} + x^3 \right)^4 \left[ \frac{1}{2} (2x - 1)^{\frac{1}{2}-1} \cdot \frac{d}{dx} (2x - 1) + 3x^2 \right]\\
\\
&= 5 \left( \sqrt{2x - 1} + x^3 \right)^4 \left[ \frac{1}{2}(2x - 1)^{-\frac{1}{2}} (2) + 3x^2 \right]\\
\\
&= 5 \left( \sqrt{2x - 1} + x^3 \right)^4 \left[ \frac{1}{(2x - 1)^{\frac{1}{2}}} + 3x^2 \right]\\
\\
&= 5 \left( \sqrt{2x - 1} + x^3 \right)^4 \left[ \frac{1 + 3x^2 (2x - 1)^{\frac{1}{2}}}{(2x - 1)^{\frac{1}{2}}} \right]
\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.

So we can say that both functions agree.