Calculus and Its Applications, Chapter 1, Review Exercises, Section Review Exercises, Problem 44

Differentiate $\displaystyle g(x) = (5 - x)^2 (2x - 1)^5$.

By using Product Rule and Chain Rule,

$
\begin{equation}
\begin{aligned}
g'(x) &= (5 - x)^2 \cdot \frac{d}{dx} (2x - 1)^5 + (2x - 1)^5 \cdot \frac{d}{dx}(5 - x)^2\\
\\
g'(x) &= (5 - x)^2 \cdot 5(2x - 1)^{5 - 1} \cdot \frac{d}{dx} (2x - 1) + (2x - 1)^5 \cdot 2(5 - x)^{2 - 1} \cdot \frac{d}{dx} (5 - x)\\
\\
g'(x) &= (5 - x)^2 \cdot 5(2x - 1)^4 (2) + (2x - 1)^5 \cdot 2 (5 - x)(-1)\\
\\
g'(x) &= 10(5 - x)^2 (2x - 1)^4 - 2(5 - x)(2x - 1)^5\\
\\
g'(x) &= 2(5 - x)(2x - 1)^4 [5 (5 - x) - (2x - 1)]\\
\\
g'(x) &= 2(5 - x)(2x - 1)^4 [ 25 - 5x - 2x + 1]\\
\\
g'(x) &= 2(5 - x)(2x - 1)^4 (26 - 7x)\\
\\
g'(x) &= 2(5 - x)(26 - 7x)(2x - 1)^4
\end{aligned}
\end{equation}
$