Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 92

Differentiate $F(x) = \left[ 6x (3 -x )^5 + 2 \right]^4$

By using Chain Rule and Product Rule, we get

$
\begin{equation}
\begin{aligned}
F'(x) &= 4 \left[ 6x (3x -5)^5 + 2 \right]^{4 - 1} \cdot \frac{d}{dx} \left[ 6x (3 - x)^5 + 2 \right]\\
\\
F'(x) &= 4 \left[ 6x (3x - 5)^5 + 2 \right]^3 \left[ 6x \cdot \frac{d}{dx} (3 - x)^5 + (3 - x)^5 \cdot \frac{d}{dx} (6x) \right]\\
\\
F'(x) &= 4 \left[ 6x (3x - 5)^5 + 2 \right]^3 \left[ 6x \cdot 5 (3 - x)^{5 - 1} \cdot \frac{d}{dx} (3 - x) + (3 - x)^5 (6) \right]\\
\\
F'(x) &= 4 \left[ 6x (3x - 5)^5 + 2 \right]^3 \left[ 30x (3 - x)^4 ( -1 )+ 6(3 -x)^5 \right]\\
\\
F'(x) &= 4 \left[ 6x (3x - 5)^5 + 2 \right]^3 \left[ -30x (3 - x)^4 + 6(3 -x)^5 \right]\\
\\
F'(x) &= 4 \left[ 6 (3 - x)^4 \right] \left[ 6x (3x - 5)^5 + 2 \right]^3 [(3 - x) - 5x]\\
\\
F'(x) &= 24(3-x)^4 \left[ 6x (3x - 5)^5 + 2 \right]^3 [3 - 6x]\\
\\
F'(x) &= 72(3 - x)^4 (1 - 2x) \left[ 6x (3x - 5)^5 + 2 \right]^3
\end{aligned}
\end{equation}
$

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