Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 39

Determine the derivative of the function $y = \left[ x^2+1(1-3x)^5 \right]^3$


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left[ x^2+1(1-3x)^5 \right]^3\\
\\
y' &= 3\left[ x^2+1(1-3x)^5 \right]^2 \frac{d}{dx} \left[ x^2+(1-3x)^5 \right]\\
\\
y' &= 3\left[ x^2+1(1-3x)^5 \right]^2 \left[ \frac{d}{dx} (x^2) + \frac{d}{dx} (1-3x)^5 \right]\\
\\
y' &= 3\left[ x^2+1(1-3x)^5 \right]^2 \left[ 2x + 5 (1-3x)^4 \frac{d}{dx} (1-3x)\right]\\
\\
y' &= 3\left[ x^2+1(1-3x)^5 \right]^2 \left[ 2x + 5 (1-3x)^4 (-3) \right]\\
\\
y' &= 3\left[ x^2+1(1-3x)^5 \right]^2 \left[ 2x - 15 (1-3x)^4\right]
\end{aligned}
\end{equation}
$