Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 17

Determine the derivative of the function $g(x) = (1+4x)^5(3+x-x)^8$


$
\begin{equation}
\begin{aligned}
g'(x) &= \left[ (1+4x)^5 \cdot \frac{d}{dx} (3 + x - x^2)^8 \right] + \left[ (3+x-x^2)^8 \cdot \frac{d}{dx} (1+4x)^5\right]\\
\\
g'(x) &= \left[ (1+4x)^5 \cdot 8(3+x-x^2)^7 \frac{d}{dx} (3+x-x^2)\right] + \left[ (3+x-x^2)^8 \cdot 5(1+4x)^4 \frac{d}{dx} (1+4x) \right]\\
\\
g'(x) &= \left[ (1+4x)^5 \cdot 8(3+x-x^2)^7(1-2x)\right] + \left[ (3+x-x^2)^8 \cdot 5(1+4x)^4 (4)\right]\\
\\
g'(x) &= \left[8(1+4x)^5(3+x-x^2)^7(1-2x)\right] + \left[20 (3+x-x^2)^8(1+4x)^4 \right]\\
\\
g'(x) &= (3+x-x^2)^7(1+4x)^4 \left[ 8 (1+4x)(1-2x)+20(3+x-x^2) \right]\\
\\
g'(x) &= (3+x-x^2)^7(1+4x)^4 \left[ 8 (1-2x+4x-8x^2) + 20(3+x-x^2)\right]\\
\\
g'(x) &= (3+x-x^2)^7(1+4x)^4 \left[ 8 (1 + 2x - 8x^2) + 20 (3+x-x^2)\right]\\
\\
g'(x) &= (3+x-x^2)^7(1+4x)^4 ( 8 + 16x-64x^2 + 60 + 20x -20x^2)\\
\\
g'(x) &= (3+x-x^2)^7(1+4x)^4 (68 + 36x - 84x ^2)\\
\\
g'(x) &= 4 (3+x-x^2)^7(1+4x)^4 ( 17+9x-21x^2)
\end{aligned}
\end{equation}
$