Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 19

Determine the derivative of the function $y = (2x-5)^4(8x^2-5)^{-3}$


$
\begin{equation}
\begin{aligned}
y' &= \left[ (2x-5)^4 \cdot \frac{d}{dx} (8x^2-5)^{-3} \right] \left[ (8x^2-5)^{-3} \cdot \frac{d}{dx} (2x-5)^4 \right]\\
\\
y' &= \left[ (2x-5)^4 \cdot -3(8x^2-5)^{-4} \frac{d}{dx} (8x^2-5) \right] + \left[ (8x^2-5)^{-3} \cdot 4(2x-5)^3 \frac{d}{dx} (2x-5)\right]\\
\\
y' &= \left[ -3(2x-5)^4(8x^2-5)^{-4}(16x)\right] + \left[ 4(8x^2-5)^{-3}(2x-5)^3(2)\right]\\
\\
y' &= \left[(-48x)(2x-5)^4(8x^2-5)^{-4} \right] + \left[ (8)(8x^2-5)^{-3}(2x-5)^3\right]\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-3} \left[ -6x(2x-5)(8x^2-5)^{-1}+1 \right]\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-3} \left( \frac{-12x^2+30x}{8x^2-5} + 1 \right)\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-3} \left( \frac{-12x^2+30x+8x^2-5}{8x^2-5}\right)\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-3} \left( \frac{-4x^2+30x-5}{8x^2-5}\right)\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-3} (-4x^2+30x-5)(8x^2-5)^{-1}\\
\\
y' &= 8(2x-5)^3(8x^2-5)^{-4}(-4x^2+30x-5)
\end{aligned}
\end{equation}
$

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