Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 42

Differentiate $\displaystyle f(x) = \frac{(5x - 4)^7}{(6x + 1)^3}$.



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\begin{equation}
\begin{aligned}

f'(x) =& \frac{\displaystyle (6x + 1)^3 \cdot \frac{d}{dx} (5x-4)^7 - (5x-4)^7 \cdot \frac{d}{dx} (6x + 1)^3 }{[(6x + 1)^3]^2}
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f'(x) =& \frac{\displaystyle (6x + 1)^3 (7) (5x-4)^6 \cdot \frac{d}{dx} (5x-4) - (5x-4)^7 (3) (6x + 1)^2 \cdot \frac{d}{dx} (6x+1) }{(6x+1)^6}
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f'(x) =& \frac{(6x+1)^3 (7) (5x-4)^6 (5) - (5x-4)^7 (3) (6x+1)^2 (6) }{(6x+1)^6}
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f'(x) =& \frac{35 (6x+1)^3 (5x - 4)^6 - 18 (5x - 4)^7 (6x+1)^2}{(6x+1)^6}
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f'(x) =& \frac{(6x+1)^2 (5x - 4)^6 [35 (6x+1) - 18 (5x - 4)]}{(6x+1)^6}
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f'(x) =& \frac{(5x - 4)^6 (210x + 35 - 90x + 72)}{(6x+1)^4}
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f'(x) =& \frac{(5x-4)^6 (120x + 107)}{(6x+1)^4}

\end{aligned}
\end{equation}
$