Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 16

Differentiate $\displaystyle y = \frac{7x^3}{(4-9x)^5}$.

Apply Quotient Rule


$
\begin{equation}
\begin{aligned}

y' =& \frac{\displaystyle (4-9x)^5 \cdot \frac{d}{dx} (7x^3) - 7x^3 \cdot \frac{d}{dx} (4-9x)^5 }{[(4-9x)^5]^2}
\\
\\
y' =& \frac{\displaystyle (4-9x)^5 (21x^2) - (7x^3)(5)(4-9x)^4 \cdot \frac{d}{dx} (4-9x)}{(4-9x)^{10}}
\\
\\
y' =& \frac{(4-9x)^5 (21x^2) - (35x^3)(4-9x)^4 (-9)}{(4-9x)^{10}}
\\
\\
y' =& \frac{(4-9x)^5 (21x^2) - 315x^3 (4-9x)^4}{(4-9x)^{10}}
\\
\\
y' =& \frac{(4 - 9x)^4 [(4 - 9x) (21x^2) + 315x^3]}{(4-9x)^{10}}
\\
\\
y' =& \frac{21x^4 (4-9x + 15x)}{(4 - 9x)^6}
\\
\\
y' =& \frac{21x^2 (4 + 6x)}{(4 - 9x)^6}


\end{aligned}
\end{equation}
$