Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 34

Differentiate $f(x) = 6x^{-4} (6x^3 + 10x^2 -8x + 3)$

$
\begin{equation}
\begin{aligned}
f(x) &= \left( 6x^{-4} \right) \left( 6x^3 \right) + \left( 6x^{-4} \right) (10x^2) - \left( 6x^{-4} \right) (8x) + 3 \left( 6x^{-4} \right)
&& \text{Apply Distributive property}\\
\\
f(x) &= 36x^{-4 + 3} + 60x^{-4 + 2} - 48x^{-4 +1} + 18x^{-4}
&& \text{Multiply variables with same bases by adding their exponents}\\
\\
f(x) &= 36x^{-1} + 60x^{-2} - 48x^{-3} + 18 x^{-4}
\end{aligned}
\end{equation}
$

Then by taking the derivative, we get

$
\begin{equation}
\begin{aligned}
f'(x) &= 36 \cdot \frac{d}{dx} (x^{-1}) + 60 \cdot \frac{d}{dx} (x^{-2}) - 48 \cdot \frac{d}{dx} (x^{-3}) + 18 \cdot \frac{d}{dx} (x^{-4})\\
\\
f'(x) &= 36 \cdot (-1) x^{-1 -1} + 60 \cdot (-2) x^{-2-1} -48 \cdot (-3) x^{-3 - 1}+ 18 \cdot (-4) x^{-4 -1 }\\
\\
f'(x) &= -36x^{-2} - 120x^{-3} + 144 x^{-4} - 72 x^{-5}\\
\\
f'(x) &= \frac{-36}{x^2} - \frac{120}{x^3} + \frac{144}{x^4} - \frac{72}{x^5}
\end{aligned}
\end{equation}
$