Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 62

Consider

$\displaystyle g(x) = \left( \frac{6x + 1}{2x - 5} \right)^2$

a. Determine $g'(x)$ using the Extended Power Rule.


$
\begin{equation}
\begin{aligned}

g'(x) =& 2 \left( \frac{6x + 1}{2x - 5} \right) \cdot \frac{d}{dx} \left( \frac{6x + 1}{2x - 5} \right)
\\
\\
=& 2 \left( \frac{6x + 1}{2x - 5} \right) \left[ \frac{\displaystyle (2x - 5) \cdot \frac{d}{dx} (6x + 1) - (6x + 1) \cdot \frac{d}{dx} (2x-5) }{(2x - 5)^2} \right]
\\
\\
=& 2 \left( \frac{6x + 1}{2x - 5} \right) \left[ \frac{(2x-5)(6) - (6x + 1)(2) }{(2x - 5)^2} \right]
\\
\\
=& 2 \left( \frac{6x + 1}{2x - 5} \right) \left[ \frac{12x - 30 - 12x - 2}{(2x-5)^2} \right]
\\
\\
=& 2 \left( \frac{6x + 1}{2x - 5} \right) \left[ \frac{-32}{(2x-5)^2} \right]
\\
\\
=& \frac{-64 (6x + 1)}{(2x - 5)^3}


\end{aligned}
\end{equation}
$


b. Note that

$\displaystyle g(x) = \frac{36x^2 + 12x + 1}{4x^2 - 20x + 25}$

Determine $g'(x)$ using the Quotient Rule.


$
\begin{equation}
\begin{aligned}

g'(x) =& \frac{\displaystyle (4x^2 - 20x + 25) \cdot \frac{d}{dx} (36x^2 + 12x + 1) - (36x^2 + 12x + 1) \cdot \frac{d}{dx} (4x^2 - 20x + 25) }{(4x^2 - 20x + 25)^2}
\\
\\
=& \frac{(4x^2 - 20x + 25) (72x + 12) - (36x^2 + 12x + 1) (8x - 20)}{(4x^2 - 20x + 25)^2}
\\
\\
=& \frac{12 (2x - 5)^2 (6x + 1) - 4(6x + 1)^2 (2x - 5)}{(4x^2 - 20x + 25)^2}
\\
\\
=& \frac{4 (6x + 1) (2x - 5) [3 (2x - 5) - (6x + 1)] }{[(2x - 5)^2]^2}
\\
\\
=& \frac{4 (6x + 1) (6x - 15 - 6x - 1) }{(2x - 5)^3}
\\
\\
=& \frac{4(6x + 1) (-16)}{(2x - 5)^3}
\\
\\
=& \frac{-64 (6x + 1)}{(2x - 5)^3}

\end{aligned}
\end{equation}
$


c. Compare your answers to parts (a) and (b). Which approach was easier, and why?

The extended power rule is easier to use because it has shorter solution.