Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 60

Determine an equation for the tangent line to the graph of $\displaystyle y = \left( \frac{2x + 3}{x - 1} \right)^3$ at the point $(2,343)$.

We take the first derivative of the equation to find the slope,


$
\begin{equation}
\begin{aligned}

y' = m =& 3 \left( \frac{2x + 3}{x - 1} \right)^2 \cdot \frac{d}{dx} \left( \frac{2x + 3}{x - 1} \right)
\\
\\
=& 3 \left( \frac{2x + 3}{x - 1} \right)^2 \left[ \frac{\displaystyle (x-1) \cdot \frac{d}{dx} (2x+3) - (2x+3) \cdot \frac{d}{dx} (x-1) }{(x-1)^2} \right]
\\
\\
=& 3 \left( \frac{2x + 3}{x - 1} \right)^2 \left[ \frac{(x-1) (2) - (2x+3) (1)}{(x-1)^2} \right]
\\
\\
=& 3 \left( \frac{2x + 3}{x - 1} \right)^2 \left[ \frac{2x - 2 - 2x - 3}{(x-1)^2} \right]
\\
\\
=& \frac{-15 (2x+3)^2}{(x-1)^4}
\\
\\
=& \frac{-15 [2(2) + 3]^2}{(2-1)^4}
\\
\\
=& \frac{-15 (7)^2}{1^4}
\\
\\
=& -735

\end{aligned}
\end{equation}
$



Using Point Slope Form, the equation of the tangent line is


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
y - 343 =& -735 (x-2)
\\
y - 343 =& -735x + 1470
\\
y =& -735x + 1470 + 343
\\
y =& -735x + 1813

\end{aligned}
\end{equation}
$