Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 40

Determine the equation of the tangent line to the curve $\displaystyle y = 3 \arccos \left( \frac{x}{2} \right)$ at the point $(1, \pi)$.
If $\displaystyle y = 3 \arccos \left( \frac{x}{2} \right)$, then...

$
\begin{equation}
\begin{aligned}
y' &= -3 \left( \frac{\frac{d}{dx}\left(\frac{x}{2}\right) }{\sqrt{1-\left(\frac{x}{2}\right)^2}} \right)\\
\\
y' &= -3 \left( \frac{\frac{1}{2}}{\sqrt{1 - \frac{x^2}{4}}} \right)\\
\\
y' &= \frac{-3}{2} \left( \frac{1}{\sqrt{\frac{4-x^2}{4}}} \right)\\
\\
y' &= \frac{-3}{2} \left( \frac{\frac{1}{\sqrt{4-x^2}}}{2}\right)\\
\\
y' &= \frac{-3}{\sqrt{4-x^2}}
\end{aligned}
\end{equation}
$

Recall that the first derivative is equatl to the slope of the tangent line at some point. So at point $(1,\pi)$
$\displaystyle y' = \frac{-3}{\sqrt{4-(1)^2}} = \frac{-3}{\sqrt{3}} = m$


Thus, the equation of the tanget line can be determined by using the point slope form.

$
\begin{equation}
\begin{aligned}
y-y_1 &= m(x-x_1)\\
\\
y - \pi &= \frac{-3}{\sqrt{3}} (x - 1)\\
\\
y &= \frac{-3}{\sqrt{3}} + \frac{3}{\sqrt{3}} + \pi \qquad \text{ or } \qquad y = -\sqrt{3} x + \sqrt{3} + \pi
\end{aligned}
\end{equation}
$