Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 55

a.) Determine the equation of the tangent line to the curve $\displaystyle y= \tan \left( \frac{\pi x^2}{4} \right)$ at the point $(1, 1)$.

Solving for the slope


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{d}{dx} \left[ \tan \left( \frac{\pi x^2}{4} \right) \right]
\\
\\
m =& \sec^2 \left( \frac{\pi x^2}{4} \right) \cdot \frac{d}{dx} \left( \frac{\pi x^2}{4} \right)
\\
\\
m =& \sec ^2 \left( \frac{\pi x^2}{4} \right) \cdot \left( \frac{\pi }{4} \right) \frac{d}{dx} (x^2)
\\
\\
m =& \sec ^ 2 \left( \frac{\pi x^2}{4} \right) \cdot \left( \frac{\pi}{4} \right) (2x)
\\
\\
m =& \sec ^2 \left( \frac{\pi x^2}{4} \right) \cdot \frac{\pi x}{2}
\\
\\
m =& x (5 + x^2) ^{\frac{-1}{2}}
\\
\\
m =& \frac{\displaystyle \pi x \sec ^2 \left( \frac{\pi x^2}{4} \right)}{2}
\\
\\
m =& \frac{\displaystyle \pi 1 \sec ^2 \left( \frac{\pi (1)^2}{4} \right)}{2}
\\
\\
m =& \pi


\end{aligned}
\end{equation}
$



Using the Point Slope Form



$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - 1 =& \pi (x - 1)
\\
\\
y - 1 + 1=& \pi (x - 1) + 1
\\
\\
y =& \pi (x - 1) + 1
\qquad \qquad \text{Equation of the tangent line at $(1,1)$}

\end{aligned}
\end{equation}
$


b.) Graph the curve and the tangent line on the same screen.