Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 56

a.) Determine the equation of the tangent line to the curve $\displaystyle y= \frac{|x|}{\sqrt{2 - x^2}}$ which is called a Bullet-nose Curve at the point $(1, 1)$.

Solving for the slope


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{d}{dx} \left( \frac{x}{\sqrt{2 - x^2}} \right)
\\
\\
m =& \frac{d}{dx} \left[ \frac{x }{( 2 -x^2)^{\frac{1}{2}}} \right]
\\
\\
m =& \frac{\displaystyle \left[ (2 - x^2)^{\frac{1}{2}} \cdot \frac{d}{dx} (x) \right] - \left[ (x) \cdot \frac{d}{dx} (2 - x^2)^{\frac{1}{2}} \right] }{\left[(2 - x^2)^{\frac{1}{2}}\right]^2 }
\\
\\
m =& \frac{\displaystyle (2 - x^2)^{\frac{1}{2}} (1) - (x) \left(\frac{1}{2} \right) (2 - x^2)^{\frac{-1}{2}} \cdot \frac{d}{dx} (2 - x^2) }{2 - x^2}
\\
\\
m =& \frac{\displaystyle (2 - x^2)^{\frac{1}{2}} - (x) \left(\frac{1}{2} \right) (2 - x^2)^{\frac{-1}{2}}(0 - 2x) }{2 - x^2}
\\
\\
m =& \frac{\displaystyle (2 - x^2)^{\frac{1}{2}} - \left( \frac{- \cancel{2} x^2}{\cancel{2}} \right) (2 - x^2)^{\frac{-1}{2}} }{2 - x^2 }
\\
\\
m =& \frac{(2 - x^2)^{\frac{1}{2}} + (x^2) (2 - x^2)^{\frac{-1}{2}}}{2 - x^2}
\\
\\
m =& \frac{\displaystyle (2 - x^2)^{\frac{1}{2}} + \frac{x^2}{(2 - x^2)^{\frac{1}{2}}}}{2 - x^2}
\\
\\
m =& \frac{\displaystyle \frac{ 2 - \cancel{x^2} + \cancel{x^2}}{(2 - x^2)^{\frac{1}{2}}}}{2 - x^2}
\\
\\
m =& \frac{2}{(2 - x^2)^{\frac{1}{2}} (2 - x^2)}
\\
\\
m =& \frac{2}{(2 - x^2)^{\frac{3}{2}}}
\\
\\
m =& \frac{2}{[2 - (1)^2]^{\frac{3}{2}}}
\\
\\
m =& \frac{2}{(1)^{\frac{3}{2}}}
\\
\\
m =& \frac{2}{1}
\\
\\
m =& 2

\end{aligned}
\end{equation}
$



Using the Point Slope Form



$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - 1 =& 2 (x - 1)
\\
\\
y - 1 =& 2x - 2
\\
\\
y - 1 + 1=& 2x - 2 + 1
\\
\\
y =& 2x - 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.