Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 51

a.) $\displaystyle y = \frac{1}{1 + x^2}$ is a curve called Witch of Maria Agnesi. Find the equation of the tangent line to this curve at $P \left(\displaystyle -1, \frac{1}{2}\right)$

Required:

Equation of the tangent line to the curve at $P \left(\displaystyle -1, \frac{1}{2}\right)$

Solution:

Let $y' = m$ (slope)


$
\begin{equation}
\begin{aligned}

\qquad y' = m =& \frac{\displaystyle (1 + x^2) \frac{d}{dx} (1) - \left[ (1) \frac{d}{dx} (1 + x^2) \right]}{(1 + x^2)^2}
&&
\\
\\
\qquad y' = m =& \frac{(1 + x^2) (0) - (2x)}{(1 + x^2)^2} = \frac{-2x}{(1 + x^2)^2}
&&
\\
\\
\qquad m =& \frac{-2x}{(1 + x^2)^2}
&& \text{Substitute the value of $x$ which is $-1$}
\\
\\
\qquad m =& \frac{-2(-1)}{[1 + (-1)]^2}
&& \text{Simplify the equation}
\\
\\
\qquad m =& \frac{2}{4}
&& \text{Reduce to lowest term}
\\
\\
\qquad m =& \frac{1}{2}
&&

\end{aligned}
\end{equation}
$


Solving for the equation of the tangent line:


$
\begin{equation}
\begin{aligned}

\qquad y - y_1 =& m(x - x_1)
&& \text{Substitute the value of the slope $(m)$ and the given point}
\\
\\
\qquad y - \frac{1}{2} =& \frac{1}{2} [x - (-1)]
&& \text{Distribute $\large \frac{1}{2}$ in the equation}
\\
\\
\qquad y - \frac{1}{2} =& \frac{x + 1}{2}
&& \text{Add $\large \frac{1}{2}$ to each side}
\\
\\
\qquad y =& \frac{x + 1 + 1}{2}
&& \text{Combine like terms}
\\
\\
\qquad y =& \frac{x + 2}{2}
&& \text{Equation of the tangent line to the curve at $P \left(\displaystyle -1, \frac{1}{2}\right)$}


\end{aligned}
\end{equation}
$


b.) Graph the curve and the tangent line on part (a).