Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 23

a.) Suppose that $\displaystyle F(x) = \frac{5x}{ 1 + x^2}$, find $F'(2)$ and use it to find an equation of the tangent line to the curve $\displaystyle y = \frac{5x}{1 + x^2}$ at the point $(2,2)$

Using the definition of the derivative of a function $F$ at a number $a$, denoted by $F'(a)$, is

$\qquad \displaystyle \qquad F'(a) = \lim_{h \to 0} \frac{F(a + h) - F(a)}{h}$

We have,


$
\begin{equation}
\begin{aligned}

\qquad F'(a) =& \lim \limits_{h \to 0} \frac{\displaystyle \frac{5(a + h)}{1 + (a + h)^2} - \frac{5a}{1+ a^2}}{h}
&& \text{Substitute $F'(a + h)$ and $F(a)$}\\
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{(5a + 5h)(1 + a^2)- 5 a[1 + (a + h)^2]}{(h)(1 + a^2)[1 + (a + h)^2]}
&& \text{Get the LCD of the numerator}\\
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{5a + 5a^3 + 5h + 5a^2 h - 5a (a^2 + 2ah + h^2 + 1)}{(h)(1 + a^2)[1 + (a + h)^2]}
&& \text{Expand the equation}
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{\cancel{5a} + \cancel{5a^3} + 5h + 5a^2 h - \cancel{5a} - \cancel{5a^3} - 10a^2 h - 5ah^2}{(h)(1 + a^2)[1 + (a + h)^2]}
&& \text{Combine like terms and simplify}\\
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{5h - 5a^2 h - 5ah^2}{(h)(1 + a^2)[1 + (a+ h)^2]}
&& \text{Factor the numerator}\\
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{\cancel{h}(5 - 5a^2 - 5ah)}{\cancel{(h)}(1 + a^2)[1 + (a + h)^2]}
&& \text{Cancel out like terms}\\
\\
\qquad F'(a) =& \lim \limits_{h \to 0} \frac{5 - 5a^2 - 5ah}{(1 + a^2)[1 + (a + h)^2]} = \frac{5 - 5a^2 - 5a(0)}{1 + (a + 0)^2}
&& \text{Evaluate the limit}\\
\\
\qquad F'(a) =& \frac{5 - 5a^2}{(1 + a^2)^2}
&& \text{Substitute the value of $(a)$}\\
\\
\qquad F'(2) =& \frac{5 - 5(2)^2}{(1 + (2)^2)^2}
&& \text{Simplify}

\end{aligned}
\end{equation}
$



$\qquad \fbox{$F'(2) \displaystyle = \frac{-15}{25} = \frac{-3}{5}$} \qquad$
Slope of the tangent line at $(2,2)$

Using Point Slope Form where the tangent line $y = F(x)$ at $(a, F(a))$


$
\begin{equation}
\begin{aligned}

y - F(a) =& F'(a) (x -a)\\
&& \\
\\

y =& \frac{-3x + 6}{5} + 2
&& \text{Substitute the value of $a, F(a)$ and $F'(a)$}\\
\\

y =& \frac{-3x + 6 + 10}{5}
&& \text{Combine like terms}


\end{aligned}
\end{equation}
$


$\qquad \fbox{$ y = \displaystyle \frac{ -3x + 16}{5}$} \qquad$ Equation of the tangent line at $(2,2)$

b.) Draw a graph of the curve and the tangent line on the same screen.