Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 49

Find the equation of the tangent line of the curve $\displaystyle y = \frac{2x}{x + 1}$ at Point $(1,1)$

Required:

Equation of the tangent line to the curve at $P(1,1)$

Solution:

Let $y' = m$ (slope)


$
\begin{equation}
\begin{aligned}

\qquad y' = m =& \frac{\displaystyle (x + 1) \frac{d}{dx} (2x) - \left[ (2x) \frac{d}{dx} (x + 1) \right]}{(x + 1)^2}
&& \text{Apply Quotient Rule}
\\
\\
\qquad m =& \frac{(x + 1) (2) - (2x)(1)}{(x + 1)^2}
&& \text{Simplify the equation}
\\
\\
\qquad m =& \frac{\cancel{2x} + 2 - \cancel{2x}}{(x + 1)^2}
&& \text{Combine like terms}
\\
\\
\qquad m =& \frac{2}{(x + 1)^2}
&& \text{Substitute value of $x$ which is 1}
\\
\\
\qquad m =& \frac{2}{(1 + 1)^2}
&& \text{}
\\
\\
\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 - 1 =& \frac{1}{2} (x - 1)
&& \text{Add 1 to each side}
\\
\\
\qquad y =& \frac{x - 1}{2} + 1
&& \text{Get the LCD}
\\
\\
\qquad y =& \frac{x - 1 + 2}{2}
&& \text{Combine like terms}
\\
\\
\qquad y =& \frac{x + 1}{2}
&& \text{Equation of the tangent line to the curve at $P(1, 1)$}
\\
\\

\end{aligned}
\end{equation}
$