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}$