Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 8

Determine the equation of the tangent line to the curve $\displaystyle y = \frac{2x}{(x + 1)^2}$ at the point (0, 0)

$\text{Using the definition (Slope of the tangent line)}$

$\displaystyle m = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

We have $a = 0$ and $\displaystyle f(x) = \frac{2x}{(x + 1)^2}$, so the slope is


$\begin{equation} \begin{aligned} \displaystyle m =& \lim \limits_{x \to 0} \frac{f(x) - f(0)}{x - 0} && \\ \\ \displaystyle m =& \lim \limits_{x \to 0} \frac{\frac{2x}{(x + 1)^2} - \left[ \frac{2(0)}{(0 + 1)^2} \right]}{x} && \text{Substitute value of $a$ and $x$}\\ \\ \displaystyle m =& \lim \limits_{x \to 0} \frac{2\cancel{x}}{\cancel{x}(x + 1)^2} && \text{Cancel out like terms}\\ \\ \displaystyle m =& \lim \limits_{x \to 0} \frac{2}{(x + 1)^2} = \frac{2}{(0+1)^2} && \text{Evaluate the limit}\\ \\ m =& 2 && \end{aligned} \end{equation}$


Using point slope form


$\begin{equation} \begin{aligned} y - y_1 &= m ( x - x_1) && \\ \\ y - 0 &= 2 ( x - 0 ) && \text{ Substitute value of $x, y$ and $m$ and simplify}\\ \\ y & = 2x \end{aligned} \end{equation}$



Therefore,
The equation of the tangent line at (0,0) is $y = 2x$.