Chapter 3.2, Problem 53E

To determine

To find:

The equation of the tangent and parallel to the line

Answer to Problem 53E

The equation of the line is $y=\frac{x-1}{{\left(x+1\right)}^2}+2$

Explanation of Solution

Given:

The tangent to the curve

  $y=\frac{x}{x+1}$

Concept used:

The equation is in slope −intercept form, $y=mx+c$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=\frac{x}{x+1}\mathrm{.......................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}=m\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}y=\frac{d}{dx}\left(\frac{x}{x+1}\right)\\ \frac{dy}{dx}=\frac{x+1\frac{d}{dx}\left(x\right)-\left(x\right)\frac{d}{dx}x+1}{{\left(x+1\right)}^2}\\ \frac{dy}{dx}=\frac{x+1-\left(x\right)}{{\left(x+1\right)}^2}\\ \frac{dy}{dx}=\frac{1}{{\left(x+1\right)}^2}\end{array}$

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-2=\frac{1}{{\left(x+1\right)}^2}\left(x-1\right)\\ y=\frac{x-1}{{\left(x+1\right)}^2}+2\end{array}$