Chapter 2.3, Problem 42E

To determine

To calculate: The equation of the tangent.

Answer to Problem 42E

The equation of the tangent at $\left(2,1\right)$ is $y=1-\frac{1}{2}\left(x-1\right)$

Explanation of Solution

Given information:

The equation is $y=\frac{8}{4+x^2}$ , and the point is $\left(2,1\right)$ .

Concept used:

The formulae used are $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and quotient rule is $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}\left(u\right)+u.\frac{d}{dx}.\left(v\right)}{v^2}$

Calculation:

Slope of tangent line.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(\frac{8}{4+x^2}\right)\\ =\frac{\left(4+x^2\right)\frac{d}{dx}\left(8\right)-8\frac{d}{dx}\left(4+x^2\right)}{{\left(4+x^2\right)}^2}\\ =\frac{0-8\left(2x\right)}{{\left(4+x^2\right)}^2}\end{array}$

Solve further.

  $m=-\frac{16x}{{\left(4+x^2\right)}^2}$

Solve further by substituting $2$ for $x$ in $m=-\frac{16x}{{\left(4+x^2\right)}^2}$

  $\begin{array}{l}m=-\frac{16\times 2}{{\left(4+{\left(2\right)}^2\right)}^2}\\ =-\frac{32}{{\left(8\right)}^2}\\ =-\frac{32}{64}\\ =-\frac{1}{2}\end{array}$

The slope of tangent line is $-\frac{1}{2}$

For the point $\left(2,1\right)$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}y-1=-\frac{1}{2}\left(x-1\right)\\ y=1-\frac{1}{2}\left(x-1\right)\end{array}$

The equation of tangent line at $\left(2,1\right)$is $y=1-\frac{1}{2}\left(x-1\right)$.

Conclusion: The equation of tangent line at $\left(2,1\right)$ is $y=1-\frac{1}{2}\left(x-1\right)$ .