Chapter 2.3, Problem 41E

To determine

To calculate: The equation of the tangent.

Answer to Problem 41E

The equation of the tangent at $\left(0,0\right)$ is $y=4x$and at $\left(1,2\right)$ is $y=2$ .

Explanation of Solution

Given information:

The equation is $y=\frac{4x}{x^2+1}$ , and the two points are $\left(0,0\right)$ and $\left(1,2\right)$ .

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.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(\frac{4x}{x^2+1}\right)\\ =\frac{\left(x^2+1\right)\frac{d}{dx}\left(4x\right)-4x\frac{d}{dx}\left(x^2+1\right)}{{\left(x^2+1\right)}^2}\\ =\frac{4\left(x^2+1\right)-4x\left(2x\right)}{{\left(x^2+1\right)}^2}\end{array}$

Solve further.

  $\begin{array}{c}m=\frac{4x+4-8x^2}{{\left(x^2+1\right)}^2}\\ m=\frac{4-4x^2}{{\left(x^2+1\right)}^2}\end{array}$

Solve further by substituting $0$ for $x$ in $m=\frac{4-4x^2}{{\left(x^2+1\right)}^2}$

  $\begin{array}{c}m=\frac{4-4x^2}{{\left(x^2+1\right)}^2}\\ =\frac{4-4\left(0\right)}{{\left(0+1\right)}^2}\\ =\frac{4-0}{1}\\ =4\end{array}$

The slope of tangent line is $4$

For the point $\left(0,0\right)$

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

  $\begin{array}{l}y-0=4\left(x-0\right)\\ y=4x\end{array}$

The equation of tangent line at $\left(0,0\right)$is $y=4x$.

Solve further by substituting $1$ for $x$ in $m=\frac{4-4x^2}{{\left(x^2+1\right)}^2}$

  $\begin{array}{c}m=\frac{4-4x^2}{{\left(x^2+1\right)}^2}\\ =\frac{4-4\left(1\right)}{{\left(1+1\right)}^2}\\ =\frac{4-4}{4}\\ =0\end{array}$

The slope of tangent line is $0$

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

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

  $\begin{array}{l}y-2=0\left(x-1\right)\\ y=2\end{array}$

The equation of tangent line at $\left(1,2\right)$is $y=2$.

Conclusion: The equation of tangent line at $\left(0,0\right)$ is $y=4x$ and at $\left(1,2\right)$ at $y=2$ .