Chapter 3.2, Problem 37E

(A)

To determine

To find:

The differentiation of the given function

Answer to Problem 37E

The differential equation is $f^{\prime }=\frac{4x}{{\left(x^2+1\right)}^2}$ , $f^{″}\left(x\right)=\frac{\left\{4-12x^2\right\}}{{\left(x^2+1\right)}^3}$

Explanation of Solution

Given:

The function is

  $y=\frac{\left(x^2-1\right)}{\left(x^2+1\right)}$

Concept used:

Definition of the differentiation:-Differentiation is the action of computing a derivative

The derivative of a function $y=f\left(x\right)$ of a variable $x$ is a measure of the rate at which the value y of the function changes with respect to $x$

Calculation:

The function

  $y=\frac{\left(x^2-1\right)}{\left(x^2+1\right)}\mathrm{..................}\left(1\right)$

The derivative of a function

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

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

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{\left(x^2-1\right)}{\left(x^2+1\right)}\right)\\ \frac{dy}{dx}=\frac{\left(x^2+1\right)\frac{d}{dx}\left(x^2-1\right)-\left(x^2-1\right)\frac{d}{dx}\left(x^2+1\right)}{{\left(x^2+1\right)}^2}\\ \frac{dy}{dx}=\frac{\left(x^2+1\right)\left(2x\right)-\left(x^2-1\right)\left(2x\right)}{{\left(x^2+1\right)}^2}\\ f^{\prime }=\frac{\left(2x^3+2x\right)-\left(2x^3-2x\right)}{{\left(x^2+1\right)}^2}\\ f^{\prime }=\frac{4x}{{\left(x^2+1\right)}^2}\end{array}$

Again differentiating the above equation

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{4x}{{\left(x^2+1\right)}^2}\\ f^{″}\left(x\right)=\frac{{\left(x^2+1\right)}^2\frac{d}{dx}\left(4x\right)-\left(4x\right)\frac{d}{dx}{\left(x^2+1\right)}^2}{{\left(x^2+1\right)}^4}\\ f^{″}\left(x\right)=\frac{{\left(x^2+1\right)}^2\left(4\right)-16x^2\left(x^2+1\right)}{{\left(x^2+1\right)}^4}\\ f^{″}\left(x\right)=\frac{\left\{4x^2+4-16x^2\right\}}{{\left(x^2+1\right)}^3}\\ f^{″}\left(x\right)=\frac{\left\{4-12x^2\right\}}{{\left(x^2+1\right)}^3}\end{array}$

(B)

To determine

To draw:

The graph of the curve at the given point

Explanation of Solution

Given:

The function is

  $y=\frac{\left(x^2-1\right)}{\left(x^2+1\right)}$

Concept used:

An equation of a line $l$ through a point $P_1\left(x_1,y_1\right)$ with slope $m$ . If $P\left(x,y\right)$ is any point with $x\ne x_1$

Then P is on $l$ if and only if the slope of the line through $P_1$ and P

The equation is $y-y_1=m\left(x-x_1\right)$

Calculation:

Draw the table

  $f\left(x\right)=\frac{\left(x^2-1\right)}{\left(x^2+1\right)}$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$-1$	$1$	$0$	$2$
$y-axis$	$0$	$0$	$1$	$0.6$