Chapter 3.3, Problem 20E

To determine

To calculate: The value of $\frac{dy}{dx}$ .

Answer to Problem 20E

The required value of $\frac{dy}{dx}$ is $\frac{x^2-2x-1}{{\left(1+x^2\right)}^2}$

Explanation of Solution

Given information:

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

Formula used:

Differentiation formula:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$

Calculation:

The given expression be $y=\left(1-x\right){\left(1+x^2\right)}^{-1}$ .

Rewrite the above expression as:

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

To find the value of the given expression, use the differentiation formula $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-v^{\prime }u}{v^2}$ .

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

Simplify the above expression further.

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

Hence, the required value of $\frac{dy}{dx}$ is $\frac{x^2-2x-1}{{\left(1+x^2\right)}^2}$ .