Chapter 4.5, Problem 16E

i.

To determine

To find: To find the $dy$ for the given function $y=\frac{2x}{1+x^2}$ .

Answer to Problem 16E

The $dy$ for the given function $y=\frac{2x}{1+x^2}$ is $d(y)=\frac{2-2x^2}{{(1+x^2)}^2}dx$ .

Explanation of Solution

Given: The function $y=\frac{2x}{1+x^2}$

Formula used: $d(\frac{u}{v})=\frac{vu\text{'}-uv\text{'}}{v^2}$

Calculation:

The given function is $y=\frac{2x}{1+x^2}$

By applying the differentiation formula,

  $dy=d(\frac{2x}{1+x^2})$

  $\begin{array}{l}u=2x\\ u\text{'}=2\\ v=1+x^2\\ v=2x\end{array}$

  $\begin{array}{l}d(\frac{2x}{1+x^2})=\frac{(1+x^2)2-(2x)(2x)}{{(1+x^2)}^2}dx\\ d(\frac{2x}{1+x^2})=\frac{2+2x^2-4x^2}{{(1+x^2)}^2}dx\\ d(y)=\frac{2-2x^2}{{(1+x^2)}^2}dx\end{array}$

Conclusion:

The $dy$ for the given function $y=\frac{2x}{1+x^2}$ is $d(y)=\frac{2-2x^2}{{(1+x^2)}^2}dx$ .

ii.

To determine

To find: The value of $dy$ for the function $y=\frac{2x}{1+x^2}$ when $x=-2$ and $dx=0.1$ .

Answer to Problem 16E

The value of $dy$ for the function $y=\frac{2x}{1+x^2}$ when $x=-2$ and $dx=0.1$ is $dy=-0.024$ .

Explanation of Solution

Given: The function $y=\frac{2x}{1+x^2}$

Formula used: $d(x^n)=n{x}^{n-1}$

Calculation:

The given function is $y=\frac{2x}{1+x^2}$ .

The differentiation of the function is,

  $d(y)=\frac{2-2x^2}{{(1+x^2)}^2}dx$

Applying $x=-2$ and $dx=0.1$ ,

  $\begin{array}{l}d(y)=\frac{2-2x^2}{{(1+x^2)}^2}dx\\ dy=\frac{2-2{(-2)}^2}{{(1+{(-2)}^2)}^2}(0.1)\\ dy=\frac{2-8}{{(1+4)}^2}(0.1)\\ dy=\frac{-6}{25}(0.1)\\ dy=-0.024\end{array}$

Conclusion:

The value of $dy$ for the function $y=\frac{2x}{1+x^2}$ when $x=-2$ and $dx=0.1$ is $dy=-0.024$ .