Chapter 5.5, Problem 22E

a.

To determine

To find $dy$

Answer to Problem 22E

  $dy=\frac{x^2+4x}{{\left(2+x\right)}^2}dx$

Explanation of Solution

Given information:

The given equation is $2y=x^2-xy$

Separate x and y in the equation

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

Differentiate each side

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

b.

To determine

To evaluate $dy$ using the value of $x\text{ and }dx$

Answer to Problem 22E

The value of $dy$ is $-0.0375$ at the given values

Explanation of Solution

Given information:

The value of x is 2 and $dx$ is $-0.05$

The equation is

  $\begin{array}{l}dy=\frac{x^2+4x}{{\left(2+x\right)}^2}dx\\ dy=\frac{{\left(2\right)}^2+4\cdot 2}{{\left(2+2\right)}^2}\times -0.05\left[\text{ substitute the value of }x\text{ and }dx\right]\\ dy=-0.0375\left[\text{simplify}\right]\end{array}$