Chapter 4.1, Problem 79E

To determine

To calculate: To show that $\frac{dG}{dx}=\frac{A}{G}$ .

Answer to Problem 79E

Hence, $\frac{dG}{dx}=\frac{A}{G}$

Explanation of Solution

Given Information: $G=\sqrt{uv},A=\frac{u+v}{2},u=x,v=x+c$

Calculation:

The value of G in terms of x and y is,

  $\begin{array}{c}G=\sqrt{uv}\\ =\sqrt{x\left(x+c\right)}\\ =\sqrt{x^2+xc}\end{array}$

The value of A in terms of x and y is,

  $\begin{array}{c}A=\frac{u+v}{2}\\ =\frac{x+x+c}{2}\\ =\frac{2x+c}{2}\\ =x+\frac{c}{2}\end{array}$

The value of $\frac{dG}{dx}$ is,

  $\begin{array}{c}\frac{dG}{dx}=\frac{d}{dx}\left(\sqrt{x^2+xc}\right)\\ =\frac{1}{2\sqrt{x^2+xc}}\times \frac{d}{dx}\left(x^2+xc\right)\\ =\frac{1}{2\sqrt{x^2+xc}}\times \left(2x+c\right)\\ =\frac{2x+c}{2\sqrt{x^2+xc}}\\ =\frac{2\left(x+\frac{c}{2}\right)}{2\sqrt{x^2+xc}}\\ =\frac{x+\frac{c}{2}}{\sqrt{x^2+xc}}\\ =\frac{A}{G}\end{array}$