Chapter A.3, Problem 67E

Solution

Simplify the compound fraction.

  $\frac{-(h+2x)}{x^2{(x+h)}^2}$ .

Given:

The given compound fraction is $\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}$ .

Concept Used:

• Compound fraction is a fraction with fractions in the numerator or denominator.
• First multiply the denominator of the fraction by the whole number. 
• Then, add that number to the numerator of the fraction, and keep the original denominator the same.

Calculation:

The given compound fraction is $\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}$ .

  $\begin{array}{c}\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}=\frac{\frac{x^2-{(x+h)}^2}{x^2{(x+h)}^2}}{h}\\ =\frac{\frac{x^2-(x^2+h^2+2xh)}{x^2{(x+h)}^2}}{h}\\ =\frac{x^2-x^2-h^2-2xh}{x^2{(x+h)}^{2}h}\\ =\frac{-h^2-2xh}{x^2{(x+h)}^{2}h}\\ =\frac{-h(h+2x)}{h{x}^2{(x+h)}^2}\\ =\frac{-(h+2x)}{x^2{(x+h)}^2}\end{array}$

Thus by simplifying the compound fraction $\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}$ the solution is $\frac{-(h+2x)}{x^2{(x+h)}^2}$ .