Chapter 1.4, Problem 69E

To determine

To calculate: The expression $\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}$

Answer to Problem 69E

  $\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}=\frac{-1}{\left(1+x+h\right)\left(1+x\right)}$

Explanation of Solution

Given information:

Given expression is $\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}$

Consider the expression as $\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}$

Now,

  $\begin{array}{c}\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}=\frac{\frac{1+x-\left(1+x+h\right)}{\left(1+x+h\right)\left(1+x\right)}}{h}\\ =\frac{\frac{1+x-1-x-h}{\left(1+x+h\right)\left(1+x\right)}}{h}\\ =\frac{-h}{h\left(1+x+h\right)\left(1+x\right)}\\ =\frac{-1}{\left(1+x+h\right)\left(1+x\right)}\end{array}$

Therefore, simplest form of the expression $\frac{\frac{1}{1+x+h}-\frac{1}{1+x}}{h}=\frac{-1}{\left(1+x+h\right)\left(1+x\right)}$