Q. 1TB - Calculus

Question

Simplifying quotients: Simplify and rewrite the following expressions until you can cancel a common factor in the numerator and the denominator.

$\frac{{\left(x+h\right)}^4-x^4}{h}$

$\frac{{\left(x+h\right)}^{-2}-x^{-2}}{h}$

$\frac{z^4-x^4}{z-x}$

$\frac{z^{-2}-x^{-2}}{z-x}$

Step-by-Step Solution

Answer

The simplified solutions are:$4x^3+6x^{2}h+4x{h}^2+h^3$

                                               :$\frac{(-2x-h)}{{\left(x\right)}^2{(x+h)}^2}$

                                               :$(z+x)(z^2+x^2)$

                                               :$\frac{-(z+x)}{x^2{z}^2}.$

part(a) Step 1:

Given equation:$\frac{{\left(x+h\right)}^4-x^4}{h}$.

Step 2.The solution

The solution $$\begin{gathered} =\frac{{\left(x+h\right)}^4-x^4}{h} \\ =\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4x{h}^3+h^4-\cancel{x^4}}{h} \\ =\frac{\cancel{h}(4x^3+6x^{2}h+4x{h}^2+h^3)}{\cancel{h}} \\ =4x^3+6x^{2}h+4x{h}^2+h^3. \end{gathered}$$

Part(b) step1.Given Information

Given equation:$\frac{{\left(x+h\right)}^{-2}-x^{-2}}{h}.$

Step2.The solution

The solution is $$\begin{gathered} =\frac{{\left(x+h\right)}^{-2}-x^{-2}}{h} \\ =\frac{{\displaystyle \frac{1}{{\left(x+h\right)}^2}}-{\displaystyle \frac{1}{x^2}}}{h} \\ =\frac{{\displaystyle \frac{x^2-{\left(x+h\right)}^2}{{\left(x\right)}^2{(x+h)}^2}}}{h} \\ =\frac{{\displaystyle \frac{x^2-x^2-2hx-h^2}{{\left(x\right)}^2{(x+h)}^2}}}{h} \\ =\frac{{\displaystyle \frac{\cancel{h}(-2x-h)}{{\left(x\right)}^2{(x+h)}^2}}}{\cancel{h}} \\ =\frac{(-2x-h)}{{\left(x\right)}^2{(x+h)}^2}. \end{gathered}$$

part(c) Step1.Given Information

Given equation:$\frac{z^4-x^4}{z-x}$

Step2.The solution

The solution is:$$\begin{gathered} =\frac{z^4-x^4}{z-x} \\ =\frac{{\left(z^2\right)}^2-{\left(x^2\right)}^2}{z-x} \\ =\frac{\left(z^2-x^2\right)\left(z^2+x^2\right)}{z-x} \\ =\frac{\cancel{\left(z-x\right)}}{\cancel{\left(z-x\right)}}(z+x)(z^2+x^2) \\ =(z+x)(z^2+x^2) . \end{gathered}$$

Part(d) Step1.Given Information

Given equation:$=\frac{z^{-2}-x^{-2}}{z-x}$.

Step2.The solution

The solution:$$\begin{gathered} =\frac{z^{-2}-x^{-2}}{z-x} \\ =\frac{{\displaystyle \frac{1}{z^2}}-{\displaystyle \frac{1}{x^2}}}{z-x} \\ =\frac{{\displaystyle \frac{x^2-z^2}{x^2{z}^2}}}{z-x} \\ =\frac{{\displaystyle \frac{-(z+x)\cancel{\left(z-x\right)}}{x^2{z}^2}}}{\cancel{\left(z-x\right)}} \\ =\frac{-(z+x)}{x^2{z}^2}. \end{gathered}$$