Absolutely! Let's derive the difference of the quotient [tex]\(\Delta\left[\frac{f(x)}{g(x)}\right]\)[/tex] step by step.
Firstly, we start with the definition of the difference operator [tex]\(\Delta\)[/tex] applied to a general function [tex]\(h(x)\)[/tex]:
[tex]\[
\Delta h(x) = h(x+h) - h(x)
\][/tex]
Given two functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex], we want to find [tex]\(\Delta\left[\frac{f(x)}{g(x)}\right]\)[/tex]. Note that:
[tex]\[
\Delta\left[\frac{f(x)}{g(x)}\right] = \left(\frac{f(x)}{g(x)}\right)(x+h) - \left(\frac{f(x)}{g(x)}\right)(x)
\][/tex]
Let's denote [tex]\(F(x) = \frac{f(x)}{g(x)}\)[/tex]. Applying the definition of [tex]\(\Delta\)[/tex], we get:
[tex]\[
\Delta F(x) = \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}
\][/tex]
To combine these fractions, we find a common denominator:
[tex]\[
\Delta F(x) = \frac{f(x+h)g(x) - f(x)g(x+h)}{g(x+h)g(x)}
\][/tex]
Next, we break this expression into parts that are easier to manage by introducing the differences [tex]\(\Delta f(x)\)[/tex] and [tex]\(\Delta g(x)\)[/tex]:
[tex]\[
\Delta f(x) = f(x+h) - f(x) \quad \text{and} \quad \Delta g(x) = g(x+h) - g(x)
\][/tex]
Using these definitions, we can rewrite [tex]\(f(x+h)\)[/tex] and [tex]\(g(x+h)\)[/tex] in terms of the differences:
[tex]\[
f(x+h) = f(x) + \Delta f(x) \quad \text{and} \quad g(x+h) = g(x) + \Delta g(x)
\][/tex]
Now substitute these into the expression for [tex]\(\Delta F(x)\)[/tex]:
[tex]\[
\Delta F(x) = \frac{(f(x) + \Delta f(x))g(x) - f(x)(g(x) + \Delta g(x))}{g(x+h)g(x)}
\][/tex]
Expanding the numerator, we obtain:
[tex]\[
\Delta F(x) = \frac{f(x)g(x) + \Delta f(x)g(x) - f(x)g(x) - f(x)\Delta g(x)}{g(x+h)g(x)}
\][/tex]
Simplifying the terms in the numerator:
[tex]\[
\Delta F(x) = \frac{\Delta f(x)g(x) - f(x)\Delta g(x)}{g(x+h)g(x)}
\][/tex]
Hence, we have derived the expression:
[tex]\[
\Delta\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)\Delta f(x) - f(x)\Delta g(x)}{g(x+h)g(x)}
\][/tex]
Thus, we've completed the proof.
