$
f(x) = x^3 \\[3ex]
f(x + h) = (x + h)^3 \\[3ex]
f(x + h) = x^3 + 3x^2h + 3xh^2 + h^3 \\[3ex]
$
Student: That was fast.
How did you get it?
Teacher: I used the Pascal's Triangle.
Student: What is it?
Teacher: Here is the video: Pascal's Triangle and the Binomial Theorem
You can expand binomials raised to any positive exponent quickly using that method.
Would you like for me to do it another way?
Student: Yes, please.
$
f(x) = -\dfrac{1}{16x} \\[5ex]
f(x + h) = -\dfrac{1}{16(x + h)} \\[5ex]
DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]
$
Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.
$
f(x) = \dfrac{x}{x + 1} \\[5ex]
f(x + h) = \dfrac{x + h}{x + h + 1} \\[5ex]
DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]
$
Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.
$
Numerator = f(x + h) - f(x) \\[3ex]
f(x + h) - f(x) = \dfrac{x + h}{x + h + 1} - \dfrac{x}{x + 1} \\[5ex]
= \dfrac{(x + h)(x + 1)}{(x + h + 1)(x + 1)} - \dfrac{x(x + h + 1)}{(x + h + 1)(x + 1)} \\[5ex]
(x + h)(x + 1) = x^2 + x + hx + h \\[3ex]
x(x + h + 1) = x^2 + hx + x \\[3ex]
= \dfrac{x^2 + x + hx + h}{(x + h + 1)(x + 1)} - \dfrac{x^2 + hx + x}{(x + h + 1)(x + 1)} \\[5ex]
= \dfrac{(x^2 + x + hx + h) - (x^2 + hx + x)}{(x + h + 1)(x + 1)} \\[5ex]
= \dfrac{(x^2 + x + hx + h - x^2 - hx - x)}{(x + h + 1)(x + 1)} \\[5ex]
= \dfrac{(x^2 + x + hx + h - x^2 - hx - x)}{(x + h + 1)(x + 1)} \\[5ex]
= \dfrac{h}{(x + h + 1)(x + 1)} \\[5ex]
Denominator = h \\[3ex]
DQ = \dfrac{Numerator}{Denominator} \\[5ex]
DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]
DQ = [f(x + h) - f(x)] \div h \\[5ex]
DQ = \dfrac{h}{(x + h + 1)(x + 1)} \div \dfrac{h}{1} \\[5ex]
DQ = \dfrac{h}{(x + h + 1)(x + 1)} * \dfrac{1}{h} \\[5ex]
DQ = \dfrac{1}{(x + h + 1)(x + 1)}
$
$
f(x) = \dfrac{x - 3}{x + 9} \\[5ex]
f(x + h) = \dfrac{(x + h) - 3}{(x + h) + 9} \\[5ex]
f(x + h) = \dfrac{x + h - 3}{x + h + 9} \\[5ex]
DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex]
$
Let us do this one at a time - piecemeal approach
Let us deal with the numerator first.
