VCE Mathematical Methods/Differentiation from First Principles

Formula
Given a function f, the rule of the derivative (sometimes called the "gradient") function is defined as $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \, $$.

Method
Remember that in order to evaluate a limit, we usually substitute the value given into the expression. However, with the above formula, substituting $$ h = 0 $$ will result in a division by zero, which is mathematically impossible. Therefore,in order to make use of this formula, you need to substitute the rules $$ f(x+h) $$ and $$ f(x) $$, then simplify to eliminate the fraction, and only then substitute $$ h = 0 $$. This is called differentiation from first principles.

For example:

Let $$ f: \mathbb{R} \to \mathbb{R}, f(x) = 2x $$

Let us differentiate f from first principles.

$$ \begin{align} f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ & = \lim_{h \to 0} \frac{2(x+h) - 2x}{h} \\ & = \lim_{h \to 0} \frac{2x + 2h - 2x}{h} \\ & = \lim_{h \to 0} \frac{2h}{h} \\ & = \lim_{h \to 0} 2 \\ & = 2 \\ \end{align} $$. Therefore, we can define the gradient function as $$ f': \mathbb{R} \to \mathbb{R}, f'(x) = 2 $$

Exercises
Question One Differentiate the following functions from first principles. (a) $$ f(x) = 4x $$ (b) $$ f(x) = 7x $$ (c) $$ f(x) = 2x + 1 $$ (d) $$ f(x) = 3x + 3 $$

Question Two Differentiate the following functions from first principles. (a) $$ g(x) = x^2 $$ (b) $$ f(x) = 5x^2 $$ (c) $$ f(x) = 2x^2 + 3 $$ (d) $$ f(x) = (x+3)(x+4) $$