# VCE Mathematical Methods/Differentiation from First Principles

## Theory

### 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{aligned}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{aligned}} .
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) $f(x)=(x+3)(x+4)$