# 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 ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\,}$.

### Method

Remember that in order to evaluate a limit, we usually substitute the value given into the expression. However, with the above formula, substituting ${\displaystyle h=0}$ 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 ${\displaystyle f(x+h)}$ and ${\displaystyle f(x)}$, then simplify to eliminate the fraction, and only then substitute ${\displaystyle h=0}$. This is called differentiation from first principles.

For example:

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

Let us differentiate f from first principles.

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

## Exercises

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

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