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 $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$.

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 $f:ℝ\to ℝ,f(x)=2x$

Let us differentiate f from first principles.

$\begin{array}{cc}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{2(x+h)-2x}{h}\\ & =\underset{h\to 0}{\lim }\frac{2x+2h-2x}{h}\\ & =\underset{h\to 0}{\lim }\frac{2h}{h}\\ & =\underset{h\to 0}{\lim }2\\ & =2\\ \end{array}$.
Therefore, we can define the gradient function as $f^{\prime }:ℝ\to ℝ,f^{\prime }(x)=2$

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) $f(x)=(x+3)(x+4)$