5.8.1 First Principles Differentiation

Differentiation from first principles uses the definition of the derivative of a function f(x)
The definition is

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$\lim_{h \to 0}$ means the 'limit as h tends to zero'
When $h = 0$ , $\frac{f (x + h) - f (x)}{h} = \frac{f (x) - f (x)}{0} = \frac{0}{0}$ which is undefined
- Instead we consider what happens as h gets closer and closer to zero
Differentiation from first principles means using that definition to show what the derivative of a function is
The first principles definition (formula) is in the formula booklet

STEP 1: Identify the function f(x) and substitute this into the first principles formula

e.g. Show, from first principles, that the derivative of 3x² is 6x

f (x) = 3 x^{2}

f' (x) = \underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h} = \underset{h \to 0}{l i m} \frac{3 {(x + h)}^{2} - 3 x^{2}}{h^{}}

STEP 2: Expand f(x+h) in the numerator

f' (x) = \underset{h \to 0}{l i m} \frac{3 (x^{2} + 2 h x + h^{2}) - 3 x^{2}}{h}

f' (x) = \underset{h \to 0}{l i m} \frac{3 x^{2} + 6 h x + 3 h^{2} - 3 x^{2}}{h}

STEP 3: Simplify the numerator, factorise and cancel h with the denominator

f' (x) = \underset{h \to 0}{l i m} \frac{h (6 x + 3 h)}{h}

STEP 4: Evaluate the remaining expression as h tends to zero

f' (x) = \underset{h \to 0}{l i m} (6 x + 3 h) = 6 x

As h \to 0, (6 x + 3 h) \to (6 x + 0) \to 6 x

∴

The derivative of

3 x^{2}

6 x

Most of the time you will not use first principles to find the derivative of a function (there are much quicker ways!)
However, you can be asked to demonstrate differentiation from first principles
To get full marks make sure you are are writing lim h -> 0 right up until the concluding sentence!

Prove, from first principles, that the derivative of $5 x^{3}$ is $15 x^{2}$ .

tH5zRwYR_5-8-1-ib-he-as-only-1st-princ-we-soltn

DP IB Maths: AA HL

5.8.1 First Principles Differentiation