Differentiation from first principles

In a nutshell

Differentiation from first principles is a method of calculating the gradient function $f'(x)$ from an original function $f(x)$ . The gradient function $f'(x)$ can be found by considering the gradient formula, together with two pairs of coordinates which are a small distance, $h$ , away from each other.

The gradient formula

When two points are joined up to make a chord, the gradient of the chord can be calculated by using the coordinates of the two points $A(x_1,y_1)$ and $B(x_2,y_2)$ .

Maths; Differentiation I; KS5 Year 12; Differentiation from first principles

The gradient is given by:

$m= \dfrac {\Delta y}{\Delta x} = \dfrac {y_2-y_1}{x_2-x_1}$

Consider the function $f(x)$ , where a point $A$ has co-ordinates $\big (x, f(x) \big)$ and point $B$ has coordinates a small distance, $h$ , away, giving $B\big ((x+h),f(x+h)\big )$ .

Maths; Differentiation I; KS5 Year 12; Differentiation from first principles

The gradient is given by:

$m=\dfrac {f(x+h)-f(x)}{h}$

To find the gradient at $A$ , consider bringing the point $B$ closer and closer to $A$ . As $B$ approaches $A$ , the gradient of the chord will get closer to the gradient of the curve at $A$ . It is possible to bring $B$ closer to $A$ by making $h$ smaller until it approaches $0$ . The gradient of the curve at $A$ is defined as the limiting value of the gradient of $AB$ as $h$ approaches $0$ , and is written as:

$\boxed{f'(x)=\underset{h\rightarrow 0}{lim} \space\dfrac {f(x+h)-f(x)}{h}}$

Differentiating from first principles

Using this rule is called differentiating from first principles. It can be used to find the derivative or gradient function in algebraic terms, or it can be applied to find the numerical value of the gradient at a particular point.

PROCEDURE

$1.$	Use the function $f(x)$ to find the value of $f(x+h)$ .
$2.$	Substitute $f(x)$ and $f(x+h)$ into $f'(x)=\underset{h\rightarrow 0}{lim} \space \dfrac {f(x+h)-f(x)}{h}$ .
$3.$	Simplify the numerator and factorise by taking out $h$ .
$4.$	Cancel $h$ and find $f'(x)$ in terms of $x$ and $h$ .
$5.$	Find the limiting value of $f'(x)$ as $h \rightarrow 0$ .

Example 1

Use differentiation from first principles to find the derivative of $f(x)=x^3$ .

For the function $f(x)=x^3$ , find $f(x+h)$ .

$f(x+h) = (x+h)^3$

Multiply out and simplify.

$f(x+h)=x^3+3x^2h+3xh^2+h^3$

Note: $(x+h)^3$ can be evaluated by expanding triple brackets or using the binomial expansion.

Substitute $f(x)$ and $f(x+h)$ into $f'(x)=\underset{h\rightarrow 0}{lim} \dfrac {f(x+h)-f(x)}{h}$ and simplify.

$\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{lim}\space \dfrac {f(x+h)-f(x)}{h} \\\\f'(x)&=\underset{h\rightarrow 0}{lim} \space \dfrac {x^3+3x^2h+3xh^2+h^3 - x^3}{h} \\\\f'(x)&=\underset{h\rightarrow 0}{lim} \space \dfrac {3x^2h+3xh^2+h^3}{h} \\\\f'(x)&=\underset{h\rightarrow 0}{lim} \space \dfrac {h(3x^2+3xh+h^2)}{h} \\\\f'(x)&=\underset{h\rightarrow 0}{lim} \quad3x^2+3xh+h^2\\\end{aligned}$

Find the limiting value of $f'(x)$ as $h \rightarrow 0$ . As $h \rightarrow 0, \ 3xh \rightarrow 0 \space and \space h^2 \rightarrow 0$ .

$f'(x) = 3x^2$

Therefore, the derivative of $f(x)=x^3$ is $\underline{f'(x)=3x^2}$ .

Example 2

The point $A(2,9)$ lies on the curve $y=x^2+5$ . Use differentiation from first principles to find the gradient at $A$ .

Find $f(2)$ .

$\begin{aligned}f(x)& = x^2+5 \\f(2)&= 2^2+5 \\&= 9 \\\end {aligned}$

Find $f(2+h)$ .

$\begin{aligned}f(x+h) &= (x+h)^2+5 \\f(2+h)&=(2+h)^2+5 \\&= 4+4h+h^2+5 \\&=9 + 4h+h^2\end {aligned}$

Substitute $f(2)$ and $f(2+h)$ into $f'(x)=\underset{h\rightarrow 0}{lim} \dfrac {f(x+h)-f(x)}{h}$ .

$\begin{aligned}f'(2)&=\underset{h\rightarrow 0}{lim}\space \dfrac {f(2+h)-f(2)}{h} \\\\f'(2)&=\underset{h\rightarrow 0}{lim} \space \dfrac {9+4h+h^2-9}{h} \\\\f'(2)&=\underset{h\rightarrow 0}{lim} \space \dfrac {4h+h^2}{h} \\\\f'(2)&=\underset{h\rightarrow 0}{lim} \space \dfrac {h(4+h)}{h} \\\\f'(2)&=\underset{h\rightarrow 0}{lim} \space (4+h) \\\end {aligned}$

Find the limiting value of $f'(2)$ . As $h \rightarrow 0$ , $4+h \rightarrow 4$ .

$f'(2)= 4$

Therefore, the gradient of the curve at $y=x^2+5$ at $A(2,9)$ is $\underline{4}$ .