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Differentiation I

Differentiation from first principles

Differentiation from first principles

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Summary

Differentiation from first principles

In a nutshell

Differentiation from first principles is a method of calculating the gradient function f′(x)f'(x)f′(x) from an original function f(x)f(x)f(x). The gradient function f′(x)f'(x)f′(x) can be found by considering the gradient formula, together with two pairs of coordinates which are a small distance, hhh, 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(x1,y1)A(x_1,y_1) A(x1​,y1​) and B(x2,y2)B(x_2,y_2)B(x2​,y2​).


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


The gradient is given by:

​m=ΔyΔx=y2−y1x2−x1m= \dfrac {\Delta y}{\Delta x} = \dfrac {y_2-y_1}{x_2-x_1}m=ΔxΔy​=x2​−x1​y2​−y1​​​​


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


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


The gradient is given by:

​m=f(x+h)−f(x)hm=\dfrac {f(x+h)-f(x)}{h}m=hf(x+h)−f(x)​​​

​

To find the gradient at AAA, consider bringing the point BBB closer and closer to AAA. As BBB approaches AAA, the gradient of the chord will get closer to the gradient of the curve at AAA. It is possible to bring BBB closer to AAA by making hhh smaller until it approaches 000. The gradient of the curve at AAA is defined as the limiting value of the gradient of ABABAB as hhh approaches 000, and is written as:


​f′(x)=limh→0 f(x+h)−f(x)h\boxed{f'(x)=\underset{h\rightarrow 0}{lim} \space\dfrac {f(x+h)-f(x)}{h}}f′(x)=h→0lim​ hf(x+h)−f(x)​​​​


​​​

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

​1.1.1.​​

Use the function f(x)f(x)f(x) to find the value of f(x+h)f(x+h)f(x+h).​

​2.2.2.​​

Substitute f(x)f(x)f(x) and f(x+h)f(x+h)f(x+h) into f′(x)=limh→0 f(x+h)−f(x)hf'(x)=\underset{h\rightarrow 0}{lim} \space \dfrac {f(x+h)-f(x)}{h}f′(x)=h→0lim​ hf(x+h)−f(x)​.

​3.3.3.​​

Simplify the numerator and factorise by taking out hhh. 

​4.4.4.​​

Cancel hhh and find f′(x)f'(x)f′(x) in terms of xxx and hhh.

​5.5.5.​​

Find the limiting value of f′(x)f'(x)f′(x) as h→0h \rightarrow 0h→0.​


Example 1

Use differentiation from first principles to find the derivative of f(x)=x3f(x)=x^3f(x)=x3.


For the function f(x)=x3f(x)=x^3f(x)=x3,  find f(x+h)f(x+h)f(x+h).

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


Multiply out and simplify.

​f(x+h)=x3+3x2h+3xh2+h3f(x+h)=x^3+3x^2h+3xh^2+h^3f(x+h)=x3+3x2h+3xh2+h3​​


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


Substitute f(x)f(x)f(x) and f(x+h)f(x+h)f(x+h) into f′(x)=limh→0f(x+h)−f(x)hf'(x)=\underset{h\rightarrow 0}{lim} \dfrac {f(x+h)-f(x)}{h}f′(x)=h→0lim​hf(x+h)−f(x)​ and simplify.

​​

f′(x)=limh→0 f(x+h)−f(x)hf′(x)=limh→0 x3+3x2h+3xh2+h3−x3hf′(x)=limh→0 3x2h+3xh2+h3hf′(x)=limh→0 h(3x2+3xh+h2)hf′(x)=limh→03x2+3xh+h2\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}f′(x)f′(x)f′(x)f′(x)f′(x)​=h→0lim​ hf(x+h)−f(x)​=h→0lim​ hx3+3x2h+3xh2+h3−x3​=h→0lim​ h3x2h+3xh2+h3​=h→0lim​ hh(3x2+3xh+h2)​=h→0lim​3x2+3xh+h2​​​


Find the limiting value of f′(x)f'(x)f′(x) as h→0h \rightarrow 0h→0. As h→0, 3xh→0 and h2→0h \rightarrow 0, \ 3xh \rightarrow 0 \space and \space h^2 \rightarrow 0h→0, 3xh→0 and h2→0.​​

​f′(x)=3x2f'(x) = 3x^2f′(x)=3x2​​

​

Therefore, the derivative of f(x)=x3f(x)=x^3f(x)=x3 is f′(x)=3x2‾\underline{f'(x)=3x^2}f′(x)=3x2​.​


Example 2

The point A(2,9)A(2,9)A(2,9) lies on the curve y=x2+5y=x^2+5y=x2+5. Use differentiation from first principles to find the gradient at AAA.​


Find f(2)f(2)f(2)​.

​f(x)=x2+5f(2)=22+5=9\begin{aligned}f(x)& = x^2+5 \\f(2)&= 2^2+5 \\&= 9 \\\end {aligned}f(x)f(2)​=x2+5=22+5=9​​​


Find f(2+h)f(2+h)f(2+h).

​f(x+h)=(x+h)2+5f(2+h)=(2+h)2+5=4+4h+h2+5=9+4h+h2\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}f(x+h)f(2+h)​=(x+h)2+5=(2+h)2+5=4+4h+h2+5=9+4h+h2​​​


Substitute f(2)f(2)f(2) and f(2+h)f(2+h)f(2+h) into f′(x)=limh→0f(x+h)−f(x)hf'(x)=\underset{h\rightarrow 0}{lim} \dfrac {f(x+h)-f(x)}{h}f′(x)=h→0lim​hf(x+h)−f(x)​.

​​

​f′(2)=limh→0 f(2+h)−f(2)hf′(2)=limh→0 9+4h+h2−9hf′(2)=limh→0 4h+h2hf′(2)=limh→0 h(4+h)hf′(2)=limh→0 (4+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}f′(2)f′(2)f′(2)f′(2)f′(2)​=h→0lim​ hf(2+h)−f(2)​=h→0lim​ h9+4h+h2−9​=h→0lim​ h4h+h2​=h→0lim​ hh(4+h)​=h→0lim​ (4+h)​​​


Find the limiting value of f′(2)f'(2)f′(2). As h→0h \rightarrow 0h→0, 4+h→44+h \rightarrow 44+h→4.​

​f′(2)=4f'(2)= 4f′(2)=4​​


Therefore, the gradient of the curve at y=x2+5y=x^2+5y=x2+5 at A(2,9)A(2,9)A(2,9) is 4‾\underline{4}4​.​

​​​



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FAQs - Frequently Asked Questions

What is the limiting value of the gradient?

The limiting value of the gradient is the value where f'(x) approaches as h approaches 0.

What is the gradient formula?

The gradient formula calculates the gradient of a line segment AB from the co-ordinates of the two points A(x1,y1) and B(x2,y2). The gradient formula is m = (y2-y1)/(x2-x1).

What is differentiation from first principles?

Differentiation from first principles involves a method of calculating the derived function f'(x) from the original function f(x). The gradient formula is used with two pairs of co-ordinates which are a small distance, h, away from each other.

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