Rules of Differentiation
What is differentiation?
Differentiation is the process of finding the derivative of a function. This derivitive can be defined as the rate of change of the function based on one variable (typically x, but sometimes y).
The derivitive is itself another function, that will take a variable to produce what is called the 'instantaneous rate of change', i.e. the rate of change for the funcion at any given point.
It's fine if this dosn't make sense at the moment, keep reading. Like much of this course, it is best understood graphically:
The derivitive can be defined as the gradient of a tangent to some function f(x) at some point A. For instance,
\(y = x^2\)
Will have it's derivitive function, when \(x = 2\), equal to the gradient of the straight, red line. We can, in this case find the value of the derivitive function at \(x = 2\) by calculatng the gradient (\(m\)) of the red line:
\(m = \cfrac{rise} {run} \)
\(m = \cfrac{4} {1} \)
\(m = 4\)
Therefore, the value of the derivitive function at \(x = 2\) is \(4\).
\(y\) can be any function \(f(x)\). Instead of saying 'derivative function', you must use the proper notation.
These are two ways for derivitive notation:
\(\cfrac { d } { dx } y\): Leibniz notation, where \(y\) is the function you are deriving (aka \(\cfrac { dy } { dx } \)), or:
\(f'(x)\): Langrange's notation.
There are more forms of notation, but just these are fine. Leibniz notation is best used when the function is stated as an expression of \(y\), whilst Langrange's best for \(f(x)\).
The process of differentiation
Differentiation is a process, and this process is govorned by several rules. This process does not involve drawing tangents like above, but that is the best way to visualize it.
Differentiation by first principles
The first principles approach is a very innefficient methods of differentiation, but has been formally used to prove many other rules that are much easier to apply.
This approach places two points on a given curve, the gradient of the curve approximates the tangent line, thus the value of the derivative function.
This method then uses a limit to bring the points closer and closer together until the distance between them is infinitely small. At this moment the line reflects \(f'(x)\) when \(x = 2\).
And we can see that the final gradient \(m = 2\).
So how do we calculate the first principles method mathematically?
This is the formula:
\(f'(x) = \displaystyle \lim_ { h \to 0 } \cfrac { f(x + h) - f(x)} { h } \)
\(h\) defines the distance between the target point A and some point B, which will gradually approach A as \(h\) approaches 0.
Applying the formula:
Using first principles, derive \(y = x^2\)
\(f'(x) = \displaystyle \lim_ { h \to 0 } \cfrac { f(x + h) - f(x) } { h } \)
Now substitute \(f(x) = x^2\) and simplify:
\(\begin{aligned} f'(x) &= \displaystyle \lim_ { h \to 0 } \cfrac { (x + h)^2 - x^2 } { h } \\[10pt] &= \displaystyle \lim_ { h \to 0 } \cfrac { x^2 + 2xh + h^2 - x^2 } { h } \\[10pt] &= \displaystyle \lim_ { h \to 0 } \cfrac { 2xh + h^2 } { h } \\[10pt] &= \displaystyle \lim_ { h \to 0 } { 2x + h } \end{aligned} \)
Then when the denominator is removed, we can substitute \(h = 0\), which we couldn't do before because that would result in dividing by 0: which is always undefined.
\(f'(x) = 2x\)
Cool, right?
#1: Constant Rule
As cool as first principles are, it's best avoided unless you are being tested on it. The first principles were used to prove many rules of differentiation that result in a much cleaner, quicker and less confusing solution. It may be overwhelming at first seeing all these rules, but trust me, with practise YOU WILL UNDERSTAND IT. Maths is easest to study for, just pratise and it will be as easy as counting to ten.
The first rule is the easiest, all constants (numbers that are not multiplied by a variable) become \(0\).
Calculate \(\cfrac{dy}{dx}\) when \(y = 89\)
\(\cfrac{dy}{dx} = 0\)
#2: Rule Of Multiple Terms (aka linearity of differentiation)
When deriving an expression with multiple terms seperated by a \(+\) or \(-\), the derivitive function is equal to the derivitive of each term seperated by the same symbol that was in the original function: \(+\) or \(-\).
Derive \(f(x) = a(x) - b(x) + c(x) - d(x)\)
\(f'(x) = a'(x) - b'(x) + c'(x) - d'(x)\)
#3: Power Rule
Given \(y = ax^n\) then
\(y' = anx^{n-1}\)
1) Calculate \(\cfrac{d}{dx}(x^2)\)
\(\begin{aligned} \frac{d}{dx}(x^2) &= 2x^{2-1} \\ &= 2x \ \ \ \ \color{#18C652}{\text{This is way easier then using first principles!!}} \end{aligned} \)
2) Derive \(y = 12x^3 + 3x^2 - x + 20\)
\(\begin{aligned} \cfrac{dy}{dx} &= (12\times 3\times x^{3-1}) + (3\times 2\times x^{2-1}) - (1) \\ &= 36x^2 + 6x - 1 \end{aligned} \)
#4: Product Rule
This is one of three FAT rules that are on the Methods Formula sheet:
Given \(y = f(x)g(x)\) then
\(y' = f'(x)g(x) + f(x)g'(x)\) or \(y' = f(x)g'(x) + g(x)f'(x) \) : using the mneumonic, left de right plus right de left
1) Find \(\cfrac{dy}{dx}\) when \(y = (2x+2)(3-5x)\)
\(\begin{aligned} \cfrac{dy}{dx} &= (2x+2)(-5) + (3-5x)(2) \\[5pt] &= -10x - 10 + 6 - 10x \\[5pt] &= -20x - 4 \end{aligned} \)
2) Find \(\cfrac{dy}{dx}\) when \(y = (3x+1)(2x^2-5x)\)
\(\displaystyle \begin{aligned} \cfrac{dy}{dx} &= (3x+1)(4x-5) + (2x^2-5x)(3) \\[5pt] &= 12x^2-15x+4x-5+6x^2-15x \\[5pt] &= 18x^2-26x-5 \end{aligned} \)
#5: Chain Rule
Another FAT rule on the formula sheet:
Given \(y = [f(x)]^n\) then
\(y' = n[f(x)]^{n-1}[f'(x)]\)
1) Derive \(y = (3x-7)^3\)
\(\begin{aligned} \cfrac{dy}{dx} &= 3(3x-7)^{3-1}(3) \\[5pt] &= 9(3x-7)^{2} \end{aligned} \)
2) Derive \(y = \sqrt{2x^2 - x}\)
\(\begin{aligned} y &= (2x^2-x)^{\frac{1}{2}} \\[5pt] \cfrac{dy}{dx} &= \frac{1}{2}(2x^2-x)^{\frac{1}{2} - \frac{2}{2}} (4x-1) \\[5pt] y &= (2x^2-x)^{\frac{1}{2}} \\[5pt] \cfrac{dy}{dx} &= \frac{1}{2}(4x-1)(2x^2-x)^{- \frac{1}{2}} \end{aligned} \)
#6: The Quotient Rule:
Given \(y = \cfrac{f(x)}{g(x)}\) then
\(y= \cfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\)
The best way to appy this rule is to define f(x) and g(x)
\(y= \cfrac{2x}{x^3}\)
\(f(x)=2x \ \ \ f'(x)=2 \\ g(x)=x^3 \ \ \ \ g'(x)=3x^2 \)
\(\begin{aligned} \\[10pt] y &= \frac{2(x^3) - 2x(3x^2)}{[x^3]^2} \\[10pt] y &= \frac{2x^3 - 6x^3}{[x^3]^2} \\[10pt] &= \frac{2 - 6}{x^3} \\[10pt] &= - \frac{4}{x^3} \end{aligned} \)
That's it
That's it! With practise you'll get the hang of it, eventually you'll get to a point where the rules become intuition and 'roll off the mind' like oil on water. These are not all the rules, but the basic ones, you'll learn how to differentiate functions involving e and trig later on.
The Increments Formula and Small Change
The increments formula, \(\delta y \approx \cfrac{dy}{dx} \times \delta x\) can be used to estimate the change in the dependant variable \(f(x)\) as a result of a change in the independent variable \(x\).
This formula arises from the whole two points on a line to estimate the tangent thing. Remember this?
This is essentially stating that \(\cfrac{dy}{dx} \approx \cfrac{\Delta y }{\Delta x}\), thus:
\(\\[10pt] \cfrac{dy}{dx} \approx \cfrac{\Delta y }{\Delta x} \\[10pt] \Delta y \approx \cfrac{dy}{dx}\Delta x\)
And this is how we get our increments formula.
This is fairly simple to apply, typically questions will simply just ask you to find the approximate change in \(y\) between ranges of \(x\).
1) Let \(y = 3x^2 + 2x\), find the change in \( \ y \ \) between \(x = 1.99\) and \(x = 2\)
1. find \(\cfrac{dy}{dx} \ \ \ \)
\(\begin{aligned} \\ \cfrac{dy}{dx} &= 6x + 2 \end{aligned} \)
2. find \(\Delta x\) (last x value - first x value: )
\(\begin{aligned} \\[5pt] \Delta x &= 2-1.99 \\[5pt] &= 0.01 \end{aligned} \)
3. Substitute into formula where x = first x value in range:
\(\begin{aligned} \\[5pt] \Delta y &\approx (6(1.99) + 2)(0.01) \\[5pt] &\approx 0.1394 \end{aligned} \)
Note that this method is increasingly unnacurate as the function becomes larger, or range becomes wider. The true value can be determined by \(f(x_{2}) - f(x_{1})\), the true value for the example was \(0.1397\)
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