Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
a) Simplify the following expressions
(i) \(\left(x + h\right)^{2} - x^{2} \)
(ii) \(\left(x+h\right)^{3}-x^{3}\)
b) Find \(\lim_{h\to 0} (2x + h)\)
c) For each function, determine \(f\left(x+h\right)\)
(i) \(f\left(x\right)=3x^{2}\)
(ii) \(f\left(x\right)=2x^{4}\)
(iii) \(f\left(x\right)=x^{2}-2x\)
(iv) \(f\left(x\right)=2x^{3}+x^{2}\)
Solutions
Get Ready
Questions
Look at the graph of \(y=x^{2}\) shown. The function is shown in red.
Let’s look at the gradient of \(y=x^{2}\) at the point \(P\left(3,9\right)\). The tangent is shown in blue.
Consider the point \(Q\left(4,16\right)\) (shown in orange).
The line segment \(PQ\) (shown in green) is a good approximation for the gradient of the tangent.
Work out the gradient \(PQ\):$$m=\frac{\Delta y}{\Delta x}=$$
Now move the orange point to \(\left(3.5,\qquad\right)\).
What do you notice about this gradient in relation to the tangent?
Work out the new gradient \(PQ\):$$m=\frac{\Delta y}{\Delta x}=$$
Now move the orange point to \(\left(3.1,\qquad\right)\).
What do you notice about this gradient in relation to the tangent?
Work out the new gradient \(PQ\):$$m=\frac{\Delta y}{\Delta x}=$$
Now consider the point \(Q\left(3+h, \left(3+h\right)^{2}\right)\).
Work out the new gradient \(PQ\):$$m=\frac{\Delta y}{\Delta x}=$$What happens to this gradient as \(h\to 0\)?
Now suppose that \(P\left(x,x^{2}\right)\) and \(Q\left(x+h, \left(x+h\right)^{2}\right)\).
Work out the new gradient \(PQ\):$$m=\frac{\Delta y}{\Delta x}=$$What happens to this gradient as \(h\to 0\)?
We say that $$\frac{\text{d}y}{\text{d}x}=\lim_{h\to 0} 2x+h=2x$$
Notes
The derivative of a function \(f\left(x\right)\) is given by$$f’\left(x\right)=\lim_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$$provided this limit exists.
There are two notations that we use for the derivative:
If the function is defined as \(y=\dots\) then the derivative is given by \(\frac{\text{d}y}{\text{d}x}\)
If the function is defined as \(f\left(x\right)=\dots\) then the derivative is given by \(f’\left(x\right)\)
A function is said to be differentiable at a point if the derivative exists at the point.
A function is differentiable if the derivative exists at all points in the domain of the function.
The process of finding the derivative from this formula is known as differentiation from first principles.
Examples and Your Turns
Example
From first principles, find the derivative of \(y=x^{3}\).
Your Turn
From first principles, find the derivative of \(y=2x^{2}+x\).
Your Turn
From first principles, find the derivative of \(f\left(x\right)=3x+1\).
Your Turn
From first principles, find the derivative of \(y=\sqrt{x}\).
Your Turn
From first principles, find the derivative of \(y=\frac{1}{x}\).
Your Turn
From first principles, find the derivative of \(f\left(x\right)=\frac{1}{x^{2}}\).
Theorem
If a function is differentiable at \(c\), then it is continuous at \(c\).
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Let \(f\left(x\right)\) be differentiable at \(x=c\).
We want to show that \(\lim_{x\to c}f\left(x\right)=f\left(c\right)\)
Consider $$\begin{align}\lim_{x\to c}\left(f\left(x\right)-f\left(c\right)\right)&=\lim_{x\to c}\left(\frac{f\left(x\right)-f\left(c\right)}{x-c}\left(x-c\right)\right)\\&=\lim_{x\to c}\left(\frac{f\left(x\right)-f\left(c\right)}{x-c}\lim_{x\to c}\left(x-c\right)\right)\\&=f’\left(x\right)\times 0 = 0\end{align}$$
But we also have that$$\lim_{x\to c}\left(f\left(x\right)-f\left(c\right)\right)=\lim_{x\to c}f\left(x\right) - \lim_{x\to c}f\left(c\right)$$
Putting these two results together gives that $$\lim_{x\to c}f\left(x\right) - \lim_{x\to c}f\left(c\right)=0$$Thus$$\lim_{x\to c}f\left(x\right)=\lim_{x\to c}f\left(c\right)=f\left(c\right)$$which means that \(f\left(x\right)\) is continuous at \(x=c\).
Notes
Contrapositive
A function that is not continuous at \(x=c\) is not differentiable at \(x=c\).
Converse
A function that is continuous at \(x=c\) is not necessarily differentiable at \(x=c\)
For example, consider the function $$f\left(x\right)=\left|x\right|$$Draw a sketch of \(y=f\left(x\right)\).
Is the function continuous?
Now calculate$$\lim_{h\to 0^{-}}\frac{f\left(x+h\right)-f\left(x\right)}{h}=$$and $$\lim_{h\to 0^{+}}\frac{f\left(x+h\right)-f\left(x\right)}{h}=$$So we conclude that the limit does not exist at \(x=0\) and hence it is not differentiable.
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
Explain where the meaning of \(h\) in the formula \(\frac{f\left(x+h\right)-f\left(x\right)}{h}\). Why must \(h\) approach zero?
How is the derivative related to the values calculated when finding the gradient function?
Common Mistakes / Misconceptions
The most common error is a simple algebraic error. Be careful to expand correctly, and only simplify expressions that can be simplified.
Another common error is substituting \(h=0\) too early in the process, when you cannot evaluate because you end up with \(0\) in the denominator.
A common notation error is to forget to include the \(\lim_{h\to 0}\) in each step of the working until the final step.
Connecting This to Other Skills
This is the formal way to find the Gradient Function (4.3), which builds upon the ideas of Tangents (4.1). We require the use o Limits and Continuity (4.2) to work this out.
As this is the formal definition of the derivative, we will need it to find other results throughout this unit, such as Differentiating Polynomials (4.5), the Chain (4.7), Product (4.8) and Quotient Rules (4.9), as well as differentiating Exponentials and Logarithms (4.11) and Trig functions (4.12).
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?