Calculus

How much tutoring
does your child need?

You must have heard about the word ‘Calculus’ and wondered what it really is. In this article, we will try to give you a fair bit of an idea of Calculus and the concepts you will come across in your year 11 HSC maths.

 

Calculus is the study of functions and their behaviour, and a large chunk of your year 11 maths is focused on differentiation or derivatives. So, that’s what we will try and understand here.

 

Learning Outcomes

By the time you finish reading this page, you will not only have an idea of what Calculus is about, but you will also have learned about the following concepts:

• Understand that the slope of non-linear functions is not constant
• Be able to find the slope at any point in an Algebraic power function
• Be able to apply the product rule, quotient rule, and chain rule to differentiate functions

 

So, what exactly is Differentiation?

To understand differentiation, let’s take a step back and think about a linear function given by the equation \(f(x)=mx+c\)

 

You must know that the constant m in the equation above is called its slope. It gives us an idea of the steepness of the straight line; as it’s a straight line, the slope is the same at every point on the line.

 

Now think about the function \(f(x)=x^4-4x^3+x^2+7x\). As the power of x isn’t 1 here, the graph is not a straight line, as shown below. Notice that the slope or the steepness of the graph is not fixed, as shown by the lines at points A, B, C, and D.

The slope at point A is negative as the graph ‘falls’ as we move from left to right at A. Likewise, the slope at point C is positive, as it rises from left to right at that point. Also, notice that the slopes at points B and D are zero, and the graph neither falls nor rises at those two points. Such points on a graph where the slope is nil are called stationary points.

 

The bottom line is that the slopes of non-linear graphs keep changing at different points. We use a technique called differentiation to find the slope at a particular point.

 

Differentiating an Algebraic Power Function

Let’s say we want to find the slope of the non-linear function \(f(x)=ax^n\), where a and n are constants. The slope at a point for such a graph is given as \(anx^{n-1}\).

 

We can write this either as \(\frac{d(ax^{n})}{dx}=anx^{n-1}\) or as \(f'(x)=anx^{n-1}\).

 

Don’t let the notations intimidate you. You will encounter these notations so often that they will no longer be strangers. The ‘d’ stands for delta, which means a really small increase.

So when we write \(\frac{d[f(x)]}{dx},\) we refer to the fraction with a very small increase in f(x) or y in the numerator and a very small increase in x in the denominator. Doesn’t this sound similar to the basic definition of slope as Rise/Run? This is the same as Rise/Run at a particular point whose x-coordinate is the ‘x’ in our equation above. As ‘x’ keeps changing from point to point on the graph, we have different slopes at different points. Let’s see an example at work.

 

Example:
Find the slope of the graph given by the equation \(f(x)=5x^{3}+3x\) at the point x = 1.

 

Solution:
Applying the derivative formula, we have \(f'(x)=5(3)x^{3-1}+3(1)x^{1-1}=15x^{2}+3\)

Note that x is the same as \(x^1\)

 

The equation above tells us that the slope of the graph of the given function is different at different points and can be found by plugging in the x-coordinate for a particular point into the equation \(15x^2+3\)

 

For example, where x = 0 on the graph, the slope is \(15×0^2 + 3 = 3\) and where x = 1, the slope is \(15×1^2+3=18\). See how the slope depends on the value of x and keeps changing? That’s the point!

 

The only thing we’d like to caution you about is the formula \(f'(x)=anx^{n-1}\) applies only when the function is of the form \(f(x)=ax^n\). You can’t use it to find the derivative (slope) of a trigonometric or logarithmic function (or any other function that isn’t in this form). This is similar to saying the divisibility test for 2 – that the last digit in a number must be even – can’t be applied to test the divisibility of a number by 3, remember?

 

Product Rule

The product rule of differentiation can be summarised as follows:
If \(f(x)=g(x)⋅h(x),\) then \(f'(x)=g(x)\cdot h'(x)+h(x)\cdot g'(x)\)

 

Example:

If \(f(x)=2x^{3}(1-3x^{2}),\) find \(f'(x)\)$.$

Method 1:

One way to do this is to multiply the bracket and differentiate. So we have, 

\(f(x)=2x^3 (1)+2x^3 (-3x^2 )=2x^3-6x^5\)

Therefore, \(f'(x)=2(3)x^{3-1}-6(5)x^{5-1}=6x^2-30x^4\)

Method 2:

Applying the chain rule we have,

\(f'(x)=2x^{3}\times \frac{d(1-3x^{2})}{dx}+\frac{d(2x^{3})}{dx}\times (1-3x^{2})\)

\(=2x^{3}[\frac{d(1)}{dx}-\frac{d(3x^{2})}{dx}]+\frac{d(2x^{3})}{dx}\times (1-3x^{2})\)

\(=2x^{3}(0-6x)+6x^{2}(1-3x^{2})\)

\(=-12x^{4}+6x^{2}-18x^{4}\)

\(=6x^{2}-30x^{4}\)

This is the same answer we got using the other method above. If this one takes more steps than the previous method, then why should we care to use this method? The answer is, in some situations, you MUST use this method, and there’s no other option.

Food for thought: why is the derivative of a constant (like \(\frac{d(1)}{dx}\)) equal to 0?

Quotient Rule

Just like the product rule, there’s a quotient rule that states if \(f(x)=\frac{g(x)}{h(x)}\) then the derivative of \(f(x)\) can be found as \(f'(x)=\frac{g'(x)\cdot h(x)-g(x)\cdot h'(x)}{[h(x)]^2}\)

We encourage you to create an example and apply the quotient rule. Then, solve the same problem using the traditional method and match the two answers.

Get in touch with us today for assistance with your Year 11 Calculus!

 

Chain Rule

How do you find the derivative of the function given by \(f(x)=(x^3+2)^4?\) This isn’t in the form of \(f(x)=ax^n,\) right?

Here, we let \(x^3+2=u,\) which means we need to find \(\frac{d(u^4)}{dx}\)

Note that \(\frac{d(x^{4})}{dx}=4x^{3}\) but \(\frac{d(u^{4})}{dx}\neq 4u^{3}\) as u and x are different variables.

We need to use the chain rule here as follows (the term ‘chain’ is intuitive once you understand this):

\(\frac{d(u^4)}{dx}=\frac{d(u^4)}{du}\times \frac{du}{dx}\)

 

You can think of the du in the numerator cancelling out with the du in the denominator, leaving behind \(\frac{d(u^4)}{dx}\)

Now we can solve this, as we know, \(\frac{d(u^4)}{du}=4u^3=4(x^3+2)^3\) and

\(\frac{du}{dx}=\frac{d(x^3+2)}{dx}=\frac{d(x^3)}{dx}+\frac{d(2)}{dx}=3x^2\)

So we have \(f'(x)=4(x^3+2)^3\times 3x^2=12x^2(x^3+2)^3\)

 

This may take some time and practice to sink in, so be sure to get some practice right away.

 

Wrap Up

Differentiation or derivative finds much use in Maths and Physics. As a starter, the velocity of a body in non-uniform motion is the derivative of its position function, and acceleration is the derivative of its velocity function.

 

Now that you know the basics of derivatives, get some practice!