The two links give the centre's home page, a topic list.

https://www.cimt.org.uk/projects/mepres/alevel/

These are first class resources available for download in pdf.

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a) y = kx it is direct proprortionality

b) y = -kx please state indirect or direct proportionality

it is common knowledge that when there is an increase of y1 to y2 there is a corresponding increase from x1 to x2.....this is direct proportionality.

the above definition of direct proportionality does is not applicable on the straight lines as follows

a) y = -2x + 10

b) y = -3x

c) y = -x + 8

for the function y =x2 (x squared)

a) for the domain x equal to zero or greater than zero (and also less than positive infinity) the curve is an increasing function

b) for the domain x equal to zero or greater than zero (also greater than negative infinity) the curve is a decreasing function based on the quadratic curve y = x2.

please could you confirm everything as stated above.

]]>There's only so much I can do: I'm assuming that you've got a solid basis in algebra, and I will start from about the level of maths GCSE. I assume that you will understand the concept of a function (e.g. [math]f(x) = x^2[/math]) and understand various concepts such as graphing techniques. For the later stages, I assume some knowledge in the area of trigonometry, mainly the sine and cosine functions. For the more advanced calculus, I will be working in radians instead of degrees for the measurement of angles.

There is one other thing:

Here's a list of the topics covered in the rest of this tutorial, with links to those posts:

- The basics of limits
- Introduction to differentiation
- Differentiation shortcuts
- The product rule
- The quotient rule
- The chain rule
- Calculus with trigonometric functions
- Logarithmic differentiation
- Applications: Finding the maximum and minimum values of a function

**Lesson 1 - The basics of limits**

So what actually *is* a limit? It's a very hard concept to define in layman's terms (although relatively easy from a strictly analytical point of view). I think the best way to think of it is in terms of sequences.

Imagine you have a sequence of numbers that goes like this: 1, 1/2, 1/3, 1/4, 1/5, ... and so on. If we call the n^{th} number a_{n}, then it's fairly clear to see that a_{1} is 1, a_{2} is 1/2, and so on. The mathematical definition for the n^{th} number is obviously a_{n} = 1/n.

Now we look at what happens as we get bigger and bigger values of n. We can notice that each term in the sequence gets progressively smaller as we increase the value of n, and it doesn't take a genius to work out that as we get really big values of n, we get excruciatingly small values for a_{n}. Eventually, with incredibly huge numbers, a_{n} will be almost 0 (but it never will actually be 0). So we can say that the "limit" of a_{n} is 0 as n gets really big (i.e. as n tends to infinity).

Don't start crying just yet over how complex this all is; it's an abstract concept to understand, and it'll take some time just to understand the idea, let alone how it all works. A quick remark on this: we won't be using limits that tend to infinity much in calculus at all, I just used it as an example.

A very important idea to understand is the fact that we're not actually saying that the sequence will ever hold a value of 0 - what we are saying is that if you were to go on and extend the sequence forever, then you'd be continually getting closer and closer to zero.

Quickly, some notation. You won't be using this every day, but it's handy to know. The situation described above could be represented like this:

[math]\lim_{x \to \infty} \frac{1}{x} = 0[/math]

meaning that as we put bigger and bigger numbers in to [imath]\frac{1}{x}[/imath], the answer approaches 0.

Remember, if you need help understanding any of this, you can just ask in our calculus forum.

I know quite a few people were confused by the method I employed to find an exact value for the gradient of a curve at a certain point. I hope you've managed to un-confuse yourself - as always, if you have problems, post them in the thread, or alternatively send me a PM or e-mail. I've also posted the solution for the first question in the previous thread, and started the second one. Here's the full solution to the second one:

Question 2: Find the exact value of the gradient for the curve [math]y=\frac{1}{x^2}[/math] when x = 5.

In this case, (for y=1/x

^{2}), we have P being (5, 1/25) and Q being (5+h, 1/(5+h)^{2}).

So our gradient function will be [math]\frac{\frac{1}{(5+h)^2} - \frac{1}{25}}{h}.[/math]

Simplifying this down is by no means a simple task. The best way to approach this kind of thing is to combine the numerator into a single fraction. To do this, we simply do the same thing that you'd do when you're doing, say, 1/2 - 1/3: you convert them both into sixths by multiplying the first fraction by 3 and the second fraction by 2. So we get:

[math]\frac{\frac{1}{25+10h+h^2} - \frac{1}{25}}{h} = \frac{25 - (25+10h+h^2)}{25h(25+10h+h^2)}[/math].

And now the continuation:

We have that the gradient function is [math]\frac{25 - (25+10h+h^2)}{25h(25+10h+h^2)} = -\frac{10h + h^2}{25h(25+10h+h^2)}[/math]

The trick with this fraction is to now notice that a h will cancel from the top and bottom, giving us [math]-\frac{10 + h}{625 + 250h - h^2}[/math]. Now to find our limit, we simply put h = 0 into this, giving us the final answer of -2/125.

(for those that wanted the answer to question 1, the correct value is 75).

Now onto the new stuff

I'm going to show this week that we can extend the method to a general value of x. What's the advantage of this? Well, instead of having to go through the entire process over and over again for different values of x, we can just find our specific formula for the gradient and then put our different values in to find the gradient at those points. But not only that, it allows us to then go on and define a set of basic rules for differentiation - something which you will undoubtedly use a lot if you go into any kind of job involving mathematics.

Notice that the method is *exactly the same*, and I'm going to apply it for the curve y = x^{2}. I'm also going to go through it a bit quicker and without a lot of the intermediate explanation, although I'll elaborate further if anyone is overly confused. We're going to pick a point, P, at which we want to know the gradient. But this time, instead of picking a specific point (such as (1, 1)), we're going to pick the general point (x,y) - i.e. (x,x^{2}).

Now let's pick Q, the point we're going to move towards P. Again, we'll do exactly the same thing, and choose it to be a little bit away from P. So the x-coordinate is going to be x+h, meaning our point will hold the co-ordinates (x+h, (x+h)^{2}).

Before I go further, I think this was the step that confused most people. They wondered why I'd picked x+h as the co-ordinate - the simple reason is, we want to make Q some distance away from P, and then gradually reduce the distance to obtain our value. The way to do this is to pick any value and add it onto the x-coordinate, making it a bit further down the number line from P. Then we can gradually reduce the value of this number until it's incredibly small, getting a better and better approximation of the gradient as we do so. I hope this helps a bit - I know it's confusing, but you'll get your head around it over time

Now we're going to find our gradient function in the same manner as before. So we have:

[math]\frac{(x+h)^2 - x^2}{(x+h) - x}[/math] as the gradient of the line PQ. Expanding and simplifying gives:

[math]\frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h[/math].

Now, as h tends to zero, we can see that the gradient is going to tend to 2x.

And that's it! We now know that the gradient of the curve y = x^{2} at *any point on the curve* is going to be 2x. If you don't believe me, go and try it for a few values of x and do the method over again. It definately works. As you can imagine, in a historical context this was truly amazing - a method for finding the gradient of a function* at any point on the curve in a matter of seconds without drawing tangents is quite a tremendous achievement.

(*for the interested reader only - don't read if you think it might muddle things up! This method doesn't work for every imaginable function. It is dependent on the fact that a limit is defined. For the extremely interested reader, this limit is just what we've been doing, only in a slightly different notation. If you have a function y = f(x), then the limit is defined as:

[math]\lim_{h \to 0} \left( \frac{f(x+h) - f(x)}{h} \right)[/math]

Which you might recognise if you put f(x) = x^{2}. This limit is also dependent on the function being continous at the point x. By "continous", we basically mean that the function doesn't have any gaps around the local area of x, although it has quite a simple analytical definition. If you want an example of a function that can't be differentiated at a certain point, look at y = |x|, and try and differentiate it at x = 0. As I said, ignore all of this if you think it will confuse you, it's only meant as a sidenote to show you that not all functions can be differentiated at all points).

So now onto the interesting stuff, and this is where things start happening quite quickly It turns out that if you look at functions of the form y = x^{n} (i.e. x^{4}, x^{-5}, etc), then a pattern starts to emerge when you differentiate things. When you differentiate y = x^{3}, you get the limit coming out as 3x^{2}, and when you differentiate y = x^{-2}, you get the limit as -2x^{-3}. Some of you might be starting to see the pattern, but for those who aren't, look at the power of x before you differentiate it and after, and also look at the number before the x when you differentiate it. Got it?

If not, don't worry. This is the one formula you're going to use a lot from now on, so I'm going to try and make it plain - I won't try and prove it, however, because it takes some more complex mathematics that's out of the scope of this tutorial.

The differential of a function y = x^{n} for **any value of n** is n*x^{n-1}.

Remember this forever! It's an extremely important result, and one you should never, ever forget I'll cover some examples in a moment, but first...

I said that I'd be introducing some important notation this time around, and so I am. There's various notations that you can use for differentiation, but the most popular one is the Leibniz notation. You'll wonder why we write it like this, but you'll thank the man down the line when we get to the chain rule. The name probably won't strike a bell, but this might:

[math]\frac{dy}{dx}[/math]

The first thing to remember is that the "d" is not a variable; it's actually an operator, so you can't simply cancel it from the fraction to give y/x (otherwise things would make no sense). The second thing to notice is the interpretation - spoken aloud it means "the differential (i.e. gradient) of y with respect to x". So what does this mean? If you have a function, y, written in terms of x, then you simply differentiate this "with respect to" x. For example, for y = x^{2} implies that:

[math]\frac{dy}{dx} = 2x[/math]

Also, please don't go around writing [math]y = x^2 = 2x[/math]. This makes sense, but it probably doesn't mean what you want it to mean. The proper way of writing things is:

[math]y = x^2 \Rightarrow \frac{dy}{dx} = 2x[/math].

where the arrow means "implies that" - so in real terms you're saying that if you have a function y, this implies that the derivative is dy/dx. It can be produced with the \Rightarrow command in LaTeX. I hope this is making sense to you all, but if it's a bit much to digest, then re-read it and try some examples for yourself. If you have questions on the notation, don't hesitate to ask - it's better that you get it right now rather than 3 years down the line

One last comment, and that's on a slight variation of the notation. If we know y to be, for example, 57x, then we can write this in a slightly different way:

[math]y=57x \Rightarrow \frac{dy}{dx} = \frac{d(57x)}{dx} = \frac{d}{dx}(57x)[/math]. I'll be using this notation in a second.

Now back to some examples of some actual differentiation. Remember that the differential for y=x^{n} for *any* value of n (i.e. 3, 7, -12/65, 0.1762172) is, y = nx^{n-1}. Or, written in our new notation:

[math]y = x^n \Rightarrow \tfrac{dy}{dx} = nx^{n-1}[/math].

An easy way to remember this is to say to yourself: "stick the index in front and reduce it by one". Or at least that's how I remember it Also note that having a constant in front of the x before you differentiate doesn't matter. We could have y = 5x^{2}, and the derivative would be 5*2x = 10x.

Maybe a simpler way to view this is like this (using our slight variation on the notation):

[math]y=5x^2 \Rightarrow \frac{d}{dx}(5x^2) = 5\frac{d}{dx}(x^2) = 5\cdot 2x = 10x[/math].

Another formula that you might want to remember is the more general case for the differential of ax^{n}, where a is a constant coming before the x. (i.e. for y = 2x^{2}, a=2). Using the same method as before, we get:

[math]y = ax^n \Rightarrow \frac{dy}{dx} = anx^{n-1}[/math].

This all might seem a lot to digest at the moment, so I'm going to post a couple of examples, and hopefully everyone will start getting the idea.

**Examples**

1. [math]y=x^2[/math]. In this case, n = 2. So we'll put this in front of the function and then reduce the index by one, giving dy/dx to be 2x^{1} = 2x.

2. [math]y=x^{\tfrac{3}{2}}[/math]. This time, we're using a fractional index. But we just apply exactly the same method to get our answer. n = 3/2, so n-1 = 3/2 - 1 = 1/2. Put this into our formula and we get dy/dx = 3/2 * x^{1/2}

3. [math]y = x^{-\tfrac{1}{3}}[/math]. This time we have a negative fractional index, but again, we just apply the same method. n = -1/3, implying n-1 = -4/3. This gives the answer as -1/3*x^{-4/3}.

4. [math]y = 2x[/math]. Here's a slightly more tricky example. What's the index in this case? Well, x is simply equal to x^{1}, so n = 1. We then know that n-1=0, which then implies that our differential is 2*1*x^{0}, and x^{0} = 1 (by the laws of indexes), giving our final answer as 2.

5. [math]y=7[/math]. Hang on, we haven't got an x-term! Or so we think. Notice before that I said x^{0} = 1, so saying y = 7 is the same as saying y=7x^{0}. Applying our rule again, we get n = 0, and n-1 = -1. This gives us a final answer of the gradient being 0. Thinking about it, this makes sense - look at a line with a constant value. Obviously it's not going to have a gradient since there's no change in the value of y.

**Questions**

Differentiate the following functions with respect to x, taking care to use the proper notation for each.

i) y = x^{3}.

ii) y = 2x^{13}.

iii) y = x^{-2}.

iv) y = 5x^{1/5}.

v) y = 1/2*x^{3}.

vi) y = 5x^{-1/2}.

vii) [math]y= \tfrac{5}{x^9}[/math]

viii) [math]y= \tfrac{2}{x^{1/2}}[/math]

ix) [math]y = \tfrac{4x}{x^{7/6}}[/math]

x) [math]y= \tfrac{x^a}{ax^n}[/math], where a and n are both constant.

**Summary**

I know I've covered quite a lot in one lesson - after all, we've gone from being able to find the gradient at a single point on a curve to being able to differentiate any function of the type y=x^{n}. Undoubtedly there will be a lot of questions, so make sure you check to see whether your question has been asked before.

Please rate the thread and give me some feedback, as usual, so I can improve for the next lesson we do

Cheers.

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**Lessons Syllabus**

I'll attempt to cover the following topics over a 10-lesson period (notice I'm not constricted by the number of lessons, they're pretty flexible):

- Background to differentiation, first principles and an introduction to limits (2 lessons)

- Basic rules of differentiation, essential notation and examples (1 lesson)

- Maxima and Minima (1 lesson)

- Applied differentiation (2 lesson)

- An introduction to integration (2 lessons)

- The chain rule (1 lesson - maybe 2)

- Rounding it all off (however many lessons )

**Some Information**

Here's some basic information for you all to read:

- Each lesson will be posted mid-afternoon every 3-4 days, depending on how much time I have.

- I'll post an introductory post at the beginning of the thread with the core knowledge required, and then some questions for you to attempt.

- If you have problems with anything to do with a particular day's topic, then please post it in the same thread rather than creating a new one. Please read other people's replies to save duplication of questions

- Please try and use the LaTeX facilities to typeset your mathematics properly if you have any questions.

- Please rate each thread so I can see what the response is.

- If you're really confident of the answer and get it within 2 seconds or you're quite advanced already, please give people a hand and don't just post the full solution to a problem.

- Finally, the most important thing:
**spread the word!**These things are only so successful if we have a limited number of people.

Please not that this course *is not intended to be a replacement for real teaching* - it is only intended as a supplement.

**Prerequisites**

There's only so much I can do: I'm assuming that you've got a solid basis in algebra, and I will start from about the level of maths GCSE. I assume that you will understand the concept of a function (e.g. f(x) = x^{2}) and understand various concepts such as graphing techniques. For the later stages, I assume some knowledge in the area of trigonometry, mainly the sine and cosine functions. For the more advanced calculus, I will be working in radians instead of degrees for the measurement of angles.

There is one other thing: **GRADIENTS** - know that the definition of a gradient of a straight line between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is [math]\frac{y_2 - y_1}{x_2 - x_1}[/math].

Having said all of this, let's get stuck in

**Lesson 1 - A background to differentiation**

Now, a lot of people have asked me the question: "Just what the hell is calculus anyway?" And my response is: "Probably the most important area of mathematics you're likely to come across, and the basis of most modern physics." And this is not an understatement. I won't go into much more detail historywise, but needless to say the original inventor of calculus was Sir Isaac Newton, who used it to solve the classic two body problem back in the 1600's. Since then, a new notation has been commonly accepted (created by Leibniz) and many, many advances have been made. And now onto the mathematics.

Calculus encompasses three major topics: the study of limits, differentation of functions and integration. We will be starting to look at limits and differentation over the next couple of weeks, and integration will come towards the end of the course.

So what actually *is* a limit? It's a very hard concept to define in layman's terms (although relatively easy from a strictly analyitical point of view). I think the best way to think of it is in terms of sequences.

Imagine you have a sequence of numbers that goes like this: 1, 1/2, 1/3, 1/4, 1/5, ... and so on. If we call the n^{th} number a_{n}, then it's fairly clear to see that a_{1} is 1, a_{2} is 1/2, and so on. The mathematical definition for the n^{th} number is obviously a_{n} = 1/n.

Now we look at what happens as we get bigger and bigger values of n. We can notice that each term in the sequence gets progessively smaller as we increase the value of n, and it doesn't take a genius to work out that as we get really big values of n, we get excruciatingly small values for a_{n}. So we can say that the "limit" of a_{n} is 0 as n gets really big (i.e. as n tends to infinity).

Don't start crying just yet over how complex this all is; it's an abstract concept to understand, and it'll take some time just to understand the idea, let alone how it all works. A quick remark on this: we won't be using limits that tend to infinity much in calculus at all, I just used it as an example.

A very important idea to understand is the fact that we're not actually saying that the sequence will ever hold a value of 0 - what we are saying is that if you were to go on and extend the sequence forever, then you'd be continually getting closer and closer to zero.

Quickly, some notation. You won't be using this every day, but it's handy to know. Instead of writing "the limit of f(x) as x tends to infinity is 971283.2", we can write:

[math]\lim_{x\to\infty} f(x) = 971283.2[/math] - so remember this, it's a very useful shorthand!

So now onto first principles of differentiation: this is where I tell you how to actually go about differentiating something and what we actually mean by the term 'differentiation'.

A classic problem in a GCSE maths paper is to sketch a curve out (like the classic y = x^{2}) and then they say to you: "draw a tangent to the curve at the point x = 2, and hence find the gradient at that point". And you grudgingly scrawl out a quick graph between this and that, shove a quick tangent on and get an approximate value for the gradient. We all know it's dead easy to find the gradient between two points on a straight line, but on a curve? Bah, impossible.

But this is not so.

Let's draw ourselves a graph of y= x^{2}, and have a look at a better way of doing things. Have a look at the graph at the bottom of the page, and observe the points I've drawn on it. We have a point P, and a position (1,1) and then a point Q at position (1+h, (1+h)^{2}). I initially got confused here: basically, we're looking at a point where x = 1, and then a point a little bit further down the x-axis at x = 1+h, where h is some value (we don't really care all that much).

Now we're going to consider what happens to the gradient of the line PQ when we decrease the value of h - i.e. we move the point Q closer to the point P. If you look closely, you'll notice that as you draw the two points closer and closer to one another, you're going to get a better and better approximation for the value of the gradient at the point x = 2 on the curve.

And now you scream "so what?!?!?" - and then I say "Well look - we have a method for finding *the precise* value of a gradient at a point on a curve!" And then you probably say "so?!?!" again, and I don't really care by this point What we really want to know is how we find this value.

Let's look at the gradient of the line PQ. This is equal to:

[math]\frac{(1+h)^2 - 1}{(1+h) - 1} = \frac{(1+h)^2 - 1}{h}[/math]

Now just a second. We want to find the limit of this as we decrease h (i.e. as h tends to zero). But we've got a silly h lying around all by itself on the bottom, meaning that if we just stick h=0 into here, we get something divided by zero - we can't do that. So now we have to play around a bit with the fraction, and this is the key operation of this lesson. Make sure you watch very, very carefully and understand each step in the minutest of detail. First of all, notice that (1+h)^{2} = 1 + 2h + h^{2}. So now we have the gradient equal to:

[math]\frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h[/math]

Hurrah! Now we have something that we can work with. Notice now that as h gets very, very small we have that the gradient of PQ will tend to 2+0 = 2.

So we have a method for finding the exact value of the gradient at a certain point. All those hours of drawing tangents to curve wasted whilst your teacher can work out the answer in their heads

Next time, I'll extend the method to work for all values (i.e. find the gradient at *any* point) and introduce some very important notation that you're likely to carry around for a very long time.

**Questions**

1. Find the gradient of the function y= x^{3} at the point x = 5.

2. Find the gradient of the function [math]y = \frac{1}{x^2}[/math] at the point x = 5.

I may post further questions; I just can't think of any yet.

Hope you enjoyed this and found it useful,

Dave

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