# Calculus I: Lesson 1 - A background to differentation

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If it would help, I will post the application of this method for differentiating x2[/sup'] for any value of x, rather than just a specific value.

Where n is the x coordinate of the point you want to find

((d+n)^2 + n^2) / ((d+n) - n) = (d^2 +2nd + n^2 - n^2) / d

= (d^2 + 2nd) / d

= d + 2n

As d -> 0, gradient -> 2n.

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That and explaining where I went wrong would be very helpful. You're doing a great job explaining dave. I unerstand all that you're saying, it's just more difficult to ask questions in a situation like this. It's still a nice prep for next year in calc class. This is exactly the kind of stuff we'll be starting with.

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That and explaining where I went wrong would be very helpful. You're doing a great job explaining dave. I unerstand all that you're saying, it's just more difficult to ask questions in a situation like this. It's still a nice prep for next year in calc class. This is exactly the kind of stuff we'll be starting with.

Well, the main mistake is that it's not always h+1. It's h+5 in this case, as that's the x coordinate you're working with; so, the numerator should be (1/(h+x)^2) - (1/x^2), or 1/(5+x)^2 - 1/25 and the denominator should be (h+5) - 5, not (h+1)-5.

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Wouldn't the numerator be 1/(5-x)^2 - 1/25, not +. I see why what I did was wrong now, though.

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Right, let me try and sort out this problem.

What we're trying to do is find the exact value of the gradient of a curve at one particular value of x - for example, x = 2.

Again, let's consider a point P. We're going to want P to be the point at which we want to measure the gradient, so the co-ordinates of P are (2, 4) - 4 because we're using the equation y = x2. Now we have our point Q, which is a little bit up the line from P. So we're going to let P have the x-value x = 2+h, where h is very small. Now for y = x2, we're going to have the point Q at (2+h, (2+h)2).

So, as before, we're going to work out the gradient of PQ. This is exactly the same process as you'd use to work out the gradient of a line lying between two points: we simply take the difference of the y-coordinates and divide this by the difference of the x-coordinates.

We're going to get the gradient to be equal to: $\frac{(2+h)^2 - 4}{(2+h) - 2} = \frac{(4+4h+h^2) - 4}{h} = \frac{4h+h^2}{h} = 4+h$.

So this is the gradient of the line PQ. As h goes towards zero, we're going to get the gradient going towards 4, so this is our gradient at the point x=2.

I really don't know how to explain this any more clearly without someone telling me the specific problem that they have with my explanation. I'm going into my second year of maths at uni, and (without trying to sound arrogant) I find this relatively easy now which makes it fairly difficult to explain.

Replies are welcome

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Wouldn't the numerator be 1/(5-x)^2 - 1/25, not +. I see why what I did was wrong now, though.

Yeah, editing.

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I really don't know how to explain this any more clearly without someone telling me the specific problem that they have with my explanation

In general, the problems people are having are because you're assuming people are going to make logical leaps; the '1's in the first example, for instance, aren't going to be 'obviously' from the initial coords unless you really understand what's going on, which isn't the target audience. My suggestion would be to do each step in 2 stages; a generic version, followed by subbing in the values, which will give people more of a clue what goes where.

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Okay, I'm going to get you started on the second question, since it involves some tricky algebra. In this case, (for y=1/x2), we have P being (5, 1/25) and Q being (5+h, 1/(5+h)2).

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

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:

$\frac{\frac{1}{25+10h+h^2} - \frac{1}{25}}{h} = \frac{25 - (25+10h+h^2)}{25h(25+10h+h^2)}$.

I'll let you take it from here, but it should be fairly obvious now.

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So is number 2 $\frac{2}{5}$

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I will tell you that the correct answer for question 2 is -2/125.

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Hey. I got that. The post above where I guessed 2/5 I somehow canceled out the 25h at the begining of the denominator rather than just the h. This gave me the wrong answer but I see what is going on. Any more sample problems?

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My suggestion is to think of a common function you've been given and apply the method to it. xn are the easier ones to use with first principles, but I might try and come up with something else that's a little more interesting.

However, unless you like limits you can't solve (yet), don't try trig functions

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I never even thought of throwing trig functions in there. In fact, I don't even want to start thinking about it. Seems I'm decently comfortable on this subject now. I'll try a problem or two and post that with the answer I got to see if I'm on the right track (not today though).

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If you want a moderately harder challenge, then try finding the gradient of a function at any point using the method. I'll be posting a complete proof fairly soon, so you can easily check your answer. Stick to xn functions though and make sure you're comfortable in applying them. It'll do you well should you take your maths further.

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Hello Dave and others!

First of all, I would like to ask the moderator to delete my account with username as an email address (istok@bitsyu.net, a post before), i made it by mistake. That is why I made testing.

Well, I am a graduate in engineering, and recently discovered this forum, so I told to myself "Why not joining" I am mainly looking for some hot topics in physics, but maths are of some interest, too.

Anyway, I don't know what is generally level of conversation here. Basic, intermediate, or advanced...?

I am asking this because I realise many people did not understand your question, which is very basic. Here is my answer:

1) Gradient of x^3 is 3x^2, which, for x =5, makes 3*25 = 75.

2) Gradient of 1/x^2 is -2*(1/x^3), and, just like you said, it is - 2/125 for x=5.

Or maybe I didn't understand? Was I supposed to solve it some other way?

Istok

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Daves starting a tutorial from the basics, as he stated in his 1st post and the thread title "Calculus I: Lesson 1 - A background to differentation"

it maybe easy for you, but for people such as myself, it`s entirely a NEW AREA, and so some patients will be required by those that know it all

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Or maybe I didn't understand? Was I supposed to solve it some other way?

The basic idea behind the question is to solve the problem not through the application of simple differential calculus, but through an algebraic solution working from first principles.

We don't need to do anything to your account, unless you really want the email address its bound to back.

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Yeah, I'll be moving on towards that kind of differentiation next. This lesson was intended to give the inexperienced reader some knowledge of what differentiation is and how we define it.

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We don't need to do anything to your account, unless you really want the email address its bound to back.

He could have just asked to have his username changed, but he created a second account instead.

First account was therefore deleted.

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I see now. Thanks for deleting my previous account

No probs

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Just so everyone knows: the next lesson will be posted at around about 3pm tomorrow. It's all written up and ready to go.

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Perhaps we should make an "Pre-Calculus" section too?

i.e.:

* tricks for factoring quartics, trinomials, etc

* definitions / terminology

* finite differences

* polynomial behaviour

* dividing polynomials, synthetic division

* etc.

I'll be happy to head it up. I have my grade 12 intro to calc notes out, as I'm helping a friend with it atm.

In addition, I can post some more problems, as they're in my notes.

I think the practical/application parts should be shown too whenever possible. [i.e. for gradients of a non-steady function, the average speed in a position-time graph; using the derivative to find the speed @ that point in time; etc.]

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Sure; if you want to do that, go for it. I'll be on hand to answer any questions on anything that I've learnt

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Have to apologise for the lack of appearing of anything today, I've been very busy writing up some programming that had to be done for today. With any luck, I'll post tomorrow.

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