Calculus I - Lesson 2: A continuation from first principles

[Dave]

Hello again all, time for another fun-filled tutorial-esque posting from dave Before I start, I must apologise for the delay in posting - I had a job that needed to be done, and unfortunately it took longer than I had anticipated to complete it. Anyway, that said, let's press on. First off, solutions to last week's problems.

 

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/x²), 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 = x². 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,x²).

 

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 = x² 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) = x². 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 = xⁿ (i.e. x⁴, x^-5, etc), then a pattern starts to emerge when you differentiate things. When you differentiate y = x³, you get the limit coming out as 3x², 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 = xⁿ 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 = x² 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=xⁿ 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 = 5x², 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 axⁿ, where a is a constant coming before the x. (i.e. for y = 2x², 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 2x¹ = 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 x¹, so n = 1. We then know that n-1=0, which then implies that our differential is 2*1*x⁰, and x⁰ = 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 x⁰ = 1, so saying y = 7 is the same as saying y=7x⁰. 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 = x³.

ii) y = 2x¹³.

iii) y = x^-2.

iv) y = 5x^1/5.

v) y = 1/2*x³.

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=xⁿ. 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.

[jordan]

[math]3x^2[/math]

 

[math]16x^{12}[/math]

 

[math]-2x^{-3}[/math]

 

[math]x^{-\frac{4}{5}}[/math]

 

[math]\frac{3}{2}x^2[/math]

 

[math]\frac{-5}{2}x^\frac{-3}{2}[/math]

 

[math]-45x^{-8}[/math]

 

[math]-x^{\frac{-3}{2}}[/math]

 

[math]-8x^{\frac{-13}{6}}[/math]

 

[math]-nx^{(-n-1)}[/math]

[jordan]

It's seems as though I can't make the exponent more than one figure long: as in #2 and #10 which are supposed to be ^12 and ^(-n-1) respectively. How do I do that?

[Dave]

Instead of having just ^2, you need to enclose the 2 with curly braces. For example, x^{n-1}.

[Dave]

I've just edited them for you to make it a little better.

 

The last one's not right btw

[jordan]

But the others are?

 

I was hoping to use the spoiler feature but I forgot it didn't work. Hope this didn't make anyone mad but posting the answers like that.

[jordan]

How about if I made the last one [math]-nx^{2(-n-1)}[/math]?

 

I'm really out of time and really late so I'll check back later.

[Dave]

ii), vii) and 8-10 aren't right.

 

I'll just double check my answers though.

[Dave]

How about if I made the last one [math]-nx^{2(-n-1)}[/math]?

 

I'm really out of time and really late so I'll check back later.

 

Think about it; the answer has to have some mention of a in there.

 

As a hint, use the laws of indexes.

[jordan]

ii), vii) and 8-10 aren't right.

 

Typo: #2 should be 26x¹²

Simple mistake: #7 should be -45x^-10

 

 

For 8-10 the method is questionable, but all the same. I'll show number 10:

 

[math]\frac{x^a}{ax^n}*\frac{{a^{-1}}{x^{-n}}}{{a^{-1}}{x^{-n}}}[/math]

 

[math]\frac{{x^a}{a^{-1}}{x^{-n}}}{1}[/math]

 

[math]a^{-1}x^{2a-2n}[/math]

 

As I type I think I see what happened to my a.

 

[math](2-\frac{2n}{a})x^{(2a-2n)-1}[/math]

 

I'm getting myself a little confused now. Let me know if you see any errors.

[Dave]

I'm not quite sure what you've done there. As another hint, try simplifying the fraction before you differentiate.

[jordan]

You didn't explain what to do if the x is in the denominator, so I simply took it out. In this step:

 

[math]\frac{x^a}{ax^n}*\frac{{a^{-1}}{x^{-n}}}{{a^{-1}}{x^{-n}}}[/math]

 

I multiplied the donomators to get 1 and then that put everything in the numerator. I pretty sure but not positive that it is correct. From there I combined the x's in to get x^2(a-n). I'm thinking the 2 doesn't have to be there now, but I'll leave it for a second. Now, the rule says I subtract 1 from the exponent to find the new one: x^2(a-n)-1. Then, multiplying the coefficient by the exponent yeilds a^-1*[2(a-n)] or a^-1*(2a-2n). Distributing the a^-1 gives me 2-(2n/a). This is the new coefficient.

[Dave]

The 2 doesn't have to be there - I'm not quite sure where that came from.

 

A simpler way of looking at it is like this:

 

[math]\frac{x^a}{ax^n} = \frac{1}{a}\cdot\frac{x^a}{x^n}[/math].

 

Now we know from the laws of indexes that a^b/a^c = a^b-c. So we have:

 

[math]y = \frac{1}{a}\, x^{a-n} \Rightarrow \frac{dy}{dx} = \frac{1}{a} \cdot \frac{d}{dx}(x^{a-n})[/math].

 

Now you just apply the differentiation rule and get the answer [math]\frac{dy}{dx} = \left( 1-\tfrac{n}{a} \right) x^{a-n-1}[/math].

 

I put that example in there to make people think a bit; you're often confronted with problems that make you think quite hard about how to approach them (often when you're using integration and need a solution). It appears it's done that at least, so that's good

[jordan]

Alright, I got the right answer...eventualy.

 

I thought I should take the 2 out, but that ment retyping everything in post 10 so I meant to say that in the answer of post 10, make all 2's into 1's. I had the 2 from multiplying x^a by xⁿ. Somehow I got an x² in there with the a and n. Then it got all distributed.

 

Those kinds of problems are really great though, because you have to do it a couple times and thus learn the intricacies of the method and in doing so it really gets ingrained. All around, I gave this thread 5 stars. Thanks for putting up with me.

[Dave]

That's perfectly alright, just wish some other people were looking at the stuff and posting stuff.

[bloodhound]

i can't seem to show that the derivative of e^x is e^x. using the standard definition of a derivative.

[jordan]

Since I'm fairly comfortable with this now, I have a question about the aplication. The method I used (take the power to the front and then subract 1 from the old power...) works in all cases where it's f(x)=xⁿ no matter what the function is?

[Dave]

For any function of the form f(x) = xⁿ, the derivative is nx^n-1 - where n is of course constant.

[jordan]

No matter what else is added around the xⁿ? I guess what I'm getting at is the "interested reader" section. What is that equation used for?

[Dave]

I knew I shouldn't have posted that

 

It does matter what is around the xⁿ. Obviously if we do something like take f(x) = xⁿsin(x) then we can't just differentiate it using our existing rule - that comes later with the chain rule and product rule.

 

That rule for functions that are explicitly of the form f(x) = xⁿ.

[Dave]

i can't seem to show that the derivative of e^x is e^x. using the standard definition of a derivative.

 

Please ignore this unless you're really interested, I'm just gonna post the solution of this up - at least it shows a more interesting function I suppose.

 

You need to use the fact that:

 

[math]e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n[/math].

 

Take f(x) = e^x. Then you get that:

 

[math]f'(x) = e^x \lim_{h\to 0} \left( \frac{e^h - 1}{h} \right)[/math]

 

Notice that from the first equation, by putting n = 1/h,

 

[math]e = \lim_{h\to 0} (1+h)^{\frac{1}{h}}[/math].

 

Now we get that [math] e^h = \lim_{h\to 0} (1+h)[/math]

 

So by bunging this into our original equation, we have that our other bit of the limit tends to 1, which means f'(x) = e^x.

[jordan]

It does matter what is around the xⁿ. Obviously if we do something like take f(x) = x^{n[/sup']sin(x) then we can't just differentiate it using our existing rule - that comes later with the chain rule and product rule.}

^{Cool. I'll wait.}

[Dave]

Is anyone else following these threads now? Because if it's just me and jordan, there's not a whole lot of point going on with them tbh.

[Sayonara]

I'm planning to play catch-up this weekend.

 

Been busy with IHH

[Dave]

Cool, that's fair enough. I wasn't planning on writing another one until after the weekend anyway because I've got to sort out the purchase of my car - a quite enjoyable task.

 

Although I might post a few more things on this first.