The dy/dx Post, Or, Differentiation for Dunderheads

Bear in mind that I know nothing at all beyond MPC1: so this post will only cover the basic basics of differentiation because that is all I know about it. Also PLEASE correct the inevitable errors. I'm not going to do the whole equation from first principles because that would just be confusing.

To kick off, a quote from Tom Freeman over at Matt's place:

In the first term, we started on calculus. I looked at the dy/dx stuff the teacher was drawing on the whiteboard, and I said...

"Don't the 'd's cancel?"

The simple answer is no. This is because when you see dy/dx you are not looking at a fraction, but a statement: the change in y with respect to the change in x. Tom knows this of course, but I didn't until a few weeks ago so it's worth pointing out!

Let us assume you have a curve. You want to find its gradient. But you can't because it's constantly changing, and besides, y2-y1/x2-x1 that you used for straight lines doesn't quite seem to cut it, what with all that mucking about going up and down and stuff.

Instead you approximate it by picking point A, which is where you want to know the gradient of the curve. Then you pick another point, B which is some distance off from A on the curve. So the coordinates of A are (x,y) and of B, (x+n, y+p) where n and p are the differences between the two sets of x and y co-ordinates. If you've started at A, and gone on to B, B's coordinates will be a + the increase, and y + the increase.

You draw a line to join A and B, a chord. You can measure the gradient of the chord, obviously (y2-y1/x2-x1). It's a bit ineaxct but it gives an approximation of the gradient of the curve between the two points and so you're nearish to the gradient at A.



But if you think about it, it would be a closer approximation if the line was closer to A...and then closer still and then closer and then closer...




Eventually you'd have such a small line that the gradient of it would be the gradient at point A plus not very much at all, ie a really good approximation of it.

So all along your gradient has been, say, 4 + f, where f is the increase in the gradient, which gets smaller and smaller as you get closer to point A. At point A your gradient is 4 (no increase). This is also the gradient of the tangent at point A.

This is differentiation! Chopping the distance between two points on a curve into smaller and smaller bits until you get invisibly close to point A, to get the gradient at Point A. It's a pain in the neck though to do this all the time, or to draw an infinite amount of tangents with gradients of their own, so we have dy/dx to do it for us.

This is the essence of the formula they give you at A Level. I said I wasn't going to do the proof, because it is a bit confusing; but it's worth looking up because it helps you understand why you are doing this stuff.

What stuff?

Well, say you have the curve y=x². Dy/dx, the difference in y with respect to the difference in x (ie the gradient, for that is what it means), is 2x.

Reduce the power by one, and put it in front of the x. Piece of the proverbial.

What about the curve y=3x²?

Multiply the power by the number in front of the x, (6), then reduce the power as before. 6x. 6x is the gradient of the curve y=3x².

Easy!!

But what if there's no power? And you've just got 5x? Well 5x is really 5x¹. So you bring the power down as before and multiply, which makes 5x⁰. Stuff to the 0 is generally 1. So it's 5 times 1, so it's just 5.

If there's no power of the x, get rid of the x!! And you're done!

What if it's a constant? ie 8? Well constants always differentiate to make 0. Think of 8 as 8x⁰. Bring the 0 down and x it by 8, ie 0, and you get 0x-1. Ie NOWT!


But this is still a gradient formula, not the gradient itself. Well, this changes at every point on the curve, obviously. So you input the x co-ordinate of the point you want. What is the gradient at (2,3) on the curve y=3x²? Well dy/dx gives us 6x, and so put your x co-ordinate into this. 6 x 2 = 12. The gradient at (2,3) is 12.

I'll leave this for now on the question of what to do with a polynomial. The curve y= 3x⁴ - 2x³ + 6x² - 3x + 4

you just differentiate bit by bit:

dy/dx = 12x³ - 6x² + 12x - 3

Does anyone know how to do superscript in blogger? It would really help with these squares!!! (UPDATE: Yay! I've worked it out!)

Here's the proof for anyone who cares. It's come out a bit small but it's here! Click if you're bothered.