y = x -2

Because ( dy / dx ) = nxn-1, I understand that the derivative will be

( dy / dx ) = -2x

But my problem lies in how you get to that derivative.

Now I had expected the problem to be solved in the book in the following way:

y + dy = ( x + dx )-2

y + dy = ( 1 / ( x2 + 2x(dx) + (dx)2))

y + dy = x-2 + 2x-1(dx) + (dx)-2

y + dy = x-2 + 2x-1(dx) - disregard (dx)-2

dy = 2x-1(dx) - remove the function

( dy / dx ) = 2x-1

But the Book solves it in a completely different way:

y + dy = ( x + dx )-2

y + dy = x-2 ( 1 + ( x / dx ))-2

It then proceeds to expand the equation by the binomial theorem, ending up with the same answer.

Why does Thompson do this? Is it not easier to do it the first way rather than the second? I cant understand the logic behind this. He gives no reason for doing it in this way.

>Because ( dy / dx ) = nxn-1, I understand that the
>derivative will be
>
>( dy / dx ) = -2x.
This seems odd. From what i can glean you are differentiating x^(-2), and the derivative of that is -2*x^(-3). But it'seems that this is just a typo.

>y + dy = ( 1 / ( x2 + 2x(dx) + (dx)2))
>
>y + dy = x-2 + 2x-1(dx) + (dx)-2

This is where the problem seems to lie. It appears that you are saying:
y + dy = 1/(x^2 + 2*x*dx + (dx)^2)
=> y + dy = x^(-2) + 2*x^(-1) dx + (dx)^(-2)
which does not follow.

>( dy / dx ) = 2x-1
This is wrong... but it may be a typo again.

>Why does Thompson do this? Is it not easier to do it the
>first way rather than the second? I cant understand the
>logic behind this. He gives no reason for doing it in this
>way.
There are, of course, many ways to differentiate x^(-2). You can do it as in the book, or represent it as (1/x)^2 and use the chain rule, or as e^((-2) * ln x) and use the chain rule again, or say:
0 = d/dx(1) = d/dx(x^2 * x^(-2))
= d/dx(x^2) * x^(-2) + x^2 * (d/dx(x^(-2)))
= 2*x^(-1) + x^2 * d/dx(x^(-2))
and rearrange.

I'm sure there are many other ways, and all have advantages and disadvantages (for example some of the above do not work for all values of x, but only for positive or for nonzero x). I'm sure Thompson had a good reason for doing it in the way he did: It illustrates a standard way to get derivatives, and shows the power of the binomial theorem.

Thankyou

sfwc
<><

PS Part of your message got randomly turned into a smiley. This has happened to me before, too. The way to avoid it is to click the 'Check if you DO NOT wish to use emotion icons in your message' box at the bottom of the posting form.