to ask for help with some maths homework?

[ChunkysMum]

Differentiate the following wrt x1 and x2:

f=-(1-t^2)/(x1+x2t)

If A and B are constants, no you can't differentiate with respect to them.

[Weeps]

Sorry, need a bit more help (thanks so far), apparently:

T is a constant, well it's 'time' so we need to solve this at each time step, but it's a known variable.

Any help much appreciated.

I am well of any of you who have the slightest idea what this is about. Runs off to join the other dunces.

this might help.

No help with the maths, but there's a good Carol Vorderman book which is for helping your kids with maths, which is very good and very simple to understand. It's about £9 or £10 on Amazon.

I assume you mean partial differentiation with respect to the two variables x1 and x2, t being considered as a constant?

If so,

df/dx1= (1-t2)/(x1+x2t)2

df/dx2= t(1-t2)/(x1+x2t)2

I can talk you through it if you like.

Let y = (a+bt)

So now
dy/da = 1
dy/db = t

Substitute y into original
f = -(1-t2)(y)(-1)

df/dy = -(1-t2) (-1)(y)(-2)
= (1-t2) / y2
= (1-t2) / (a + bt)2

Now use chain rule
df/da = df/dy . dy/da
= (1-t2) / (a + bt)2

and

df/db = df/dy . dy/da
= t (1-t2) / (a + bt)2

Ooops, xposted with Kirsty. Glad to see we got the same answer! :)

Me too !

I should mention a slight misuse of notation in what I wrote : partial differentiation is denoted by making the "d"s in df/dx1 round and curly.

Mumsnet still doesn't do WYSIWYG LaTeX though, so I couldn't type it .

Ah HA! You had me going for a minute but now I realise it's Mornington Crescent (Maths version)

I was going to answer this, honest, but saw it was already done.

Thank you - you a truly geniuses.

Or should that be geniae