Re: Total Derivative
- To: mathgroup at smc.vnet.net
- Subject: [mg42097] Re: Total Derivative
- From: "Will Self" <wself at msubillings.edu>
- Date: Thu, 19 Jun 2003 03:59:02 -0400 (EDT)
- References: <bcp156$7tu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Joel storch" <jstorch at earthlink.net> wrote in message news:bcp156$7tu$1 at smc.vnet.net... > Dt[x y,x,y] gives 1+dx/dy dy/dx > > I would expect: 1+dx/dy dy/dx + x d/dy (dy/dx) > > Can someone explain this result ? Dt[f, x, y] is supposed to give d/dy(df/dx). Check the Help on Dt to verify this. Where I write d, you should read curly-d, the partial derivative. Then Dt[x y, x, y] should mean d/dy(d(x y)/dx) Start with the inside, d(x y)/dx. By the product rule, this is x dy/dx + y Now do the derivative of this answer with respect to y d/dy (x dy/dx + y) = x d^2y/(dydx) + dy/dx dx/dy + 1 |---------------------------| product rule --> 1st D 2nd 2nd D 1st But because mixed partials are equal, d^2y/(dydx) = d^2y/(dxdy) = d/dx(dy/dy) = d/dx(1) = 0. So all you're left with is dy/dx dx/dy + 1