Partial Differentials Identity

sebb1e
Messages
34
Reaction score
0

Homework Statement



Prove that if z=z(x,y) is invertible that:

(dz/dx)(dy/dz)(dx/dy)=-1 where the d's represent partial differentiation not total differentiation



Homework Equations





The Attempt at a Solution



I guess you start with the 6 total derivatives and substitute them into each other in someway. Beyond this I have no idea.
 
Physics news on Phys.org
so what is the definition of invertible here?
 
I think that you can write it as x=x(y,z) and y=y(x,z)?
 
try starting with the total differential
dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} dy
 
I've reduced it to:

dz=2(dz/dx)(dx/dy)(dy/dz)dz+((dz/dy)(dy/dz)+(dy/dx)(dx/dy))dz

So I need to show that (dz/dy)(dy/dz)+(dy/dx)(dx/dy)=2 where the d's are partial derivatives. If total derivatives this is obvious, am I missing something obvious here?

Also, is using this differential notation perfectly rigorous, I was expecting to have to do it using total derivatives or does that just create a lot more work.
 
so try 2 differentials
dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y} dy
dx = \frac{\partial x}{\partial z}dz + \frac{\partial x}{\partial y} dy

now try subtstituting in for dx

you'll have to make some assumptions (may need to prove) but due to the invertibility you're probably safe to assume
\frac{\partial z}{\partial x} = \frac{1}{\frac{\partial x}{\partial z}}

though i must say I haven't dealt with a function specifically defined as invertible like this

as for the differential v total derivatives, they can be derived from each other
http://en.wikipedia.org/wiki/Total_derivative
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top