# Ratio of differentials to partial derivatives of two (implicit?) functions

• Mar 15th 2011, 12:00 PM
matt.qmar
Ratio of differentials to partial derivatives of two (implicit?) functions
Hi.

So, given $\displaystyle f(x,y,z) = 0$ and $\displaystyle g(x,y,z) = 0$

Show that $\displaystyle \frac{dx}{\frac{\delta (f,g)}{\delta (y,z)}} = \frac{dy}{\frac{\delta (f,g)}{\delta (x,z)}} = \frac{dz}{\frac{\delta (f,g)}{\delta (x,y)}}$

I think it makes sense (I can "see" why the equality holds) but can't think of a way to derive it. To begin, I wrote $\displaystyle f(x,y,z) = 0$ as $\displaystyle f(x_f(y,z),y_f(x,z),z_f(x,y)) = 0$, assuming each argument was an implicit function of the others. Similarly, $\displaystyle g(x_g(y,z),y_g(x,z),z_g(x,y)) = 0$

I feel like some version of the chain rule would be useful ( does the cross product of the gradients have some relationship to differential??), and that the above equality could be easier to derive if written as $\displaystyle \frac{dy}{dx} = \frac{\frac{\delta (f,g)}{\delta (x,z)}}{\frac{\delta (f,g)}{\delta (y,z)}} = ....$. I seem to have trouble taking the gradient,or any partial derivatives of implicit functions when using the chain rule... ie, is $\displaystyle \frac{\delta f}{\delta x_f(y,z)}\frac{\delta x_f(y,z)}{\delta (y,z)}$ the correct derivative of f w.r.t x?

Any help appreciated, thanks.
• Mar 15th 2011, 04:15 PM
matt.qmar
Progress? I think that

$\displaystyle \nabla f = \frac{\delta f}{\delta x_f} \frac{\delta x_f}{\delta (y,z)} + \frac{\delta f}{\delta y_f} \frac{\delta y_f}{\delta (x,z)} + \frac{\delta f}{\delta z_f} \frac{\delta z_f}{\delta (x,y)}$

and

$\displaystyle \nabla g = \frac{\delta g}{\delta x_g} \frac{\delta x_g}{\delta (y,z)} + \frac{\delta g}{\delta y_g} \frac{\delta y_g}{\delta (x,z)} + \frac{\delta g}{\delta z_g} \frac{\delta z_g}{\delta (x,y)}$

I think that is how the gradients look, anyways.

When we take their cross products, are the numerators of their components denoted by $\displaystyle \delta (f,g)$? if so, then it looks like the solution is that the cross product of the gradients is proportional to the differentials?
• Mar 15th 2011, 04:59 PM
matt.qmar
Whoops, I meant to write the gradients as vectors, not as a sum.

$\displaystyle \nabla f = (\frac{\delta f}{\delta x_f} \frac{\delta x_f}{\delta (y,z)} , \frac{\delta f}{\delta y_f} \frac{\delta y_f}{\delta (x,z)} , \frac{\delta f}{\delta z_f} \frac{\delta z_f}{\delta (x,y)})$

and

$\displaystyle \nabla g = (\frac{\delta g}{\delta x_g} \frac{\delta x_g}{\delta (y,z)} , \frac{\delta g}{\delta y_g} \frac{\delta y_g}{\delta (x,z)} , \frac{\delta g}{\delta z_g} \frac{\delta z_g}{\delta (x,y)})$
• Mar 15th 2011, 07:54 PM
matt.qmar
Another update, I just went to double check that I had computed the gradient correctly but got something different, and the Wolfram gradient is messier but I suppose it must be correct? I guess I didn't really know the chain rule properly...