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

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March 15th 2011, 12:00 PM
matt.qmar

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

Hi.

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

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

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 $f(x,y,z) = 0$ as $f(x_f(y,z),y_f(x,z),z_f(x,y)) = 0$ , assuming each argument was an implicit function of the others. Similarly, $g(x_g(y,z),y_g(x,z),z_g(x,y)) = 0$

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 $\frac{dy}{dx} = \frac{\frac{\delta (f,g)}{\delta (x,z)}}{\frac{\delta (f,g)}{\delta (y,z)}} = ....$ . I seem to have trouble taking the gradient,or any partial derivatives of implicit functions when using the chain rule... ie, is $\frac{\delta f}{\delta x_f(y,z)}\frac{\delta x_f(y,z)}{\delta (y,z)}$ the correct derivative of f w.r.t x?

Any help appreciated, thanks.
March 15th 2011, 04:15 PM
matt.qmar

Progress? I think that

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

and

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

I think that is how the gradients look, anyways.

When we take their cross products, are the numerators of their components denoted by $\delta (f,g)$ ? if so, then it looks like the solution is that the cross product of the gradients is proportional to the differentials?
March 15th 2011, 04:59 PM
matt.qmar

Whoops, I meant to write the gradients as vectors, not as a sum.

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

and

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