Question

How do I prove the following properties of partial derivatives? (∂z/∂x)_y=〖[∂x/dz]_y〗^(-1)
and
(∂x/∂y)_z (∂y/∂z)_x (∂z/∂x)_y=-1

Answer 1

High, kindly use the equation mode.

Answer 2

Looks like
\[\begin{align*}
\left(\frac{\partial z}{\partial x}\right)_y&=\left(\left(\frac{\partial x}{\partial z}\right)_y\right)^{-1}&(1)\\[1ex]
\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y&=-1&(2)
\end{align*}\]though I'm not sure what the subscripts mean. Probably that \(\left(\dfrac{\partial x_i}{\partial x_j}\right)_{x_k}=\dfrac{\partial x_i}{\partial x_j~\partial x_k}\)?

Answer 3

Can you post a picture of the equation?

Answer 4

Sorry for the inconvenient typing, but I don't know how to use the equation mode when typing a question! HolsterEmission these equations are correct. The subscripts represent the variable that is considered a constant in a partial derivative, but I suppose they are not always necessary.

Answer 5

assuming \(z = z(x,y)\)

and if y = const you can say that

\(z = z(x)\)


so you have the usual proof that \(y' = \dfrac{1}{x'}\), using decomposition

and if it's that limited, the second bit should follow too

Answer 6

Still having trouble proving the second property!