If three variables [itex]x,y[/itex] and [itex]z[/itex] are related via some condition that can be expressed as $$F(x,y,z)=constant$$ then the partial derivatives of the functions are reciprocal, e.g. $$\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$ Is the correct way to prove this the following.(adsbygoogle = window.adsbygoogle || []).push({});

As [itex]x,y[/itex] and [itex]z[/itex] are related by [itex]F(x,y,z)=constant[/itex], at most only two of the variables can be independent (as the third can be expressed in terms of the other two). Consider the differentials $$dx=\frac{\partial x}{\partial y}dy+\frac{\partial x}{\partial z}dz$$ $$dy=\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz$$

Substituting the second expression into the first gives $$dx=\frac{\partial x}{\partial y}\left(\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz\right)+\frac{\partial x}{\partial z}dz \\ \Rightarrow\;\;\left(1-\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}\right)dx=\left(\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}+\frac{\partial x}{\partial z}\right)dz$$

Now, we choose [itex]x[/itex] and [itex]z[/itex] to be the independent variables, and as such, for this equality to be true [itex]\forall\; x,z[/itex] it must be that the terms in the brackets vanish identically. We see then, from the left-hand side of the equality, that $$\left(1-\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}\right)=0\;\;\Rightarrow\;\;\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$ and from the right-hand side $$\left(\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}+\frac{\partial x}{\partial z}\right)=0\;\;\Rightarrow\;\;\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-\frac{\partial x}{\partial z}$$ and hence together this gives the relation $$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1.$$

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# Proving the reciprocal relation between partial derivatives

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