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I am considering the following question

Transform the expression

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$

into one in $\rho$ and $\phi$.

Now my problem has arisen when considering $\frac{\partial \rho}{\partial x}$.

From considering $x=\rho cos(\phi)$, I obtain $\frac{\partial x}{\partial \rho} = cos(\phi)$ so then $\frac{\partial \rho}{\partial x} = \frac{1}{cos(\phi)}$

And yet in the book I am looking at, they instead used $\rho = (x^2+y^2)^{1/2}$ to obtain $\frac{\partial \rho}{\partial x} = \frac{x}{(x^2+y^2)^{1/2}}=\frac{\rho cos(\phi)}{\rho} = cos{\phi}$ !!!

It seems that maybe where I am going wrong is by not accounting for $y$ in the partial derivative of $\rho$. However I don't understand why I should take it into account. As I am taking the partial derivative, and I not just taking the derivative where there is an $x$ explicitly stated? (Although I do see a bit of a problem using my 'rule', because if I were to take the other expression for $\rho$ as $y=\rho sin(\theta)$, then I would get $\frac{\partial \rho}{\partial x} = 0$. Perhpas I misunderstood the partial derivative as only taking the derivative when a variable is explicitly stated....

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Meep
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1 Answers1

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It does look like you might be misunderstanding the definition of a partial derivative. To compute the partial derivative with respect to some parameter at a point, you vary that parameter while holding the others constant. In essence, you compute the rate of change of the function in the direction of one of the coordinate axes.

In general, you can’t compute partial derivatives of the inverse of a function by taking the reciprocals of the function’s partial derivatives. The problem, as explained here, is that you’re usually not holding the same variables constant when you do that. In your case, you’d need to hold, say, $y$ constant in both ${\partial\rho\over\partial x}$ and ${\partial x\over\partial\rho}$, so $\rho$ and $\theta$ are no longer independent variables when computing the latter.

The correct way to compute partial derivatives of a function’s inverse from its partial derivatives is by inverting the Jacobian matrix (a.k.a. differential or total derivative) of the function. In this case, $$J=\pmatrix{{\partial x\over\partial\rho}&{\partial x\over\partial\phi}\\{\partial y\over\partial\rho}&{\partial y\over\partial\phi}}=\pmatrix{\cos\phi&-\rho\sin\phi\\\sin\phi&\rho\cos\phi}$$ and so $$J^{-1}=\pmatrix{\cos\phi&\sin\phi\\-\frac1\rho\sin\phi&\frac1\rho\cos\phi}=\pmatrix{{\partial\rho\over\partial x}&{\partial\rho\over\partial y}\\{\partial\phi\over\partial x}&{\partial\phi\over\partial y}}.$$

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amd
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  • Can we (in general) do something like this with second derivatives? If we have all derivatives up to second order for on side. Here I mean that we have ${\partial^2 x\over\partial\rho^2}$ etc... and want the derivatives ${\partial^2 \rho \over\partial x^2}$ etc. (in terms of $r$ and $\theta$ )

    https://math.stackexchange.com/questions/2303255/inverse-partial-second-derivatives

     – Kvothe Jun 01 '17 at 21:11