I am doing a problem set and have come across the following question where I cannot get the right answer:

A variable $z$ may be expressed either as a function of $(u,v)$ or of $(x,y)$, where $u = x^2 + y^2$, $v=2xy$. Find: $$\left(\frac{\partial z}{\partial u}\right)_{v} \text{ in terms of } \left(\frac{\partial z}{\partial x}\right)_{y} \text{ and } \left(\frac{\partial z}{\partial y}\right)_{x}$$

Using the chain rule, I get:

$$\left(\frac{\partial z}{\partial u}\right)_{v}=\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial x}{\partial u}\right)_{v}+\left(\frac{\partial z}{\partial y}\right)_{x}\left(\frac{\partial y}{\partial u}\right)_{v}$$

Which by the reciprocity relation I get:

$$\begin{align*}\left(\frac{\partial z}{\partial u}\right)_{v}&=\left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{\partial u}{\partial x}\right)_{v}^{-1} +\left(\frac{\partial z}{\partial y}\right)_{x}\left(\frac{\partial u}{\partial y}\right)_{v}^{-1} \\ &= \left(\frac{\partial z}{\partial x}\right)_{y}\left(\frac{1}{2x}\right) + \left(\frac{\partial z}{\partial y}\right)_{x}\left(\frac{1}{2y}\right)\end{align*}$$

However the answer given is:

$$\left(\frac{\partial z}{\partial u}\right)_{v}=\frac{1}{2(x^2 - y^2)}\left\{x\left(\frac{\partial z}{\partial x}\right)_{y}-y\left(\frac{\partial z}{\partial y}\right)_{x}\right\}$$


1 Answer 1


The mistake is that reciprocal relation here doesn't hold. Note in your derivation, you have assumed that $$\frac{\partial x}{\partial u}\frac{\partial u}{\partial x}=1$$ which is not true in two variables. Instead, you shall use inverse function theorem to compute it. $$\begin{pmatrix}\frac{\partial z}{\partial u}\\\frac{\partial z}{\partial v}\end{pmatrix}=\begin{pmatrix}\frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}\\\frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}\end{pmatrix}\begin{pmatrix}\frac{\partial z}{\partial x}\\\frac{\partial z}{\partial y}\end{pmatrix}=\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}\\\frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}\end{pmatrix}^{-1}\begin{pmatrix}\frac{\partial z}{\partial x}\\\frac{\partial z}{\partial y}\end{pmatrix}$$

  • $\begingroup$ Could you please expand further on how to compute the partial derivative using the Inverse Function Theorem, because I can only find statements of it for functions: $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ $\endgroup$ Nov 5, 2013 at 21:44
  • $\begingroup$ @Shaktal See my addition plz. $\endgroup$
    – Shuchang
    Nov 6, 2013 at 0:41

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