# What does the partial derivative notation $\frac{\partial (f,g)}{\partial (x,y)}$ mean?

I am currently reading an old math book which contains the following unexplained notation:

Let $f(x,y)$ and $g(x,y)$ be functions $\mathbb{R}^2\rightarrow \mathbb{R}$. The notation

$$\frac{\partial(f,g)}{\partial(x,y)}$$

apparently refers to a function of the form $\mathbb{R}^2\rightarrow \mathbb{R}$. I am not sure, but I suspect it may be defined as

$$\frac{\partial(f,g)}{\partial(x,y)} \equiv \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}.$$

Is this notation/definition common in any particular field?

And especially if so, could there be an obvious interpretation of the following notation, which this book also uses without explanation?

$$\frac{\partial[f,g]}{\partial(x,y)}$$

Edit: I believe @Fred is correct that the parentheses are used to denote the Jacobian. Here is the notation, as used in a simplified excerpt of Calculating Curves by Ron Doerfler and others:

$$\left\{ \begin{array}{c} 0 = \frac{\partial u}{\partial y} f_1(x) + \frac{\partial v}{\partial y}\\ 0 = \frac{\partial u}{\partial x} f_2(y) + \frac{\partial v}{\partial x}\\ \end{array}\right.$$

Let us assume that $\frac{\partial(u,v)}{\partial(x,y)}=0$. Then the above equations yield that

$$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}[f_1(x) - f_2(y)] = 0.$$

Thus we can posit that

$$\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial x}\frac{\partial u}{\partial y} = e^\theta.$$

I am still unsure about the meaning of the square bracket notation. The square brackets are a little more difficult to place in context, but here is an attempt:

$$g_3(z) = u f_3(z) + v$$

We clearly have $\frac{\partial[g_3(z), z]}{\partial(x,y)} = 0$. By substituting the above equation and observing that $\frac{\partial[f_3(z), z]}{\partial(x,y)} = 0$, we obtain

$$f_3(z)\frac{\partial(u,z)}{\partial(x,y)} + \frac{\partial(v,z)}{\partial(x,y)} = 0.$$

Possibly the square brackets are simply an alias for round brackets which are used to avoid potentially visually-noisy nested round brackets(?).

• Never used the above notation. It might be the Lie bracket $[f,g] = fg - gf$. – Wuestenfux Jun 12 '17 at 7:51
• What book? And how does the book use this notation? Even a single example should clarify a lot. – Chris Culter Jun 12 '17 at 8:01

In vector calculus, we said, for the generalization of the chain rule, that $$\frac{\partial \mathbf f}{\partial \mathbf x}=\frac{\partial \mathbf f}{\partial \mathbf u}\frac{\partial \mathbf u}{\partial \mathbf x}$$ Where $\mathbf x, \mathbf f, \mathbf u$ are vectors (of functions).

Thus, $$\frac{\partial (f, g)}{\partial (x, y)}= \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial g}{\partial x} \end{bmatrix} \begin{bmatrix} \frac{\partial x}{\partial x}& \frac{\partial x}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{bmatrix}$$

Well, it is true that sometimes books use this notation for the determinant of this matrix (However I didn't). Then we get, $$\begin{vmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{vmatrix} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}- \frac{\partial f}{\partial y} \frac{\partial g}{\partial x}$$ Which is the Jacobian of $h(x,y)=(f(x,y), g(x,y))$.

Let $h: \mathbb R^2 \to \mathbb R^2$ be defined by $h(x,y) =(f(x,y),g(x,y))$.

Then $\frac{\partial(f,g)}{\partial(x,y)}$ is the Jacobian of $h$.

$$\frac{\partial(f,g)}{\partial(x,y)} = \left(\begin{matrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{matrix}\right)$$

However, I think that sometimes the notations is used for the determinant if this matrix.