I have the following question

$$f(x,y) = (x + \sin y)(\cos(πx^2 + y)) $$ Consider the following change of variables for $$\int \int f(x,y) dxdy \\ u = f(x, y), v = xy$$ What is the Jacobian of this change of variables at the point $(x, y) = (2, 0)$?

I did the following:

$J = \frac{\partial (u,v)}{\partial (x,y)} = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$

Evaluating $J$ at $(2,0)$ gives $2$. However this was not the correct answer. The solution stated:

The Jacobian matrix at $(2,0)$ is $\frac{\partial (u,v)}{\partial (x,y)} = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ which gives the “Jacobian” 2. However, we are interested in the Jacobian of $\frac{\partial (x,y)}{\partial (u,v)}$ which by the chain rule is 1/2.

So it makes sense that I should have calculated $\frac{\partial (x,y)}{\partial (u,v)}$. That being said, I do not see how I can go about doing this. How does using the chain rule give us the answer?

  • 1
    $\begingroup$ For most people $\frac{\partial(x,y)}{\partial(u,v)}$ denotes the Jacobian determinant, not matrix. But it's all a matter of convention. However, the solution is wrong. If you're converting the $uv$ integral you started with to an $xy$ integral, then you need to multiply by $\frac{\partial(u,v)}{\partial(x,y)}$ when you do the $xy$-integral. $\endgroup$ Nov 19, 2019 at 17:44
  • $\begingroup$ Sorry, the conversion is from an xy integral to a uv integral. The question is poorly written, I will edit. $\endgroup$ Nov 19, 2019 at 21:58

1 Answer 1


For any diffeomorphism $g$, the Chain Rule gives $$J(g^{-1}\circ g)(x)=J(g^{-1})(g(x))Jg(x),$$ but $g^{-1}\circ g$ is the identity map, so its Jacobian is the identity matrix. Therefore, the Jacobian of the inverse transformation is the inverse matrix of the Jacobian of the original transformation.

So we have $\frac{\partial (x,y)}{\partial (u,v)}$ is the inverse matrix of $\frac{\partial (u,v)}{\partial (x,y)}$, and therefore its determinant is the reciprocal of $\det\frac{\partial (u,v)}{\partial (x,y)}$.


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