A derivative in one variable is a single number, the best linear multiplier near a point. When a map sends several inputs to several outputs, the best linear multiplier is a matrix. That matrix is the Jacobian. It is not a new kind of derivative. It is the derivative, written so that every first partial derivative has a seat.

If you already like the sentence $f(x)\approx f(a)+f'(a)(x-a)$, the multivariable cousin is

$$\mathbf{f}(\mathbf{a}+\mathbf{h})\approx \mathbf{f}(\mathbf{a})+D\mathbf{f}(\mathbf{a})\,\mathbf{h}.$$

Here $D\mathbf{f}(\mathbf{a})$ is the Jacobian matrix at $\mathbf{a}$, and $\mathbf{h}$ is a small input step. Matrix times vector is the linear map. The error, meaning the true output minus this prediction, should shrink faster than the length of $\mathbf{h}$ as $\mathbf{h}$ shrinks to $\mathbf{0}$. That is the same demand as in one variable, told on what a derivative actually measures. If that sentence still feels empty, repair it there and come back. The geometry of the matrix itself lives on linear algebra and in eigenvalues in plain language. The calculus setting is the multivariable pillar. The study order this site uses is the linear algebra first course, then this page.

How the entries are arranged

Write $\mathbf{f}(x,y)=(u(x,y),v(x,y))$ as a map from the plane to the plane. The Jacobian is

$$D\mathbf{f}=\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}.$$

Row $i$ is the gradient of the $i$th output, written as a row. Column $j$ is the velocity of the output when you move only the $j$th input and hold the others fixed. Both readings are useful. Gradients tell you how each coordinate function tilts. Columns tell you where a small step along each axis is sent by the linear approximation.

That column picture is the same story as "a matrix is what it does to basis vectors," which you already use for eigenvalues. The Jacobian is the matrix of the derivative at one point. It is allowed to change when you move the point. A constant matrix is a linear map on the whole plane. A Jacobian is usually only the best linear stand in near the point where you computed it.

A polar picture, including the area factor

Example Polar coordinates send $(r,\theta)$ to $(x,y)=(r\cos\theta,\, r\sin\theta)$. The Jacobian is

$$D=\begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}.$$

The first column is the unit radial direction. Increasing $r$ a little moves you straight out from the origin, and the speed of that motion does not depend on $r$. The second column is a tangential direction of length $r$. Increasing $\theta$ a little walks you along a circle, and a larger radius makes the same angle change sweep a longer arc. The determinant is $r\cos^2\theta+r\sin^2\theta=r$. Local area in the $(r,\theta)$ plane is scaled by $|r|$ when you look in the $(x,y)$ plane. That is why a polar integral picks up a factor of $r$. For the usual chart with $r\ge 0$, the absolute value is $r$ itself.

At $r=0$ the determinant vanishes. Many angles collapse to one point, the origin. The linear map crushes the $\theta$ direction. Polar coordinates are a poor chart exactly there. The derivative of the coordinate map still exists. What fails is local invertibility. You cannot solve uniquely for $(r,\theta)$ in a full neighborhood of the origin in the plane, because $\theta$ is not a single valued function around that point.

A map you can check with a matrix

Example Let $\mathbf{f}(x,y)=(x^2-y^2,\, 2xy)$. This is the real and imaginary parts of $(x+iy)^2$, but you do not need complex numbers to use it. The partials are $u_x=2x$, $u_y=-2y$, $v_x=2y$, $v_y=2x$, so

$$D\mathbf{f}(x,y)=\begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}.$$

The determinant is $4x^2+4y^2=4(x^2+y^2)$. At $(1,0)$ the matrix is $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$, the determinant is $4$, and both eigenvalues are $2$. The linear approximation doubles every small step. Check a number. $\mathbf{f}(1,0)=(1,0)$. A step $(0.1,0)$ is predicted to land at $(1,0)+(0.2,0)=(1.2,0)$. The true value is $\mathbf{f}(1.1,0)=(1.21,0)$. The error is $0.01$ in the first coordinate, much smaller than the step $0.1$. A wrong matrix, for example the identity, would leave an error about the size of the step. Only this Jacobian makes the miss shrink faster than the step.

At the origin the Jacobian is the zero matrix, so the determinant is $0$. The prediction says a small step barely moves the output. In fact $\mathbf{f}(h,0)=(h^2,0)$, which is quadratic. The error of the zero prediction, divided by $|h|$, still goes to $0$, so the derivative really is the zero matrix. The map is not one to one near the origin, because $\mathbf{f}(x,y)=\mathbf{f}(-x,-y)$. A singular Jacobian warned you. It did not say the derivative failed to exist.

What the determinant is saying

For a map from $\mathbb{R}^n$ to $\mathbb{R}^n$, $\det D\mathbf{f}(\mathbf{a})$ is the signed volume scaling of the best linear approximation at $\mathbf{a}$. Positive means orientation is kept. Negative means it is flipped. Zero means some direction is crushed, so that linear map is not invertible. Area in the plane is the $n=2$ case of the same sentence. The polar factor $r$ and the factor $4(x^2+y^2)$ above are both this number.

The inverse function theorem, in undergraduate language, says that if $D\mathbf{f}(\mathbf{a})$ is invertible, then $\mathbf{f}$ itself is invertible on small neighborhoods of $\mathbf{a}$ and of $\mathbf{f}(\mathbf{a})$, and the derivative of the local inverse is the inverse matrix of $D\mathbf{f}(\mathbf{a})$. You do not need the full proof to use the slogan. A nonzero Jacobian determinant is a green light for a local inverse. A zero determinant is a place to slow down. It is not an automatic disaster at every nearby point. In the example above, the determinant vanishes only at the origin. Nearby, $4(x^2+y^2)$ is positive and a local inverse exists.

Eigenvalues of the Jacobian, when you compute them, are stretch factors of that linear approximation along special directions. They describe $D\mathbf{f}(\mathbf{a})$, not the full nonlinear map. A large eigenvalue says the approximation stretches hard in one direction. It does not say what $\mathbf{f}$ does on the other side of the domain. Read eigenvalues in plain language if "stretch factor" is the foggy half of that sentence.

The chain rule is matrix multiplication

If you compose maps, you multiply Jacobians, in the order that the chain rule already uses in one variable. A small case is enough to see it. Let $\mathbf{g}(t)=(t,t^2)$, a curve, and let $f(x,y)=x^2+y$, a scalar. The composition is $f(\mathbf{g}(t))=t^2+t^2=2t^2$, whose ordinary derivative is $4t$.

The Jacobian of $f$ is the gradient row $\langle 2x,1\rangle$. Along the curve that row is $\langle 2t,1\rangle$. The velocity $\mathbf{g}'(t)$ is the column $\langle 1,2t\rangle$. Their product is $2t\cdot 1+1\cdot 2t=4t$. Same number. The gradient is a $1$ by $2$ Jacobian. The velocity is a $2$ by $1$ Jacobian. The chain rule did not invent a new operation. It multiplied the matrices. When both maps are plane to plane, the product is $2$ by $2$ times $2$ by $2$, and the order matters, because matrix multiplication is not commutative.

Maps that change dimension

If $\mathbf{f}:\mathbb{R}^n\to\mathbb{R}^m$, the Jacobian is an $m$ by $n$ matrix. It is still the derivative. It often cannot be square, so it has no single determinant. Rank becomes the useful word. Full rank means the linear approximation hits a subspace of the largest possible dimension. For a scalar function, $m=1$, the Jacobian is a single row, the gradient, and the linear approximation is the tangent line or tangent plane story.

For a curve $\mathbf{r}(t)$ in space, $n=1$, the Jacobian is a single column $\mathbf{r}'(t)$, the velocity. Nothing here conflicts. The definition specializes cleanly. A surface parametrization $\mathbf{r}(u,v)$ has a $3$ by $2$ Jacobian. Its two columns are $\mathbf{r}_u$ and $\mathbf{r}_v$, and their cross product is the normal that shows up in a surface integral. That cross product is the bridge to Stokes theorem intuition. You do not need Stokes theorem to compute a Jacobian. You do need the columns when a flux integral asks for $d\mathbf{S}$.

Change of variables in integrals

When you rewrite a double or triple integral in new coordinates, you push a region through a map $\mathbf{g}$. Little squares or little boxes in the parameter domain become little parallelograms in space. Their volumes differ by $|\det D\mathbf{g}|$. That factor is the Jacobian determinant, taken in absolute value because the usual integral uses unsigned volume. Forgetting it is the most common computational error after a substitution looks correct on the integrand alone.

Polar coordinates are the drill. The integrand gains a factor $r$, and the region has to be described in $r$ and $\theta$, not left in $x$ and $y$. Both changes are part of the substitution. A correct factor on the wrong region is still the wrong integral. The determinant tells you the local scale. It does not draw the region for you.

Pitfall. The Jacobian matrix and the Jacobian determinant are related and not identical. People say "the Jacobian" for either. In a sentence about invertibility, eigenvalues, or a chain rule product, you want the matrix. In a sentence about area or volume after a substitution, you want the absolute value of the determinant. A zero matrix is a determinant of zero. A determinant of zero does not tell you the whole matrix.

How to know you understood

  1. Write the Jacobian of $(r\cos\theta, r\sin\theta)$ and say why the determinant equals $r$.
  2. At $(1,0)$ for $(x^2-y^2, 2xy)$, state the matrix, the determinant, and the error of the linear prediction on the step $(0.1,0)$.
  3. Explain why a zero Jacobian at the origin of that map does not mean "no derivative."
  4. Multiply the gradient of $x^2+y$ by the velocity of $(t,t^2)$ and match it with the ordinary derivative of $2t^2$.
  5. Say, in one sentence, when you insert $|\det D\mathbf{g}|$ into an integral and when you leave it out.

Practice on the matrix tools

Compute every partial by hand before you open a calculator. The tools on this site do not differentiate a vector formula. They audit a matrix you already wrote down.

Try it. On the 2 by 2 matrix calculator, enter the Jacobian at $(1,0)$ from the example above: $a=2$, $b=0$, $c=0$, $d=2$. The determinant should be $4$, and the inverse should be $(1/2)$ times the identity. Then open the 2 by 2 eigenvalue calculator on the same four numbers. Both eigenvalues should be $2$, matching a pure stretch by $2$. For a single partial, hold $y$ fixed at $0$ so $u(x,0)=x^2$, and check the slope at $x=1$ on the numerical derivative calculator. You should see about $2$, which is the first entry of the first row. The calculator never saw $y$. You held it fixed.

A cross product of two columns, when the Jacobian is $3$ by $2$, belongs on the vector operations calculator. Swap the order of the cross product and the normal flips. That is an orientation choice, not a new derivative.

After you have written a 2 by 2 Jacobian by hand, check its determinant and its stretch factors.

Open the 2 by 2 matrix calculator

Two gradient rows side by side are the Jacobian of a pair $(f,g)$. They are parallel exactly when that determinant is zero, which is the tangency condition on Lagrange multipliers intuition. The multiplier page assumes you can already read a gradient. This page is the place those rows become a matrix.

Citations and further reading