How this calculation works
Hold the other input fixed for each partial derivative. Place output derivatives in rows and input derivatives in columns, then evaluate the matrix at the base point. The determinant describes signed local area scaling; the finite-displacement comparison shows what the first-order approximation omits.
J=[[Pₓ,Pᵧ],[Qₓ,Qᵧ]]; det J=PₓQᵧ−PᵧQₓ; F(a+h)≈F(a)+J(a)hA worked example
For P=x²+y and Q=xy at (2,1), J=[[4,1],[1,2]], whose determinant is 7. A displacement (0.1,0.2) predicts map value (5.6,2.5), compared with actual (5.61,2.52).
Good to know
Two-input, two-output real polynomial maps, degree ≤6 only. A nonzero local determinant does not establish global injectivity. The near-zero diagnostic uses floating-point scaling, not exact symbolic rank or a condition-number computation.
Reference: OpenStax Calculus: partial derivatives (opens in a new tab) ↗
Rows belong to outputs and columns to inputs
For a map F=(P,Q), the first row contains P’s x and y partial derivatives. The second row contains Q’s derivatives. Keeping that order explicit avoids transposing the derivative matrix while still obtaining a superficially familiar determinant.
At (2,1), P=x²+y and Q=xy give J=[[4,1],[1,2]]. The determinant is 4×2−1×1=7. Both output components and all derivative entries are shown so the map convention can be checked.
The vector field calculator summarizes related partial derivatives as divergence and scalar curl. A polynomial Jacobian calculator instead keeps the complete two-output derivative matrix.
The determinant describes a local map
A Jacobian determinant measures signed local area scaling for this two-dimensional map. Its sign records orientation at the point. It does not provide a single global scale factor for a nonlinear transformation over a large region.
A nonzero local determinant does not by itself prove the whole map is globally one-to-one. Conversely, a near-zero floating-point diagnostic is not an exact algebraic-rank result. Read the polynomial derivatives and the base point alongside that diagnostic.
Compare a finite displacement with its linearization
The first-order prediction is F(a+h)≈F(a)+J(a)h. For displacement (0.1,0.2) in the example, it predicts (5.6,2.5), while evaluating the map at the displaced input gives (5.61,2.52). The discrepancy comes from nonlinear terms.
The change of basis matrix calculator handles an exact linear coordinate conversion under its own rational-input scope. A polynomial Jacobian instead changes with location. The vector field calculator uses related partial derivatives to describe divergence and scalar curl, which are different summaries from this determinant.
A few questions, answered
Does the calculator support a three-by-three Jacobian?
No. It is scoped to two inputs and two polynomial outputs.
Is the determinant a global area ratio?
It is local signed area scaling at the entered point. A nonlinear map can vary across a finite region.
Why does the displaced value differ from the prediction?
The Jacobian captures first-order change only. Higher-order terms contribute to a finite displacement.