Calculus III treated partial derivatives, the Jacobian, and Green's theorem as tools for real-valued functions of two real variables. This course is about to reuse every one of them almost at once. It uses the same x and y, now relabeled as the real and imaginary parts of a single complex number.
Chapter 3 asks whether a function of z is differentiable in the strong complex sense. Its whole test is a pair of partial-derivative equations. Chapter 7's central theorem has two proofs. One of them applies Green's theorem directly to the real and imaginary halves of a contour integral.
A partial derivative means this. Pretend the other variable is a fixed number for a moment. Then take the ordinary derivative you already know how to take.
The Jacobian packs four separate partial derivatives into one matrix. Its determinant, a single number computed from those four, answers one question. When the transformation is applied to a tiny square, does the square get bigger or smaller, and by what factor?
Green's theorem trades a walk around the edge of a shape for a sum taken over its whole interior. Which side of that trade is easier depends on the problem.
Here is a slip this invites. You might compute f_y by differentiating with respect to x by mistake. Freezing the wrong variable is the most common way to get a correct-looking partial derivative that answers the wrong question.
The partial derivative f_x freezes every other variable. It differentiates in x alone, by the ordinary single-variable rules. The partial derivative f_y does the same, freezing x instead.
Take a transformation u=u(x,y), v=v(x,y). The Jacobian collects all four partial derivatives into one matrix, \begin{pmatrix}u_x&u_y\\v_x&v_y\end{pmatrix}. Its determinant is a single number. It reports how much a tiny patch of area is stretched or shrunk right at that point.
Green's theorem, stated above, converts a line integral into a double integral. The line integral is walked once around a closed curve C. The double integral adds up Q_x-P_y over the flat piece of the plane that C closes off. It turns one kind of computation into a very different one that is often easier.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For f(x,y,z) = x^2yz^3, find f_z(x,y,z).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Freeze x and y; the power rule on z^3 gives 3z^2.
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Why
This works the same way with three variables -- freeze all but the one you're differentiating with respect to.
From Calculus III — Multivariable Calculus
For f(x,y) = \dfrac{x}{x+y}, find f_x(1,1).
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Why
Quotient rule with y fixed: f_x = \dfrac{(x+y)(1) - x(1)}{(x+y)^2} = \dfrac{y}{(x+y)^2}, so f_x(1,1) = \dfrac{1}{4} = 0.25.
Correct answer
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Why
Use the quotient rule treating y as a constant.
f_x = y / (x+y)^2.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For u=x-y,\ v=x+y (solved as x=(u+v)/2, y=(v-u)/2), find the Jacobian \left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|.
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Why
\dfrac{\partial x}{\partial u}=\tfrac12,\ \dfrac{\partial x}{\partial v}=\tfrac12,\ \dfrac{\partial y}{\partial u}=-\tfrac12,\ \dfrac{\partial y}{\partial v}=\tfrac12, giving determinant \tfrac14+\tfrac14=\tfrac12.
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Why
Compute the four partials of x and y with respect to u and v, each equal to plus or minus 1/2.
The determinant is (1/2)(1/2) - (1/2)(-1/2).
From Calculus III — Multivariable Calculus
The Jacobian is the multivariable generalization of which single-variable concept?
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Why
Just as single-variable substitution trades dx for g'(u) du, the Jacobian is the exact factor a coordinate change trades dA or dV for.
Recall what single-variable u-substitution replaces dx with.
Both formulas trade one differential for another, multiplied by a derivative-based factor.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Using Green's Theorem, find \oint_C \mathbf{F}\cdot d\mathbf{r} for \mathbf{F}(x,y)=(y^2,\,x^2) around the square with vertices (0,0),(2,0),(2,2),(0,2) (counterclockwise).
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Why
Q_x-P_y = 2x-2y. Integrating over the square: \iint_0^2\int_0^2 (2x-2y)\,dx\,dy. By symmetry (swap x\leftrightarrow y flips the sign of the integrand but the square is symmetric under that swap), this integral is 0.
Correct answer
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Why
Set up the double integral of 2x-2y over the square and look for symmetry before grinding through the algebra.
From Calculus III — Multivariable Calculus
Using Green's Theorem, find the circulation of \mathbf{F}(x,y)=(-y,x) around a square of side length 4 (counterclockwise), centered anywhere.
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Why
Q_x-P_y=2 everywhere. The square's area is 4^2=16. Circulation =2\cdot16=32 — the same regardless of where the square is centered, since curl is constant.
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Why
The curl is constant, so circulation is just curl times enclosed area, regardless of the region's exact position or even its exact shape.
Theorem
Green's Theorem
Recalled from Calculus III — Multivariable Calculus · Green's theorem — first taught there.
Let C be a positively oriented (counterclockwise), piecewise-smooth, simple closed curve in the plane, and let R be the region it encloses. If P and Q have continuous first partial derivatives on an open region containing R, then \oint_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.
Equivalently, writing \mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j}, the circulation form is \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_R \operatorname{curl} \mathbf{F}\, dA, trading a circulation integral around the boundary for a curl integral over the whole enclosed region.
Sketch for a region R that is both x-simple and y-simple (the general case patches finitely many such regions together, with the shared interior edges canceling since each is traversed once in each direction). Write R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Then \iint_R -\frac{\partial P}{\partial y}\, dA = -\int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\, dx, by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly \oint_C P\,dx, splitting C into its bottom piece y=g_1(x) (traversed left to right) and top piece y=g_2(x) (traversed right to left, contributing the minus sign). The same argument with R described as y-simple gives \iint_R \frac{\partial Q}{\partial x}\, dA = \oint_C Q\,dy. Adding the two established identities gives the theorem.
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Zero curl, everywhere, on this field
Drop a loop anywhere on this field, \mathbf F=(x,y) — Green's theorem gives zero circulation around it, since Q_x-P_y=1-1=0 at every point, not just on average.
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Worked example
Both partial derivatives, evaluated at a point
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1
For f(x,y)=x^2y^3-4xy+y, find f_x(1,2) and f_y(1,2). Freeze y and differentiate in x: f_x = 2xy^3-4y.
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2
Freeze x and differentiate in y: f_y = 3x^2y^2-4x+1.
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3f_x(1,2) = 2(1)(8)-4(2) = 8, \qquad f_y(1,2) = 3(1)(4)-4(1)+1 = 9\boxed{f_x(1,2)=8,\quad f_y(1,2)=9}
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Worked example
A Jacobian determinant, and what it measures
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1
For u=x^2y, v=x+y^2, find the Jacobian determinant at (x,y)=(1,2). u_x=2xy,\ \ u_y=x^2,\ \ v_x=1,\ \ v_y=2y
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2
At (1,2): u_x=4, u_y=1, v_x=1, v_y=4. \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix}4&1\\1&4\end{vmatrix} = (4)(4)-(1)(1)
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3
A tiny square at (1,2) is stretched to 15 times its original area once u and v are applied to it. \boxed{\dfrac{\partial(u,v)}{\partial(x,y)}\Big|_{(1,2)} = 15}
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Where this goes next
What Chapter 3 and Chapter 7 both build on directly
Chapter 3's Cauchy-Riemann equations are a pair of partial-derivative equations, u_x=v_y and u_y=-v_x. They tie a complex function's real and imaginary parts together.
Chapter 7's central theorem gets one full proof. This proof assumes those same partial derivatives are continuous. It writes the real and imaginary parts of a closed contour integral as line integrals. It then applies Green's theorem to each. The resulting double integral vanishes because the Cauchy-Riemann equations force Q_x-P_y=0.
Check your understanding (counts toward your progress)
For f(x,y)=3x^2y-5y^3, find f_x.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Differentiating 3x^2y in x (with y frozen) gives 6xy; the -5y^3 term has no x, so it contributes 0.
Correct answer
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Why
Treat y as a fixed constant.
What does the Jacobian determinant's absolute value measure at a point?
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The four partial derivatives are combined by a determinant, not an average — the specific combination is what makes it measure area.
Linearity is a global property of the whole map — the Jacobian determinant is a purely local, point-by-point number.
It's a strictly LOCAL number at one point, not a total accumulated anywhere else.
Why
The Jacobian determinant reports how much a tiny patch of area at that specific point is stretched or shrunk once the transformation is applied.
Think about what a determinant measures for an ordinary linear map, then apply that same idea pointwise.
For u=xy^2, v=x^2+y, find the Jacobian determinant \partial(u,v)/\partial(x,y) at (x,y)=(1,1).
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Why
At (1,1): u_x=1, u_y=2, v_x=2, v_y=1. Determinant =1(1)-2(2)=-3.
Correct answer
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Why
u_x=y^2, u_y=2xy, v_x=2x, v_y=1.
Using Green's theorem, \oint_C(0\,dx+x\,dy) around the boundary of the square [0,2]\times[0,2] equals \iint(Q_x-P_y)\,dA over that square. Since Q_x-P_y=1 everywhere and the square has area 4, what is the value of the line integral?
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Why
\iint_{[0,2]\times[0,2]} 1\,dA = 1\cdot(\text{area}) = 1\cdot4 = 4.
Correct answer
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Why
A constant integrand of 1 integrates to exactly the area.
Practice: Freezing one variable at a time (5 problems · doesn’t affect your progress)
To find f_y(x,y), you...
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That procedure computes f_x, not f_y — the roles of the two variables are swapped here.
A partial derivative only ever moves in ONE direction at a time — combining both isn't how f_y is defined.
Setting y=0 throws away the exact information a derivative in y needs to measure.
Why
f_y freezes x and applies the ordinary derivative rules in y alone.
Treat x as constant and differentiate with respect to y
For f(x,y)=4xy^2-y, compute the partial derivative f_x.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Freezing y, the derivative of 4xy^2 in x is 4y^2; -y has no x and contributes 0.
Correct answer
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Why
For f(x,y)=4xy^2-y, find the partial derivative f_y instead of f_x.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Freezing x: the derivative of 4xy^2 in y is 8xy; the derivative of -y is -1.
Correct answer
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Why
Freeze x and differentiate each term in y separately.
For u=x+y^2, v=xy, find the Jacobian determinant at (x,y)=(2,1).
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Why
At (2,1): u_x=1, u_y=2, v_x=1, v_y=2. Determinant =1(2)-2(1)=0.
Correct answer
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Why
u_x=1, u_y=2y, v_x=y, v_y=x.
A Jacobian determinant comes out to exactly 0 at a point. What does that indicate?
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A zero determinant is a perfectly meaningful, computable result — it doesn't mean the transformation itself breaks down at that input.
A determinant of exactly 0 is a valid, common outcome — it's not a sign of an arithmetic mistake.
Area-preserving means the determinant has absolute value exactly 1, not 0 — a determinant of 0 is the opposite extreme, total collapse.
Why
A Jacobian determinant of 0 means a tiny patch of area gets crushed to zero area at that point. So the transformation fails to be locally invertible there.
The transformation locally collapses area to zero there — it fails to be locally invertible at that point