The
derivative
of a one-variable function measures one rate of change. A surface
z = f(x, y) has a slope in every direction, so we begin
with the two simplest: along x, and along
y. The partial derivative with respect to
x freezes y as a constant and
differentiates the rest as an ordinary one-variable function,
\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x + h,\, y) - f(x,\, y)}{h}.
The curly \partial ("partial") signals that other variables are
being held fixed. Likewise \dfrac{\partial f}{\partial y} freezes
x. We also write f_x and
f_y for short. The rule in practice is just this:
to get f_x, treat y as a number and
differentiate normally.
Both partials, step by step
Take
f(x, y) = x^2 y + \sin(xy).
Step 1 — \partial f/\partial x, treating
y as constant. The first term x^2 y has
constant factor y, so it differentiates to
2xy. For \sin(xy) the
chain rule gives
\cos(xy) times the derivative of the inside
xy with respect to x, which is
y:
f_x = 2xy + y\cos(xy).
Step 2 — \partial f/\partial y, treating
x as constant. Now x^2 y has constant
factor x^2, differentiating to x^2. For
\sin(xy) the inside xy differentiates to
x in y:
f_y = x^2 + x\cos(xy).
Step 3 — go to second order. A partial is itself a function of
x and y, so we can differentiate again.
The mixed partial f_{xy} means "first
x, then y". Differentiate
f_x = 2xy + y\cos(xy) with respect to y.
The first term gives 2x. The second is a product
y \cdot \cos(xy): the product rule gives
\cos(xy) + y \cdot (-\sin(xy)) \cdot x, so
f_{xy} = 2x + \cos(xy) - xy\sin(xy).
Step 4 — the other order. Now f_{yx}: differentiate
f_y = x^2 + x\cos(xy) with respect to x.
The first term gives 2x. The second is
x \cdot \cos(xy): the product rule gives
\cos(xy) + x \cdot (-\sin(xy)) \cdot y, so
f_{yx} = 2x + \cos(xy) - xy\sin(xy).
Step 5 — compare. The two mixed partials are identical:
f_{xy} = f_{yx} = 2x + \cos(xy) - xy\sin(xy).
That was no accident. For any function whose second partials are continuous, the order of
differentiation does not matter — a beautiful symmetry called Clairaut's
theorem.
The slope of a slice
Geometrically, fixing y = b slices the surface
z = f(x, y) with the vertical plane
y = b. The cut edge is a one-variable curve
z = f(x, b), and f_x(a, b) is just its
ordinary slope at x = a. Move the slice below (change
b) and watch the slice-curve — and its slope — change.
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The partial derivative
f_x = \partial f / \partial x is the ordinary derivative of
f in x with all other variables held
constant; likewise f_y. Geometrically
f_x(a, b) is the slope of the slice
z = f(x, b) at x = a.
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Higher and mixed partials iterate this:
f_{xx}, f_{yy},
f_{xy} = (f_x)_y and
f_{yx} = (f_y)_x.
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Clairaut's theorem. If f_{xy} and
f_{yx} are continuous on a region, then they are equal there:
f_{xy} = f_{yx}.
The order of mixed differentiation does not matter.
Clairaut's theorem has a hypothesis — continuity of the second partials — and it earns its
keep. The textbook counterexample is
f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2} \quad (f(0,0) = 0).
Away from the origin it is a perfectly ordinary rational function, but a careful
computation of the mixed partials at the origin gives
f_{xy}(0, 0) = -1 \ne 1 = f_{yx}(0, 0).
The two orders genuinely disagree. The escape clause is precisely Clairaut's hypothesis:
here f_{xy} is discontinuous at the origin, so the
theorem never applied. The moral is the recurring one of multivariable calculus — the
symmetric, well-behaved answer is the rule, but only when a continuity condition holds, and
the pathologies hide exactly where it fails.