Source: Stewart's Calculus, 8th ed., Sections 14.1, 14.3, 14.4, 14.5
Difficulty: Intermediate | Prerequisites: Comfort with first-order partial derivatives, single-variable chain rule, and basic multivariable function notation (Sections 14.1, 14.3).
Partial derivatives extend the idea of a derivative to functions of several variables, and they underpin nearly everything in the second half of a multivariable calculus course. This set of notes covers second partial derivatives (including mixed partials and Clairaut's theorem), implicit differentiation for surfaces defined by equations in x, y and z, and linear approximation (the tangent plane at a point). If you are comfortable taking first-order partials and applying the chain rule, you are ready for this material. These topics feed directly into optimisation (critical points, Lagrange multipliers) and double integrals, both of which appear on the same midterm.
Second partial derivatives measure how a rate of change itself changes. Implicit differentiation lets you find partial derivatives of z even when you cannot solve for z explicitly. Linearization gives you the tangent-plane approximation to a surface at a point, which is the multivariable version of the tangent-line approximation from Calc I.
Tags: second partial derivative, mixed partial, Clairaut's theorem, implicit differentiation, linearization, tangent plane, differential, local linear approximation
Second partial derivative
The derivative of a partial derivative. For a function f(x, y), f_xx means you differentiate with respect to x twice. f_yy means you differentiate with respect to y twice.
Think of it as: how the slope in one direction is itself changing as you move further in that direction.
Mixed partial derivative
A second partial derivative where you differentiate with respect to two different variables, e.g. f_xy (differentiate first with respect to x, then with respect to y).
In simple terms, this measures how the rate of change in one direction shifts as you move in the other direction.
Clairaut's theorem (equality of mixed partials)
If f_xy and f_yx are both continuous, then f_xy = f_yx. The order of differentiation does not matter.
Think of it as: you get the same answer regardless of which variable you differentiate first, provided the function is reasonably well behaved.
Implicit differentiation (multivariable)
A technique for finding dz/dx and dz/dy when z is defined implicitly by an equation F(x, y, z) = 0 rather than explicitly as z = f(x, y). You differentiate the entire equation with respect to the variable of interest, treating z as a function of x and y, then solve for the desired partial.
In simple terms, you cannot isolate z, so you differentiate the whole equation and rearrange.
Linearization (tangent plane approximation)
The linear function L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) that best approximates f near the point (a, b). It is the equation of the tangent plane to the surface z = f(x, y) at (a, b).
Think of it as: the multivariable version of the tangent line from Calc I. Close to the point, the surface looks nearly flat, and L gives you that flat approximation.
Computing second partials
Start with f(x, y). Compute f_x and f_y first.
Then differentiate each first partial again: f_xx = (f_x)_x, f_yy = (f_y)_y, f_xy = (f_x)_y, f_yx = (f_y)_x.
By Clairaut's theorem, f_xy = f_yx whenever both are continuous (which is almost always the case on exams).
Worked pattern (from Problem 1a)
Given f(x, y) = x^4 y - 2x^3 y^2:
f_x = 4x^3 y - 6x^2 y^2
f_y = x^4 - 4x^3 y
f_xx = 12x^3 y - 12xy^2
f_xy = 4x^3 - 12x^2 y (and f_yx gives the same result)
f_yy = -4x^3
Quotient-rule cases (from Problem 1b)
When the function is a ratio, e.g. z = y / (2x + 3y), use the quotient rule for each first partial, then again for each second partial. The algebra is heavier but the process is identical. Expect denominators of the form (2x + 3y)^3 in the second partials.
When to use it
Use implicit differentiation when z is defined by an equation like F(x, y, z) = 0 and you cannot (or do not want to) solve for z.
Method
Write the equation so everything is on one side.
To find dz/dx: differentiate every term with respect to x, treating y as a constant and z as a function of x. Every time you differentiate a z-term, attach a dz/dx factor (chain rule). Then isolate dz/dx.
To find dz/dy: same process, but differentiate with respect to y and treat x as constant.
Worked pattern (from Problem 2a)
Given x^2 - y^2 + z^2 - 2z = 4:
Differentiate with respect to x: 2x + 2z(dz/dx) - 2(dz/dx) = 0
Factor: (2z - 2)(dz/dx) = -2x
Solve: dz/dx = x / (1 - z)
For dz/dy by the same equation: dz/dy = y / (z - 1).
Log and product cases (from Problem 2b)
Given yz + x ln y = z^2:
dz/dx: differentiate to get y(dz/dx) + ln y = 2z(dz/dx), then dz/dx = ln y / (2z - y).
dz/dy: differentiate to get z + y(dz/dy) + x/y = 2z(dz/dy), then dz/dy = (x + yz) / (y(2z - y)).
The formula
L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)
This is the equation of the tangent plane to the surface at the point (a, b, f(a, b)).
Step-by-step process
Evaluate f(a, b).
Compute f_x and f_y, then evaluate each at (a, b).
Substitute into the formula.
Worked pattern (from Problem 3a)
f(x, y) = sqrt(xy) at (1, 4):
f(1, 4) = sqrt(4) = 2
f_x = y / (2 sqrt(xy)), so f_x(1, 4) = 4 / (2 * 2) = 1
f_y = x / (2 sqrt(xy)), so f_y(1, 4) = 1 / (2 * 2) = 1/4
L(x, y) = 2 + 1(x - 1) + (1/4)(y - 4)
Trig and inverse-trig cases (from Problems 3c, 3d)
For f(x, y) = 4 arctan(xy) at (1, 1):
f(1, 1) = 4 arctan(1) = pi
f_x = 4y / (1 + x^2 y^2), so f_x(1, 1) = 2
f_y = 4x / (1 + x^2 y^2), so f_y(1, 1) = 2
L(x, y) = pi + 2(x - 1) + 2(y - 1)
For f(x, y) = y + sin(x/y) at (0, 3):
f(0, 3) = 3 + sin(0) = 3
f_x = cos(x/y) / y, so f_x(0, 3) = 1/3
f_y = 1 - x cos(x/y) / y^2, so f_y(0, 3) = 1
L(x, y) = 3 + (1/3)(x - 0) + 1(y - 3)
Second partial derivatives of f(x, y)
f_xx = d/dx(f_x), f_yy = d/dy(f_y), f_xy = d/dy(f_x), f_yx = d/dx(f_y)
Clairaut's theorem
If f_xy and f_yx are continuous on a domain D, then f_xy = f_yx throughout D.
Implicit differentiation shortcut (using F)
If F(x, y, z) = 0 defines z implicitly, then dz/dx = -F_x / F_z and dz/dy = -F_y / F_z (provided F_z is not zero).
Linearization formula
L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)
Second partial derivatives appear in the heat equation and wave equation in physics and engineering. The sign of f_xx tells you about concavity in the x-direction, which matters when designing surfaces (car bodies, aircraft wings) that must curve in specific ways.
Linearization is the basis of error estimation in lab science. When you measure x and y with small errors dx and dy, the linearization tells you approximately how much your computed f(x, y) will be off: df is roughly f_x dx + f_y dy. Engineers use this constantly to propagate measurement uncertainty.
Students sometimes think f_xy means "differentiate with respect to x and y at the same time." It does not. It means differentiate first with respect to x, then with respect to y, in sequence.
When doing implicit differentiation, a common error is forgetting the dz/dx or dz/dy factor when differentiating z-terms. Every z-term needs the chain rule applied.
Students often confuse the linearization with the tangent line. In multivariable calculus the linearization gives a tangent plane (a flat surface), not a line.
In the implicit differentiation shortcut dz/dx = -F_x / F_z, students sometimes put the wrong sign or swap numerator and denominator. The negative sign is part of the formula, not optional.
Second partial derivatives are tested directly (compute all four) and indirectly (they feed into the second derivative test for local extrema in the next topic block).
Implicit differentiation problems almost always appear on Math 126 midterms. Expect at least one equation mixing x, y, z with products or logs.
Linearization is a staple exam question: given a function and a point, write down L(x, y). Speed matters here, so practise the three-step process (evaluate f, compute and evaluate both partials, substitute) until it is automatic.
Clairaut's theorem is rarely tested as a standalone question, but it saves time: if you have already computed f_xy, you do not need to compute f_yx separately.
True or false: f_xy always equals f_yx for any function f(x, y). (Answer: False. It requires both mixed partials to be continuous.)
Fill in the blank: To find dz/dx from F(x, y, z) = 0 using the shortcut, you compute dz/dx = ______. (Answer: -F_x / F_z)
True or false: The linearization L(x, y) is a quadratic function. (Answer: False. It is linear in x and y.)
Fill in the blank: f_xx for f(x, y) = x^4 y - 2x^3 y^2 is ______. (Answer: 12x^3 y - 12xy^2)
True or false: When computing dz/dx via implicit differentiation, you treat y as a constant. (Answer: True.)
Q: Find all second partial derivatives of f(x, y) = x^4 y - 2x^3 y^2.
A: f_xx = 12x^3 y - 12xy^2, f_xy = f_yx = 4x^3 - 12x^2 y, f_yy = -4x^3.
Q: Use implicit differentiation to find dz/dx given x^2 - y^2 + z^2 - 2z = 4.
A: Differentiate with respect to x: 2x + (2z - 2)(dz/dx) = 0, so dz/dx = x / (1 - z).
Q: Find the linearization of f(x, y) = (1 + y) / (1 + x) at (1, 3).
A: f(1, 3) = 4/2 = 2. f_x = -(1 + y)/(1 + x)^2, so f_x(1, 3) = -1. f_y = 1/(1 + x), so f_y(1, 3) = 1/2. Therefore L(x, y) = 2 - (x - 1) + (1/2)(y - 3).
Q: Find all second partial derivatives of z = y / (2x + 3y).
A: z_xx = 8y / (2x + 3y)^3, z_xy = (6y - 4x) / (2x + 3y)^3, z_yy = -12x / (2x + 3y)^3.
Q: Find dz/dy given yz + x ln y = z^2.
A: Differentiate with respect to y: z + y(dz/dy) + x/y = 2z(dz/dy). Rearrange: dz/dy = (x + yz) / (y(2z - y)).
Second partial derivatives feed directly into the second derivative test for classifying critical points (covered in the next study notes on local and absolute extrema). The discriminant D = f_xx f_yy - (f_xy)^2 uses all three second partials.
Implicit differentiation reappears when working with Lagrange multipliers, where you differentiate constraint equations.
Linearization connects to differentials and error estimation (Section 14.4), and the idea generalises to Taylor polynomials in several variables.
partial derivative, second-order partial derivative, mixed partial, Clairaut's theorem, Schwarz's theorem, equality of mixed partials, implicit differentiation multivariable, implicit partial derivative, linearization multivariable, tangent plane approximation, local linear approximation, differential, total differential, error propagation, Math 126 UW, Stewart Calculus 14.1, 14.3, 14.4, 14.5