For f(x, y) = x⁴ + y⁴ at the origin, which statements are true?

The second-derivative test (rt − s²) is inconclusive there

(0, 0) is a strict local minimum

The second-derivative test (rt − s²) is inconclusive there

(0, 0) is a strict local minimum

For f(x, y) = x² + 4xy + y², which statements are true?

The Hessian at (0, 0) is indefinite

(0, 0) is a local minimum because f_xx > 0 and f_yy > 0

Functions of Several Variables: Limits, Directional Derivatives, Taylor’s Theorem, Extrema, Lagrange Multipliers and Multiple Integrals

The second paragraph of Section 1 of the GATE Statistics paper is several-variable calculus: limit, continuity, partial and directional derivatives, the gradient, Taylor’s theorem, the total derivative, maxima, minima and saddle points, the method of Lagrange multipliers, and double and triple integrals with their applications. The shared multivariable chapter teaches partial derivatives, the second-derivative test and the evaluation of multiple integrals; this chapter adds the parts a statistician needs beyond it — why partial derivatives can exist where a function is not even continuous, the directional derivative as ∇f · u, Taylor’s theorem and the Hessian, constrained optimisation by Lagrange multipliers (the method behind maximising a likelihood under a constraint), and the Jacobian change of variables that turns a double integral into a probability.

1. Limits and continuity in several variables

lim_{(x,y)→(a,b)} f = L means |f − L| < ε whenever 0 < ‖(x, y) − (a, b)‖ < δ — along every path at once. So one path giving a different value, or a family of paths giving values that depend on the path, proves the limit does not exist. For f = xy/(x² + y²), the line y = mx gives m/(1 + m²), which changes with m: no limit at the origin. For f = x²y/(x² + y²), polar coordinates give |f| = r|cos²θ sin θ| ≤ r → 0, and the limit is 0.

⚠️ Straight lines are not enough

Agreement along every straight line does not prove a limit. f = x²y/(x⁴ + y²) tends to 0 along every line y = mx, but along the parabola y = x² it equals 1/2. To prove a limit exists, bound |f − L| by something that depends only on the distance r, as the polar argument does.

2. Partial derivatives, directional derivatives, the gradient and the total derivative

The directional derivative of f at a in the direction of a unit vector u is D_u f(a) = lim_h→0 [f(a + hu) − f(a)]/h. When f is differentiable, D_u f = ∇f · u, where ∇f = (f_x, f_y, f_z) is the gradient. It follows that f increases fastest in the direction of ∇f, at rate ‖∇f‖, and not at all in directions perpendicular to it — so ∇f is normal to the level surfaces f = c. For f = x²y + yz³ at (1, 1, 1), ∇f = (2xy, x² + z³, 3yz²) = (2, 2, 3); in the direction (1, 2, 2)/3 the derivative is (2 + 4 + 6)/3 = 4.

f is differentiable at a if f(a + h) = f(a) + L(h) + o(‖h‖) for a linear map L — the total derivative, whose matrix is the Jacobian of partial derivatives (for a scalar f, the row ∇f). Differentiable ⇒ continuous and all directional derivatives exist. Continuous partial derivatives ⇒ differentiable (a sufficient condition). The converse chain fails at every step: f = xy/(x² + y²), f(0, 0) = 0 has f_x(0, 0) = f_y(0, 0) = 0, because f vanishes on both axes, yet it is not continuous at the origin, let alone differentiable.

What implies what at a point
Statement	Implies	Does not imply
Partials continuous near a	differentiable at a	—
Differentiable at a	continuous; D_u f = ∇f · u for all u	continuous partials
Partials exist at a	nothing about other directions	continuity, differentiability

3. Taylor’s theorem, the Hessian, and maxima, minima and saddle points

Taylor’s theorem in several variables, to second order: f(a + h) = f(a) + ∇f(a) · h + ½ hᵀ H(a) h + o(‖h‖²), where H = [∂²f/∂xᵢ∂xⱼ] is the Hessian, symmetric when the second partials are continuous. For e^x cos y about (0, 0): f = 1, ∇f = (1, 0), H = [[1, 0], [0, −1]], so f ≈ 1 + x + (x² − y²)/2.

At a critical point ∇f = 0 the quadratic term decides: H positive definite ⇒ strict local minimum; negative definite ⇒ strict local maximum; indefinite (eigenvalues of both signs) ⇒ saddle point; semidefinite ⇒ the test is silent. In two variables with r = f_xx, s = f_xy, t = f_yy: rt − s² > 0 and r > 0 is a minimum, rt − s² > 0 and r < 0 a maximum, rt − s² < 0 a saddle. For f = x³ + y³ − 3xy, ∇f = 0 gives (0, 0) and (1, 1); at (0, 0) rt − s² = −9, a saddle; at (1, 1) it is 36 − 9 = 27 with r = 6, a local minimum of value −1.

⚠️ A failed test is not "neither"

x⁴ + y⁴ and x⁴ − y⁴ both have a zero Hessian at the origin, so rt − s² = 0 for both — yet the first has a strict minimum there and the second a saddle. When the Hessian is singular, look at the function itself along lines through the point.

4. The method of Lagrange multipliers

To find extrema of f subject to g = c, look for points where the level curve of f is tangent to the constraint: ∇f = λ∇g, together with g = c. With several constraints, ∇f = Σλᵢ∇gᵢ. The multiplier λ is the rate at which the optimal value changes with c. Maximising xyz subject to x + y + z = 12 (positive variables): yz = xz = xy = λ forces x = y = z = 4 and a maximum of 64. Minimising x² + y² + z² subject to x + 2y + 2z = 6: (2x, 2y, 2z) = λ(1, 2, 2) gives (x, y, z) = (2/3, 4/3, 4/3) and the minimum 4, the squared distance 6²/(1 + 4 + 4) from the origin to the plane.

ℹ️ The likelihood under a constraint

In a multinomial with counts n₁, …, n_k, maximise Σnᵢ ln pᵢ subject to Σpᵢ = 1: nᵢ/pᵢ = λ for every i, so pᵢ ∝ nᵢ and λ = n, giving p̂ᵢ = nᵢ/n. The same calculation, with a second constraint, gives the maximum-entropy distributions.

5. Double and triple integrals, the Jacobian, and applications

Fubini: for an integrable f on a region, the double integral equals either iterated integral, so the order may be changed — and changing it can make an impossible inner integral possible. ∫₀¹∫ₓ¹ e^y² dy dx has no elementary inner integral; over the same triangle 0 ≤ x ≤ y ≤ 1 in the other order it is ∫₀¹ y e^y² dy = (e − 1)/2 ≈ 0.859.

Change of variables: if (x, y) = T(u, v) is one-to-one and smooth, dx dy = |J| du dv with J = ∂(x, y)/∂(u, v). Polar: |J| = r; cylindrical: r; spherical: ρ² sin φ. The polar change evaluates ∫∫_ℝ² e^−(x²+y²) dx dy = ∫₀^2π∫₀^∞ e^−r² r dr dθ = π, so ∫_ℝ e^−x² dx = √π — the normalising constant of the normal density.

Area and volume: area = ∫∫_R dA; volume under z = f(x, y) = ∫∫_R f dA; volume of a solid = ∫∫∫ dV.
Mass and centroid: with density ρ, mass = ∫∫ρ dA and x̄ = ∫∫xρ dA / mass.
Probability: for a joint density f(x, y), P((X, Y) ∈ A) = ∫∫_A f. With f = x + y on the unit square, P(X ≤ 1/2) = ∫₀^1/2(x + 1/2) dx = 1/8 + 1/4 = 3/8.

Key takeaways

A two-variable limit must be the same along every path; two paths that disagree disprove it, and a bound depending only on r proves it.
D_u f = ∇f · u for a unit vector u; the maximum rate is ‖∇f‖ in the direction of ∇f.
Continuous partials ⇒ differentiable ⇒ continuous; mere existence of partials implies neither (xy/(x² + y²)).
At ∇f = 0: Hessian PD → minimum, ND → maximum, indefinite → saddle, singular → the test is silent.
Lagrange: ∇f = λ∇g with g = c; the multinomial MLE p̂ᵢ = nᵢ/n is this method.
dx dy = |J| du dv (polar r, spherical ρ² sin φ); the polar change gives ∫e^−x² = √π.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For f(x, y) = xy/(x² + y²), the limit of f as (x, y) → (0, 0):
1. is 0
2. is 1/2
3. does not exist
4. is 1
Show answer
Answer: C — does not exist
Along y = mx, f = m/(1 + m²): 0 along the x-axis (m = 0) and 1/2 along y = x. Two paths give different values, so there is no limit. Answering 0 checks only the axes, where f is identically zero.
The directional derivative of f(x, y, z) = x²y + yz³ at (1, 1, 1) in the direction of the vector (1, 2, 2) is ____.
Numerical answer — type the value.
Show answer
Answer: 4
∇f = (2xy, x² + z³, 3yz²) = (2, 2, 3) at (1, 1, 1). The unit vector is (1, 2, 2)/3, so D_u f = (2 + 4 + 6)/3 = 4. Forgetting to normalise gives 12, three times too large.
The maximum rate of change of f(x, y) = x² + y² at the point (3, 4) is ____.
Numerical answer — type the value.
Show answer
Answer: 10
The maximum directional derivative is ‖∇f‖ = ‖(2x, 2y)‖ = ‖(6, 8)‖ = 10, attained in the direction (3, 4)/5. Giving 5, the distance to the origin, confuses the gradient with the position vector.
Let f(x, y) = xy/(x² + y²) for (x, y) ≠ (0, 0) and f(0, 0) = 0. Which statements are true?
1. f_x(0, 0) exists and equals 0
2. f is continuous at (0, 0)
3. f is differentiable at (0, 0)
4. f_y(0, 0) exists and equals 0
Show answer
Answer: A — f_x(0, 0) exists and equals 0; D — f_y(0, 0) exists and equals 0
(A), (D) f(h, 0) = 0 and f(0, h) = 0 for all h, so both difference quotients are 0. (B) Along y = x, f = 1/2 ≠ f(0, 0): not continuous. (C) Differentiability implies continuity, so it fails too. This is the standard counterexample: partial derivatives exist, yet the function is not even continuous.
The value of f(x, y) = x³ + y³ − 3xy at its local minimum is ____.
Numerical answer — type the value.
Show answer
Answer: -1
f_x = 3x² − 3y = 0 and f_y = 3y² − 3x = 0 give (0, 0) and (1, 1). At (1, 1): r = 6x = 6, t = 6, s = −3, rt − s² = 27 > 0 with r > 0, a minimum, and f = 1 + 1 − 3 = −1. (0, 0) has rt − s² = −9 < 0 — a saddle, so the value 0 there is not the answer. Type the answer as -1.
For f(x, y) = x⁴ + y⁴ at the origin, which statements are true?
1. (0, 0) is a critical point
2. The second-derivative test (rt − s²) is inconclusive there
3. (0, 0) is a strict local minimum
4. (0, 0) is a saddle point
Show answer
Answer: A — (0, 0) is a critical point; B — The second-derivative test (rt − s²) is inconclusive there; C — (0, 0) is a strict local minimum
∇f = (4x³, 4y³) = 0 at the origin. All second partials vanish there, so rt − s² = 0 and the test says nothing. But f > 0 = f(0, 0) for every other point, so the origin is a strict minimum, not a saddle. Compare x⁴ − y⁴, which has the same zero Hessian and is a saddle.
The maximum value of xyz for positive x, y, z subject to x + y + z = 12 is ____.
Numerical answer — type the value.
Show answer
Answer: 64
∇(xyz) = λ∇(x + y + z) gives yz = xz = xy = λ, so x = y = z = 4 and xyz = 64. The AM–GM inequality confirms it: xyz ≤ ((x + y + z)/3)³ = 64. Answering 48 (= 4 × 12) multiplies the wrong quantities.
The minimum value of x² + y² + z² subject to x + 2y + 2z = 6 is ____.
Numerical answer — type the value.
Show answer
Answer: 4
(2x, 2y, 2z) = λ(1, 2, 2) gives x = λ/2, y = z = λ; the constraint gives λ/2 + 4λ = 6, λ = 4/3, so (x, y, z) = (2/3, 4/3, 4/3) and x² + y² + z² = 4/9 + 16/9 + 16/9 = 4. Geometrically it is the squared distance 6²/9 = 4 to the plane; answering 2 gives the distance, not its square.
The value of ∫∫ e^{−(x² + y²)} dx dy over the unit disc x² + y² ≤ 1, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.99
In polar coordinates: ∫₀^2π∫₀¹ e^−r² r dr dθ = 2π × [−e^−r²/2]₀¹ = π(1 − e⁻¹) = 3.1416 × 0.6321 = 1.986 → 1.99. Omitting the Jacobian r leaves ∫₀¹e^−r² dr, which has no elementary antiderivative — the r is what makes the integral doable.
The value of ∫₀¹ ∫ₓ¹ e^y² dy dx, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.86
The region is 0 ≤ x ≤ y ≤ 1. Reversing the order: ∫₀¹ ∫₀^y e^y² dx dy = ∫₀¹ y e^y² dy = (e − 1)/2 = 0.859 → 0.86. Keeping the given order stalls on ∫e^y² dy, which is the signal to swap.
Under x = r cos θ, y = r sin θ, the element dx dy becomes:
1. r dr dθ
2. dr dθ
3. r² dr dθ
4. r sin θ dr dθ
Show answer
Answer: A — r dr dθ
∂(x, y)/∂(r, θ) = det[[cos θ, −r sin θ], [sin θ, r cos θ]] = r cos²θ + r sin²θ = r. The factor r² sin φ belongs to spherical coordinates in three dimensions; dr dθ alone forgets the Jacobian entirely.
(X, Y) has joint density f(x, y) = x + y on the unit square 0 < x, y < 1. P(X ≤ 1/2), correct to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.375
Integrate y out first: ∫₀¹(x + y) dy = x + 1/2, the marginal density of X. Then ∫₀^1/2(x + 1/2) dx = 1/8 + 1/4 = 3/8 = 0.375. Answering 0.5 treats X as uniform; its marginal density x + 1/2 puts more mass near 1.
The second-order Taylor polynomial of f(x, y) = eˣ cos y about (0, 0) is:
1. 1 + x + (x² − y²)/2
2. 1 + x + (x² + y²)/2
3. 1 + x − y + (x² − y²)/2
4. 1 + x + x² − y²
Show answer
Answer: A — 1 + x + (x² − y²)/2
f(0, 0) = 1, f_x = 1, f_y = −eˣ sin y = 0, f_xx = 1, f_xy = 0, f_yy = −1. So f ≈ 1 + x + ½(x² − y²). The "+y²" option gets the sign of f_yy wrong, and dropping the ½ forgets that the quadratic term is ½hᵀHh.
For f(x, y) = x² + 4xy + y², which statements are true?
1. (0, 0) is a critical point
2. The Hessian at (0, 0) is indefinite
3. (0, 0) is a saddle point
4. (0, 0) is a local minimum because f_xx > 0 and f_yy > 0
Show answer
Answer: A — (0, 0) is a critical point; B — The Hessian at (0, 0) is indefinite; C — (0, 0) is a saddle point
∇f = (2x + 4y, 4x + 2y) = 0 only at the origin. H = [[2, 4], [4, 2]] has determinant 4 − 16 = −12 < 0, eigenvalues 6 and −2: indefinite, so a saddle — along y = −x, f = −2x² < 0. Positive diagonal entries do not make a matrix positive definite; the cross term dominates.