Multivariable calculus: practice problems with step-by-step solutions

Also called: calculus 3 · partial differentiation · double integrals · functions of several variables

Free Multivariable calculus practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: partial derivatives and double integrals, gradient, tangent plane and chain rule, critical points and integrals over regions.

Free practice problemEasy
xy3−3P

For f(x, y) = 3xy³ − x³ − 4y² + 2xy, find the partial derivative ∂f/∂x at the point (3, −3).

f(x, y) = 3xy³ − x³ − 4y² + 2xyP = (3, −3)
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What you practise

  1. Easypartial derivatives and double integrals
  2. Mediumgradient, tangent plane and chain rule
  3. Hardcritical points and integrals over regions

How to solve Multivariable calculus problems

  1. For a partial derivative, treat the other variable as a constant and use the one-variable rules.
  2. For gradient questions compute ∇f = (fx, fy) at the point; use |∇f| for the steepest slope and ∇f·u with a unit vector u for a directional derivative.
  3. For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
  4. For a double integral sketch the region, choose the order (or polar coordinates for discs) so the inner limits are the bounding curves, and integrate inside first.

Common mistakes

  • Differentiating the variable that should be held constant.
  • Forgetting to normalise the direction vector.
  • Writing D = fxx·fyy + fxy² or calling a point with D < 0 an extremum.
  • Using constant inner limits for a region bounded by curves.
  • Leaving out the factor r in dA = r dr dθ.

Key formulas

Partial derivatives
fx = ∂f/∂x (y constant), fxy = fyx
Gradient
∇f = (fx, fy), largest slope |∇f|
Directional derivative
Duf = ∇f·u, u = v/|v|
Tangent plane
z = f(a, b) + fx(x − a) + fy(y − b)
Chain rule
dz/dt = fx·dx/dt + fy·dy/dt
Second-derivative test
D = fxxfyy − fxy²: D > 0, fxx > 0 minimum; D > 0, fxx < 0 maximum; D < 0 saddle
Double integral
∬R f dA = ∫ab∫g(x)h(x) f dy dx
Polar coordinates
x = r cos θ, y = r sin θ, dA = r dr dθ

Learn Multivariable calculus from scratch

Before this: Derivatives, Integrals

1Partial derivatives

A function of two variables, z = f(x, y), is a surface above the xy-plane. On a map it is drawn with level curves, lines where f has the same value.

A partial derivative measures the slope in one direction. For ∂f/∂x (also written fx) you hold y constant and differentiate with respect to x, using the ordinary rules.

Example: f(x, y) = x²y + 3y². Treat y as a number for fx, and x as a number for fy:

fx = 2xy fy = x² + 6y

At (1, 2) this gives fx = 4 and fy = 1 + 12 = 13.

Differentiating again gives the second partials fxx, fyy and fxy. The order of the mixed derivative does not matter: fxy = fyx. Here fxy = 2x.

Self-check What is ∂f/∂y for f(x, y) = x³y²?
  1. 2x³y
  2. 3x²y²
  3. 6x²y
  4. x³y²

Answer: 2x³y. Hold x constant: x³ is a constant factor, and y² gives 2y, so fy = 2x³y. 3x²y² is ∂f/∂x.

2Gradient and directional derivative

The gradient collects both partial derivatives: ∇f = (fx, fy). It points in the direction in which f rises fastest, at right angles to the level curve.

xy∇fP

The steepest slope itself is the length |∇f| = √(fx² + fy²).

The slope in any other direction v is the directional derivative. First make v a unit vector u = v/|v|, then take the dot product:

Duf = ∇f·u

For f = x²y + 3y² at (1, 2): ∇f = (4, 13). In the direction v = (3, 4), u = (0.6, 0.8), so Duf = 4·0.6 + 13·0.8 = 12.8.

Self-check At a point ∇f = (3, 4). What is the largest rate of increase of f there?
  1. 12
  2. 25
  3. 5
  4. 7

Answer: 5. It is |∇f| = √(3² + 4²) = 5. Adding the components gives 7, and forgetting the square root gives 25.

3Tangent plane and chain rule

Near a point (a, b) the surface is almost flat. The tangent plane is

z = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b)

It gives the linear approximation f(a + Δx, b + Δy) ≈ f(a, b) + fxΔx + fyΔy.

Example: f = x²y + 3y² has f(1, 2) = 14, so near (1, 2): f(1.1, 1.9) ≈ 14 + 4·0.1 + 13·(−0.1) = 13.1. The exact value is 13.129.

When x and y both depend on t, the chain rule adds the two routes: dz/dt = fx·dx/dt + fy·dy/dt.

4Critical points

At a maximum, a minimum or a saddle point the surface is flat: fx = 0 and fy = 0. Such points are critical points.

The second-derivative test uses D = fxxfyy − fxy² at the point:

  • D > 0 and fxx > 0: local minimum.
  • D > 0 and fxx < 0: local maximum.
  • D < 0: saddle point, up in one direction and down in another.
  • D = 0: the test decides nothing.
Worked example

Find and classify the critical points of f(x, y) = x³ − 3x + y² − 4y.

1

Set both partials to zero

fx = 3x² − 3 = 0 → x = ±1 fy = 2y − 4 = 0 → y = 2

The critical points are (1, 2) and (−1, 2).

2

Second derivatives

fxx = 6x, fyy = 2, fxy = 0
3

Test (1, 2)

D = 6·2 − 0 = 12 > 0 and fxx = 6 > 0: a local minimum, with f(1, 2) = 1 − 3 + 4 − 8 = −6.

4

Test (−1, 2)

D = −6·2 − 0 = −12 < 0: a saddle point.

Answer

Local minimum at (1, 2) with f = −6; saddle point at (−1, 2).

5Double integrals

A double integral ∬R f dA adds up f over a region R; for f = 1 it is the area of R. Compute it as two single integrals, the inner one first.

Over a rectangle the limits are numbers: ∫02∫01 (x + 2y) dy dx. The inner integral gives x + 1, and ∫02 (x + 1) dx = 4.

Over a general region, the inner limits are the curves that bound it. For the region between y = x² and y = 2x, y runs from x² up to 2x, and x from 0 to 2.

xyy = 2xy = x²2
Area = ∫02 (2x − x²) dx = 4 − 8/3 = 4/3

For discs and sectors use polar coordinates: x² + y² = r² and dA = r dr dθ. Over the disc r ≤ 2: ∬(x² + y²) dA = ∫02π∫02 r³ dr dθ = 2π·4 = 8π.

Self-check In polar coordinates, what does dA become?
  1. dr dθ
  2. r dr dθ
  3. r² dr dθ
  4. θ dr dθ

Answer: r dr dθ. A small polar patch is Δr wide and r·Δθ long, so dA = r dr dθ. Forgetting the factor r is the most common mistake.

6Common mistakes and exam questions

  • Differentiating the "constant" variable as well in a partial derivative.
  • Using v instead of the unit vector u in a directional derivative.
  • Writing D = fxxfyy + fxy², or classifying a point with D < 0 as an extremum.
  • Using numbers as inner limits when the region is bounded by curves.
  • Forgetting the factor r in polar coordinates.

Easy: partial derivatives at a point and integrals over rectangles. Medium: gradient, directional derivative, tangent plane and chain rule. Hard: classifying critical points and integrals over curved regions, in polar coordinates or with the order reversed.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
xy−1−3P

For f(x, y) = 2x²y + 2x² + 4xy, find the partial derivative ∂f/∂x at the point (−1, −3).

f(x, y) = 2x²y + 2x² + 4xyP = (−1, −3)

Find fx(−1, −3)

Show hints
  1. Treat y as a constant number and differentiate with respect to x.
  2. ∂f/∂x = 4xy + 4x + 4y
  3. Substitute x = −1 and y = −3.
Show the answer and the step-by-step solution

Partial derivatives: ∂f/∂x is the ordinary derivative with respect to x while y is held constant (treat y as a number), and ∂f/∂y the other way round. The rules (power, product, chain) are the same as in one variable. Mixed second derivatives give the same result in either order: fxy = fyx. A double integral over a rectangle is computed as two single integrals, inside first: ∫∫ f dA = ∫(∫ f dy) dx.

Differentiate

y is held constant.

∂f/∂x = 4xy + 4x + 4y

Substitute

fx(−1, −3) = −4

Answer

fx(−1, −3) = −4
Example 2Medium

z = 2x²e3y with x = −t − 2 and y = t² + 2. Use the chain rule to find dz/dt at t = 0.

z = 2x²e3yx = −t − 2y = t² + 2t = 0

Find dz/dt

Show hints
  1. dz/dt = fx·dx/dt + fy·dy/dt.
  2. At t = 0: x = −2, y = 2, dx/dt = −1, dy/dt = 0.
  3. fx = 4xe3y, fy = 6x²e3y
Show the answer and the step-by-step solution

Gradient: ∇f = (fx, fy) points in the direction of steepest ascent, and |∇f| is the largest rate of change. The directional derivative along a direction v is Duf = ∇f·u with the unit vector u = v/|v|. The tangent plane at (a, b) is z = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b); it gives the linear approximation of f near that point. Chain rule: dz/dt = fx·dx/dt + fy·dy/dt.

Point and rates

x(0) = −2, y(0) = 2
dx/dt = −1, dy/dt = 2t = 0

Partial derivatives

fx = 4xe3y = −3227
fy = 6x²e3y = 9682

Chain rule

dz/dt = (−3227)·(−1) + 9682·0 = 3227

Answer

dz/dt = 3227
Example 3Hard
xy02−1012(0, 0)(1, 1)

f(x, y) = x³ + y³ − 3xy has a critical point at (1, 1). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x³ + y³ − 3xyP = (1, 1)

Find D

Show hints
  1. Find the three second derivatives fxx, fyy and fxy.
  2. At the point: fxx = 6, fyy = 6, fxy = −3.
  3. D = fxxfyy − fxy².
Show the answer and the step-by-step solution

Critical points and double integrals: a critical point has fx = fy = 0. Second-derivative test with D = fxxfyy − fxy²: D > 0 and fxx > 0 is a local minimum, D > 0 and fxx < 0 a local maximum, D < 0 a saddle point, and D = 0 decides nothing. For a double integral over a general region, let the inner limits be the curves that bound the region; for discs and sectors use polar coordinates, x = r cos θ, y = r sin θ and dA = r dr dθ (do not forget the factor r).

Second derivatives

fxx = 6, fyy = 6, fxy = −3

Discriminant

D = 6·6 − (−3)² = 27
→ local minimum

Answer

D = 27

Questions and answers

What is Multivariable calculus?

A function of two variables, z = f(x, y), is a surface above the xy-plane. On a map it is drawn with level curves, lines where f has the same value.

A partial derivative measures the slope in one direction. For ∂f/∂x (also written fx) you hold y constant and differentiate with respect to x, using the ordinary rules.

How do I solve Multivariable calculus problems step by step?
  1. For a partial derivative, treat the other variable as a constant and use the one-variable rules.
  2. For gradient questions compute ∇f = (fx, fy) at the point; use |∇f| for the steepest slope and ∇f·u with a unit vector u for a directional derivative.
  3. For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
  4. For a double integral sketch the region, choose the order (or polar coordinates for discs) so the inner limits are the bounding curves, and integrate inside first.
What are common mistakes in Multivariable calculus?
  • Differentiating the variable that should be held constant.
  • Forgetting to normalise the direction vector.
  • Writing D = fxx·fyy + fxy² or calling a point with D < 0 an extremum.
  • Using constant inner limits for a region bounded by curves.
  • Leaving out the factor r in dA = r dr dθ.
What do the Multivariable calculus exercises cover?
  • Easy: partial derivatives and double integrals
  • Medium: gradient, tangent plane and chain rule
  • Hard: critical points and integrals over regions

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Engineering Mathematics 2Calculus 2

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Printable worksheet

Multivariable calculus worksheet with answers
Practise Multivariable calculus for freePractise Multivariable calculus now5 free problems every day. No card needed.