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Partial derivatives and stationary points

Differentiate surfaces partially, use gradients, tangent planes and linear estimates, and classify maxima, minima and saddle points.

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  • GCE A-Level H2 Further Mathematics 2027
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Enlarged diagram

The height of a hillside depends on two coordinates, and so does the cost of a box with two adjustable sides. A function z = f(x,y) is a surface above the xy-plane, and the calculus of one variable extends to it one direction at a time. This lesson shows how to differentiate with one variable held fixed, how the two partial derivatives combine into a gradient, tangent plane and linear approximation, and how second derivatives classify the flat points of a surface as maxima, minima or saddles.

Differentiate with one variable fixed

The partial derivative fₓ = (∂ f)/(∂ x) is the gradient of the surface in the x-direction: differentiate with respect to x, treating y as a constant. Likewise fy treats x as a constant. Differentiating again gives the second-order partial derivatives fₓₓ, f_yy and the mixed derivatives f_xy and f_yx, which are equal for every function in this course.

For f(x,y) = x²y + y³,

fₓ = 2xy, fy = x² + 3y²,; fₓₓ = 2y, f_yy = 6y, f_xy = 2x.

All the one-variable rules still apply. In x²e^y, the factor e^y is a constant when you differentiate with respect to x; in sin(xy), the chain rule gives fₓ = y cos(xy).

Two directions through one surface point

The solid section is f(1 + s, 1) = (1 + s)² + 1. The dashed section is f(1, 1 + s) = 1 + s + (1 + s)³. Both have height 2 at s = 0, with different slopes there.

The solid section is f(1 + s, 1) = (1 + s)² + 1. The dashed section is f(1, 1 + s) = 1 + s + (1 + s)³. Both have height 2 at s = 0, with different slopes there.The solid section is f(1 + s, 1) = (1 + s)² + 1. The dashed section is f(1, 1 + s) = 1 + s + (1 + s)³. Both have height 2 at s = 0, with different slopes there.
For f(x, y) = x²y + y³, both sections pass through (1, 1, 2). The horizontal input s is displacement from that point. Holding y = 1 gives initial slope 2; holding x = 1 gives initial slope 4.
Open full-size graphOpen full-size graph
What to noticeThese are two vertical sections, not a perspective drawing of the whole surface.
Figure data

This table is included in print while it is open.

Figure data: Two directions through one surface point
Displacement sVary x; y = 1Vary y; x = 1
-0.5001.2500000.6250
-0.4751.2756250.6697
-0.4501.3025000.7164
-0.4251.3306250.7651
-0.4001.3600000.8160
-0.3751.3906250.8691
-0.3501.4225000.9246
-0.3251.4556250.9825
-0.3001.4900001.0430
-0.2751.5256251.1061
-0.2501.5625001.1719
-0.2251.6006251.2405
-0.2001.6400001.3120
-0.1751.6806251.3865
-0.1501.7225001.4641
-0.1251.7656251.5449
-0.1001.8100001.6290
-0.0751.8556251.7165
-0.0501.9025001.8074
-0.0251.9506251.9019
0.0002.0000002.0000
0.0252.0506252.1019
0.0502.1025002.2076
0.0752.1556252.3173
0.1002.2100002.4310
0.1252.2656252.5488
0.1502.3225002.6709
0.1752.3806252.7972
0.2002.4400002.9280
0.2252.5006253.0633
0.2502.5625003.2031
0.2752.6256253.3477
0.3002.6900003.4970
0.3252.7556253.6512
0.3502.8225003.8104
0.3752.8906253.9746
0.4002.9600004.1440
0.4253.0306254.3186
0.4503.1025004.4986
0.4753.1756254.6840
0.5003.2500004.8750

Check your understanding 1

Hold the other variable still

For f(x,y) = x²e^y + 3xy², find fₓ, fy and f_xy, and evaluate each at (1,0).

Show answerHide answer
fₓ = 2xe^y + 3y², fₓ(1,0) = 2,; fy = x²e^y + 6xy, fy(1,0) = 1,; f_xy = 2xe^y + 6y, f_xy(1,0) = 2.

Differentiating fy with respect to x gives the same 2xe^y + 6y.

Gradient, tangent plane and linear approximation

The two partial derivatives form the gradient vector, which points in the direction of steepest increase. The rate of change in the direction of a unit vector u is the directional derivative:

∇ f = (fₓ; fy), Dᵤf = ∇ f · u.

Since ∇ f · u = |∇ f| cos θ, the directional derivative is greatest, equal to |∇ f|, when u points along ∇ f, and zero along a contour, perpendicular to ∇ f.

Close to a point (a,b), a smooth surface is almost flat. Its tangent plane there is

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

and using this plane to estimate nearby values is local linearisation. A normal vector to the surface z = f(x,y) at the point is (fₓ; fy; -1).

Worked example 1

Tangent plane and normal line to a dome

Find the equation of the tangent plane and a vector equation of the normal line to the surface z = 4-x²-2y² at the point (1,1,1).

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  1. Evaluate the partial derivatives

    Method

    fₓ = -2x and fy = -4y, so fₓ(1,1) = -2 and fy(1,1) = -4.

    Reason

    These are the slopes of the plane in the x- and y-directions. First check that the point lies on the surface: 4-1-2 = 1.

    Working

    Answer

    z = 1-2(x-1)-4(y-1),; 2x + 4y + z = 7.

  2. Use the normal vector

    Working

    Answer

    A normal is (fₓ,fy,-1) = (-2,-4,-1), or equivalently (2,4,1), which matches the coefficients of the plane. The normal line is

    r = (1; 1; 1) + t (2; 4; 1).

Check your understanding 2

Rates and estimates at a point

For f(x,y) = x²y + y³ at (1,2): find ∇ f, the directional derivative in the direction of (3,4), and the greatest rate of increase. Use the tangent plane to estimate f(1.1,1.9) and compare with the exact value.

Show answerHide answer

∇ f(1,2) = (4,13). The direction (3,4) has length 5, so u = (1/5)(3,4) and

Dᵤf = (1/5)(4 × 3 + 13 × 4) = 64/5.

The greatest rate of increase is |∇ f| = square root of (4² + 13²) = square root of 185 ≈ 13.6.

Since f(1,2) = 10, the tangent plane gives

f(1.1,1.9) ≈ 10 + 4(0.1) + 13(-0.1) = 9.1.

The exact value is 1.21 × 1.9 + 1.9³ = 9.158; the small difference comes from the curvature that the plane ignores.

Stationary points and the second-derivative test

At a stationary point both fₓ = 0 and fy = 0, so the tangent plane is horizontal. Unlike a curve, a surface can be flat at a point that is neither a maximum nor a minimum: at a saddle point it rises in some directions and falls in others. Classify a stationary point with

D = fₓₓf_yy-f_xy²,

evaluated at the point:

Condition at the stationary pointNature
D > 0 and fₓₓ > 0local minimum
D > 0 and fₓₓ < 0local maximum
D < 0saddle point
D = 0test inconclusive

The test comes from the quadratic approximation near a stationary point (a,b), where the linear terms vanish:

f(a + h,b + k) ≈ f(a,b) + (1/2)(fₓₓh² + 2f_xyhk + f_yyk²).

Completing the square shows that the bracket keeps one sign for every small (h,k) ≠ (0,0) exactly when D > 0, with the sign of fₓₓ; when D < 0 it takes both signs.

Worked example 2

A minimum and a saddle

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

Show full solutionHide solution

  1. Solve both equations

    Method

    fₓ = 3x²-3 = 0 and fy = 2y-4 = 0.

    Reason

    A stationary point needs both partial derivatives zero at once.

    Working

    Answer

    x = ±1 and y = 2, so the stationary points are (1,2) and (-1,2).

  2. Apply the test

    Method

    fₓₓ = 6x, f_yy = 2, f_xy = 0, so D = 12x.

    Reason

    The sign of D separates extrema from saddles.

    Working

    Answer

    At (1,2), D = 12 > 0 and fₓₓ = 6 > 0: a local minimum, with f(1,2) = -6.

    At (-1,2), D = -12 < 0: a saddle point, with f(-1,2) = -2.

Two sections through the saddle point (−1, 2)

The horizontal axis is the distance s moved from (−1, 2). The solid curve f(−1 + s, 2) = s³ − 3s² − 2 rises to −2 at s = 0 and falls on both sides near it, reaching −6 at s = −1 and −4 at s = 1. The dashed curve f(−1, 2 + s) = s² − 2 dips to −2 at s = 0 and rises to −1 at s = ±1. The two curves touch at (0, −2).

The horizontal axis is the distance s moved from (−1, 2). The solid curve f(−1 + s, 2) = s³ − 3s² − 2 rises to −2 at s = 0 and falls on both sides near it, reaching −6 at s = −1 and −4 at s = 1. The dashed curve f(−1, 2 + s) = s² − 2 dips to −2 at s = 0 and rises to −1 at s = ±1. The two curves touch at (0, −2).The horizontal axis is the distance s moved from (−1, 2). The solid curve f(−1 + s, 2) = s³ − 3s² − 2 rises to −2 at s = 0 and falls on both sides near it, reaching −6 at s = −1 and −4 at s = 1. The dashed curve f(−1, 2 + s) = s² − 2 dips to −2 at s = 0 and rises to −1 at s = ±1. The two curves touch at (0, −2).
f(x, y) = x³ − 3x + y² − 4y, cut through its stationary point (−1, 2) in two directions. Moving parallel to the x-axis, f has a local maximum at the point; moving parallel to the y-axis, f has a local minimum there. Both sections have zero gradient at s = 0.
Open full-size graphOpen full-size graph
What to noticeEach curve is a vertical slice of the surface along one direction; the surface itself is not drawn. Curves are drawn from sampled values joined by short straight segments.
Figure data

This table is included in print while it is open.

Figure data: Two sections through the saddle point (−1, 2)
Distance moved, sAlong x: f(−1 + s, 2)Along y: f(−1, 2 + s)
-1.00-6.000000-1.0000
-0.95-5.564875-1.0975
-0.90-5.159000-1.1900
-0.85-4.781625-1.2775
-0.80-4.432000-1.3600
-0.75-4.109375-1.4375
-0.70-3.813000-1.5100
-0.65-3.542125-1.5775
-0.60-3.296000-1.6400
-0.55-3.073875-1.6975
-0.50-2.875000-1.7500
-0.45-2.698625-1.7975
-0.40-2.544000-1.8400
-0.35-2.410375-1.8775
-0.30-2.297000-1.9100
-0.25-2.203125-1.9375
-0.20-2.128000-1.9600
-0.15-2.070875-1.9775
-0.10-2.031000-1.9900
-0.05-2.007625-1.9975
0.00-2.000000-2.0000
0.05-2.007375-1.9975
0.10-2.029000-1.9900
0.15-2.064125-1.9775
0.20-2.112000-1.9600
0.25-2.171875-1.9375
0.30-2.243000-1.9100
0.35-2.324625-1.8775
0.40-2.416000-1.8400
0.45-2.516375-1.7975
0.50-2.625000-1.7500
0.55-2.741125-1.6975
0.60-2.864000-1.6400
0.65-2.992875-1.5775
0.70-3.127000-1.5100
0.75-3.265625-1.4375
0.80-3.408000-1.3600
0.85-3.553375-1.2775
0.90-3.701000-1.1900
0.95-3.850125-1.0975
1.00-4.000000-1.0000

Spot the mistake 1

Positive in both directions, still a saddle

About 5 min

A student says that f(x,y) = x² + 3xy + y² has a minimum at the origin because fₓₓ = 2 > 0 and f_yy = 2 > 0. Explain the error.

Write your answer, then compare it with the solution

Show full solutionHide solution

  1. Include the mixed derivative

    Method

    f_xy = 3, so D = 2 × 2-3² = -5 < 0.

    Reason

    fₓₓ and f_yy only describe the two axis directions.

    Working

    Answer

    The origin is a saddle point. Along the line y = -x, f = x²-3x² + x² = -x² < 0 for x ≠ 0, while along the axes f > 0.

Common mistakes

  • Treating the other variable as zero instead of as a constant: in (∂/(∂ x))(x²y), the answer is 2xy, not 2x or 0.
  • Using a non-unit direction vector in Dᵤf = ∇ f · u.
  • Solving fₓ = 0 and fy = 0 separately and combining every x with every y, instead of solving them simultaneously.
  • Classifying with fₓₓ and f_yy alone and ignoring f_xy.
  • Concluding “minimum” or “maximum” when D = 0. The test gives no information there.

Before you move on

Try it yourself 1

Two stationary points linked together

Minimal support

Find and classify the stationary points of f(x,y) = x³ + y³-3xy.

Try this before viewing the solution

Nature of the stationary point (0, 0)
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  1. Solve the equations together

    Method

    fₓ = 3x²-3y = 0 and fy = 3y²-3x = 0 give y = x² and x = y².

    Reason

    Each equation contains both variables, so substitute one into the other.

    Working

    Answer

    x = (x²)² = x⁴,; x(x³-1) = 0,

    so the real solutions are x = 0 and x = 1, giving the points (0,0) and (1,1).

  2. Classify both points

    Working

    Answer

    fₓₓ = 6x, f_yy = 6y and f_xy = -3, so D = 36xy-9.

    At (0,0), D = -9 < 0: a saddle point.

    At (1,1), D = 27 > 0 and fₓₓ = 6 > 0: a local minimum, with f(1,1) = 1 + 1-3 = -1.

Questions

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