Difficulty: Intermediate | Prerequisites: Doc 1 (Vectors, Planes, Curves), single-variable differentiation, chain rule from Calculus I
This material covers what happens when you start differentiating functions of two or more variables. Partial derivatives, linearisation, and the chain rule are the core computational tools of multivariable calculus. You will also need to reason about when multivariable functions are (and are not) continuous, which is subtler than the single-variable case. Nearly everything in the second half of the course (optimisation, Lagrange multipliers, multiple integrals) builds on these ideas.
TL;DR: Partial derivatives measure how a function changes when you move in one coordinate direction while holding the others fixed. Linearisation uses those derivatives to approximate the function near a known point. The chain rule extends to multiple variables via a sum of products. Continuity in several variables requires the limit to be the same along every path, not just a few.
Partial derivative
The derivative of a multivariable function with respect to one variable, treating all other variables as constants. Notation: f_x, f_y, or df/dx, df/dy. In simple terms: how fast the output changes when you nudge one input, leaving the rest alone.
Linearisation (tangent plane approximation)
L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b). The best flat approximation to f near the point (a, b). Think of it as: the multivariable version of the tangent line from Calculus I.
Mixed partial derivative
A second derivative taken with respect to two different variables, e.g. f_xy = d/dy(df/dx). By Clairaut's theorem, f_xy = f_yx when both are continuous. In simple terms: how the rate of change in one direction itself changes as you move in a different direction.
Forward / backward / centre difference estimate
Numerical methods for approximating a derivative from discrete data. Forward: (f(x + h) - f(x)) / h. Backward: (f(x) - f(x - h)) / h. Centre: (f(x + h) - f(x - h)) / (2h). The centre difference is usually the most accurate. Think of it as: estimating slope from nearby data points when you do not have an explicit formula.
Implicit differentiation
Finding dz/dx or dz/dy when z is defined implicitly by F(x, y, z) = 0. The key formulas are dz/dx = -F_x / F_z and dz/dy = -F_y / F_z. In simple terms: using the chain rule on the whole equation rather than solving for z first.
Level set / level curve
The set of all (x, y) where f(x, y) = c for a constant c. For a function of two variables, these are curves in the xy-plane. Think of it as: a contour line on a map, where every point on the line has the same elevation.
Continuity at a point (multivariable)
A function f(x, y) is continuous at (a, b) if the limit of f as (x, y) approaches (a, b) exists and equals f(a, b). The limit must be the same along every possible path. In simple terms: you cannot sneak up on the point from different directions and get different answers.
When given a contour (level curve) plot, you estimate partial derivatives by measuring how far you must move in the x or y direction to reach the next contour.
Method:
Identify the function value at your point from the contour it sits on.
To estimate f_x: move horizontally (in the x-direction) to the nearest adjacent contour. Read off the change in function value (delta f) and the horizontal distance (delta x). Then f_x is approximately delta f / delta x.
To estimate f_y: do the same vertically.
Choose forward, backward, or centre differences based on which contours are closest and most clearly readable. The centre difference is usually best when the data is evenly spaced.
Worked example (Midterm Q3a): At the point (0, 3), the contour plot shows h(0, 3) = 16. Moving to the right (forward in x), the next contour (h = 12) is reached at roughly x = 1/4, giving h_x approximately (12 - 16) / (1/4) = -16. Moving downward (forward in negative y), the contour h = 12 appears at roughly y = 2.55, giving h_y approximately (12 - 16) / (2.55 - 3) = -8.9, approximately -9.
The linearisation is then h(x, y) approximately 16 + (-16)(x - 0) + (-9)(y - 3).
For (0.1, 2.95): h approximately 16 + (-16)(0.1) + (-9)(-0.05) = 16 - 1.6 + 0.45 = 14.85.
When given a table of function values at grid points, the centre difference is usually the best estimate, provided the grid is evenly spaced.
Centre difference formulas:
g_x(a, b) approximately [g(a + h, b) - g(a - h, b)] / (2h)
g_y(a, b) approximately [g(a, b + k) - g(a, b - k)] / (2k)
where h and k are the grid spacings in x and y.
Worked example (Midterm Q3b): Table values at (1, 2) with grid spacing 1 in both directions.
g_x(1, 2) = [g(2, 2) - g(0, 2)] / (2 - 0) = (23 - 27) / 2 = -2.
g_y(1, 2) = [g(1, 3) - g(1, 1)] / (3 - 1) = (28 - 22) / 2 = 3.
Linearisation: g(x, y) approximately 25 + (-2)(x - 1) + 3(y - 2).
For (1.1, 1.95): g approximately 25 + (-2)(0.1) + 3(-0.05) = 25 - 0.2 - 0.15 = 24.65.
Note: the table in this problem was constructed so that forward, backward, and centre differences all give the same result. This is not always the case; in general, prefer the centre difference.
The mixed partial g_xy means: first differentiate g with respect to x, then differentiate the result with respect to y. Numerically, you estimate g_x at several y-values and then take the difference quotient of those estimates in the y-direction.
Worked example (Midterm Q3c): Is g_xy(1, 1) positive, negative, or zero?
First estimate g_x at three y-values using the centre difference:
g_x(1, 0) approximately [g(2, 0) - g(0, 0)] / 2 = (18 - 16) / 2 = 1
g_x(1, 1) approximately [g(2, 1) - g(0, 1)] / 2 = (19 - 25) / 2 = -3
g_x(1, 2) approximately [g(2, 2) - g(0, 2)] / 2 = (23 - 27) / 2 = -2
Now estimate (g_x)_y at y = 1 using the centre difference on these values:
g_xy(1, 1) approximately [g_x(1, 2) - g_x(1, 0)] / (2 - 0) = (-2 - 1) / 2 = -3/2 < 0.
So g_xy(1, 1) is negative.
The linearisation of f(x, y) at (a, b) is:
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 z = f(x, y) at the point (a, b, f(a, b)). It provides the best linear approximation near that point.
The same formula works for implicitly defined functions. If F(x, y, z) = 0 defines z as a function of x and y near (x0, y0, z0), and you know dz/dx and dz/dy there, then:
z approximately z0 + (dz/dx)(x - x0) + (dz/dy)(y - y0).
Worked example (Midterm Q8b): F(x, y, z) = x^3 z^2 - z^3 xy = 0 near (1, 1, 1). From part (a), dz/dx = 2 and dz/dy = -1.
At x = 1.1, y = 0.95:
z approximately 1 + 2(0.1) + (-1)(-0.05) = 1 + 0.2 + 0.05 = 1.25.
The multivariable chain rule says that if a quantity depends on several intermediate variables, each of which depends on the independent variable(s), you sum the contributions along every path from input to output:
\frac{d\theta}{dt} = \frac{\partial \theta}{\partial h}\frac{dh}{dt} + \frac{\partial \theta}{\partial b}\frac{db}{dt}In related-rates problems, you are given the rates of change of some variables and asked to find the rate of change of another.
General strategy:
Draw and label the diagram. Assign variables to every changing quantity.
Write an equation relating the variable you want to the variables whose rates you know.
Differentiate both sides with respect to time using the chain rule.
Substitute the known values and rates at the specific instant asked about.
Worked example (Midterm Q6, summary): A triangle has fixed side l = 5, height h shrinking at 2 cm/s, and base b(t) = 1 + 2t^2. Find d(theta)/dt when h = 4 and b = 9.
The solution decomposes theta = theta_1 + theta_2 (the angles at the apex to either side of the altitude), then differentiates each using the chain rule and right-triangle identities. After careful algebra:
When b = 9, t = 2, so db/dt = 8.
x = 3 (the foot of the altitude splits the base), l2 = sqrt(52).
The final answer is d(theta)/dt = 17/13 rad/s.
The key lesson: related-rates problems in several variables are algebraically heavy, but the method is always the same. Write everything in terms of known quantities, differentiate, and substitute.
When a surface is defined implicitly by F(x, y, z) = 0 and you need the partial derivatives of z with respect to x and y:
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}These come from differentiating F(x, y, z(x, y)) = 0 with respect to x (or y) and solving for dz/dx (or dz/dy). The key is that dF/dx = 0 along the level set.
Worked example (Midterm Q8a): F(x, y, z) = x^3 z^2 - z^3 xy.
F_x = 3x^2 z^2 - z^3 y. At (1, 1, 1): F_x = 3 - 1 = 2.
F_y = -z^3 x. At (1, 1, 1): F_y = -1.
F_z = 2x^3 z - 3z^2 xy. At (1, 1, 1): F_z = 2 - 3 = -1.
Then dz/dx = -F_x / F_z = -2 / (-1) = 2, and dz/dy = -F_y / F_z = -(-1) / (-1) = -1.
When z = f(u) where u = x - y, the chain rule gives:
dz/dx = f'(u) * du/dx = f'(u) * 1 = f'(u).
dz/dy = f'(u) * du/dy = f'(u) * (-1) = -f'(u).
Adding these: dz/dx + dz/dy = f'(u) + (-f'(u)) = 0.
This holds for any differentiable function f. The statement is TRUE.
This is a useful pattern: whenever z = f(ax + by), you can find relationships between the partial derivatives purely from the chain rule, without knowing the explicit form of f.
For a function f(x, y), the level set f(x, y) = c is the collection of all points in the xy-plane where f takes the value c. Different values of c give different level curves.
Worked example (Midterm Q7a): f(x, y) = 4xy / (x^2 + y^2) for (x, y) not equal to (0, 0), and f(0, 0) = 0.
f = 0: xy = 0, so x = 0 or y = 0 (the coordinate axes).
f = 1: x^2 + y^2 = 4xy. Completing the square gives (x - (2 + sqrt(3))y)(x - (2 - sqrt(3))y) = 0, so the level set is two lines through the origin.
f = -1: similarly, two lines through the origin with slopes -(2 + sqrt(3)) and -(2 - sqrt(3)).
f = 2: x^2 + y^2 = 2xy, so (x - y)^2 = 0, meaning x = y (a single line).
f = -2: (x + y)^2 = 0, meaning x = -y (a single line).
All level curves are lines through the origin. This is a radial function: its value depends only on the direction from the origin, not the distance.
A function f(x, y) is continuous at (a, b) if the limit of f(x, y) as (x, y) approaches (a, b) exists and equals f(a, b). The critical difference from single-variable calculus: there are infinitely many paths to (a, b), and the limit must be the same along every single one.
To show a function is NOT continuous at a point, it suffices to find two paths that give different limiting values.
Worked example (Midterm Q7b): f(x, y) = 4xy / (x^2 + y^2), f(0, 0) = 0.
From the level-set analysis, along x = y: f(x, x) = 4x^2 / (2x^2) = 2. Along x = -y: f(x, -x) = -4x^2 / (2x^2) = -2. Along x = 0: f(0, y) = 0.
Three different paths give three different limits (2, -2, and 0). The limit as (x, y) approaches (0, 0) does not exist, so f is not continuous at the origin.
Students often think that checking two paths is enough to prove continuity. Two matching limits only fail to disprove continuity. Proving it requires showing the limit is the same along all paths, which usually means using the squeeze theorem or converting to polar coordinates.
Students confuse f_xy with f_xx or f_yy. The mixed partial f_xy means differentiate first with respect to x, then with respect to y.
When estimating derivatives from data, students sometimes use the forward difference even when a centre difference is available and more accurate.
In implicit differentiation, students forget the minus sign: dz/dx = -F_x / F_z, not +F_x / F_z.
⚠️ Estimating partial derivatives from contour plots and tables is a guaranteed exam question. Know all three difference methods and when to use each.
⚠️ Linearisation appears both as a standalone problem and as part (b) of implicit differentiation problems. The formula is the same in both cases.
⚠️ Related rates problems are long and algebraically demanding. Practice the method so you do not waste time figuring out what to do.
⚠️ Continuity questions often pair with level-set sketching. If all level curves pass through the origin at different heights, the function cannot be continuous there.
⚠️ Implicit differentiation with the formula dz/dx = -F_x / F_z is faster and cleaner than differentiating F = 0 from scratch each time. Memorise it.
True or false: the centre difference estimate uses the two points on either side of the target point. (Answer: True.)
Fill in the blank: for F(x, y, z) = 0, dz/dx = ___. (Answer: -F_x / F_z.)
True or false: if z = f(x - y), then dz/dx + dz/dy = 0 for any differentiable f. (Answer: True.)
True or false: if the limit of f(x, y) along two different paths to (0, 0) gives the same value, the function must be continuous there. (Answer: False; you need the limit to agree along all paths.)
Fill in the blank: the linearisation of f(x, y) at (a, b) is L(x, y) = f(a, b) + ___ + ___. (Answer: f_x(a, b)(x - a) + f_y(a, b)(y - b).)
Q: Given f(x, y) = x^2 y + 3xy^2, find f_x(1, 2) and f_y(1, 2).
A: f_x = 2xy + 3y^2 = 2(1)(2) + 3(4) = 16. f_y = x^2 + 6xy = 1 + 12 = 13.
Q: Use linearisation to estimate f(1.05, 1.98) given f(1, 2) = 10, f_x(1, 2) = 16, f_y(1, 2) = 13.
A: L = 10 + 16(0.05) + 13(-0.02) = 10 + 0.8 - 0.26 = 10.54.
Q: For F(x, y, z) = x^2 + y^2 + z^2 - 9 = 0, find dz/dx at (1, 2, 2).
A: F_x = 2x = 2, F_z = 2z = 4. dz/dx = -F_x / F_z = -2/4 = -1/2.
Q: Is g(x, y) = (x^2 - y^2) / (x^2 + y^2) continuous at the origin (with g(0, 0) = 0)?
A: Along x = 0: g(0, y) = -1. Along y = 0: g(x, 0) = 1. Different paths give different limits, so g is not continuous at (0, 0).
Q: If z = f(2x + 3y), what is the relationship between dz/dx and dz/dy?
A: dz/dx = 2f'(u), dz/dy = 3f'(u) where u = 2x + 3y. So 3(dz/dx) = 2(dz/dy), or equivalently 3(dz/dx) - 2(dz/dy) = 0.
Partial derivatives are the foundation of the gradient vector, which you will use extensively for directional derivatives, optimisation, and Lagrange multipliers.
Linearisation generalises to the total differential dz = f_x dx + f_y dy, which reappears in error analysis and in setting up integrals via change of variables.
The continuity analysis using paths connects to the epsilon-delta definition of limits in higher dimensions, and foreshadows the kinds of arguments needed when proving theorems in real analysis.
partial derivative, linearisation, tangent plane, linear approximation, chain rule, related rates, implicit differentiation, mixed partial, Clairaut's theorem, level curve, level set, contour plot, continuity, multivariable limit, centre difference, forward difference, backward difference, composed function, total differential, MATH 215, Calculus III, multivariable calculus