Difficulty: Intermediate–Advanced | Prerequisites: Derivatives and Partial Derivatives study notes (Ch. 1), basic vector operations.
The multivariable chain rule lets you differentiate compositions of functions by multiplying their Jacobian matrices. Directional derivatives extend partial derivatives to any direction, not just along the coordinate axes, using the dot product of the gradient with a unit vector. Parameterised paths describe curves and motion in space, and the chain rule connects derivatives along those paths back to the partial derivatives of the underlying function.
Differentiable (multivariable)
A function f is differentiable at a point if the tangent plane provides a good local approximation, meaning the error between f and its linear approximation tends to zero faster than the distance to the point.
Think of it as: when you zoom in far enough, the surface looks perfectly flat. If it never quite flattens out, the function is not differentiable there.
Chain rule (multivariable)
If g: ℝⁿ → ℝᵐ is differentiable at x₀ and f: ℝᵐ → ℝᵖ is differentiable at g(x₀), then the derivative of the composition f ∘ g at x₀ is D(f ∘ g)(x₀) = Df(g(x₀)) · Dg(x₀).
In simple terms, you multiply the Jacobian matrices in order: outer function's Jacobian evaluated at the inner function's output, times the inner function's Jacobian.
Directional derivative
The rate of change of f at a point x₀ in the direction of a unit vector v, computed as D_v f(x₀) = ∇f(x₀) · v.
Think of it as: a partial derivative asked "how fast does f change along the x-axis?" The directional derivative asks the same question along any direction you choose.
Unit vector
A vector of length 1. The directional derivative formula requires v to be a unit vector; if it is not, you must normalise it first by dividing by its magnitude.
Parameterised path (curve)
A function E(t) mapping an interval [a, b] into ℝ² or ℝ³, describing a curve or trajectory. The derivative E'(t) gives the velocity vector along the path.
In simple terms, it is a recipe that traces out a curve as the parameter t runs through its range.
Cycloid
The curve traced by a point on the rim of a circle of radius 1 rolling along a straight line. Position = (t − sin t, 1 − cos t), velocity = (1 − cos t, sin t).
Think of it as: the path a pebble stuck in a tyre would follow as the tyre rolls.
Polar coordinates
A coordinate system where a point is described by its distance r from the origin and its angle θ from the positive x-axis. Related to Cartesian coordinates by x = r cos θ, y = r sin θ.
In simple terms, instead of "go 3 right and 4 up," you say "go 5 units at an angle of about 53 degrees."
Theorem 9 (sufficient condition): if all partial derivatives ∂fᵢ/∂xⱼ exist and are continuous in a neighbourhood of x₀, then f is differentiable at x₀. This is the test you will use most often.
Theorem 8: if f is differentiable at x₀, then f is continuous at x₀. Differentiability is a stronger condition than continuity.
The converse of Theorem 8 is false: a function can be continuous everywhere but differentiable nowhere (though such examples are unusual in a standard calculus course).
Given g: ℝⁿ → ℝᵐ differentiable at x₀ and f: ℝᵐ → ℝᵖ differentiable at g(x₀), the composition f ∘ g is differentiable at x₀.
The differential of the composition: d(f ∘ g)(x₀) = df(g(x₀)) · dg(x₀).
In matrix form: D(f ∘ g)(x₀) = Df(g(x₀)) · Dg(x₀). You multiply the Jacobian of the outer function (evaluated at the output of the inner function) by the Jacobian of the inner function.
Example: for f(x, y) = x e^(x²+y²) and a path γ(t) = (t, −t), the chain rule gives d/dt [f(γ(t))] = ∇f(γ(t)) · γ'(t).
When converting from Cartesian (x, y) to polar (r, θ), with x = r cos θ, y = r sin θ:
∂f/∂r = cos θ · ∂f/∂x + sin θ · ∂f/∂y
∂f/∂θ = −r sin θ · ∂f/∂x + r cos θ · ∂f/∂y
These follow directly from the chain rule. The Jacobian of the coordinate transformation does the heavy lifting.
This technique is essential for problems with circular or radial symmetry, where polar coordinates simplify the algebra.
The directional derivative of f at x₀ in the direction of a unit vector v is D_v f(x₀) = ∇f(x₀) · v.
The gradient ∇f points in the direction of steepest increase. Its magnitude |∇f| is the maximum rate of increase.
The directional derivative is maximised when v points in the same direction as ∇f, and minimised (most negative) when v points in the opposite direction.
The gradient is always perpendicular to the level surfaces (level curves in 2D) of f. This is a key geometric fact.
A path E(t) maps a parameter interval [a, b] into ℝ² or ℝ³.
Line segment from (x₀, y₀, z₀) to (x₁, y₁, z₁): E(t) = (x₀, y₀, z₀) + t(x₁−x₀, y₁−y₀, z₁−z₀), t ∈ [0, 1].
Ellipse x²/a² + y²/b² = 1: parameterise as x = a cos α, y = b sin α, α ∈ [0, 2π).
Parabola y = ax²: parameterise as x = t, y = at².
Cycloid (circle of radius 1 rolling on a line): position = (t − sin t, 1 − cos t), velocity = (1 − cos t, sin t).
For a cylinder with volume V = πr²h:
∂V/∂r = 2πrh (sensitivity to changes in radius)
∂V/∂h = πr² (sensitivity to changes in height)
If both r and h change over time: dV/dt = (∂V/∂r)(dr/dt) + (∂V/∂h)(dh/dt).
This is a direct application of the chain rule to a real-world quantity.
Multivariable chain rule (matrix form)
D(f \circ g)(\vec{x}_0) = Df(g(\vec{x}_0)) \cdot Dg(\vec{x}_0)Directional derivative
D_{\vec{v}} f(\vec{x}_0) = \nabla f(\vec{x}_0) \cdot \vec{v}Polar coordinate partial derivatives
\frac{\partial f}{\partial r} = \cos\theta \, \frac{\partial f}{\partial x} + \sin\theta \, \frac{\partial f}{\partial y}\frac{\partial f}{\partial \theta} = -r\sin\theta \, \frac{\partial f}{\partial x} + r\cos\theta \, \frac{\partial f}{\partial y}Total rate of change (cylinder volume example)
\frac{dV}{dt} = \frac{\partial V}{\partial r} \frac{dr}{dt} + \frac{\partial V}{\partial h} \frac{dh}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}Line segment parameterisation
\mathcal{E}(t) = (x_0, y_0, z_0) + t(x_1 - x_0,\, y_1 - y_0,\, z_1 - z_0), \quad t \in [0,1]Ellipse parameterisation
x = a\cos\alpha, \quad y = b\sin\alpha, \quad \alpha \in [0, 2\pi)Cycloid
\text{Position} = (t - \sin t,\; 1 - \cos t), \quad \text{Velocity} = (1 - \cos t,\; \sin t)The chain rule is the mathematical engine behind backpropagation in neural networks: each layer's contribution to the output error is computed by multiplying Jacobians from the output layer back to the input. Every time a deep-learning model trains, it runs the multivariable chain rule millions of times.
Directional derivatives and gradients are central to gradient descent, the optimisation algorithm used across machine learning, economics, and operations research. The gradient tells you which direction to step to reduce a cost function most quickly.
Students often forget that the directional derivative formula D_v f = ∇f · v requires v to be a unit vector. If you use a non-unit vector, you get the wrong answer. Always normalise first.
Confusing the chain rule's matrix multiplication order. It is Df(g(x₀)) · Dg(x₀), not the other way round. Matrix multiplication is not commutative.
Assuming the chain rule result for d/dt[f(γ(t))] is just "multiply the partial derivatives by the path derivatives and add." That is correct for scalar f, but for vector-valued f you need the full Jacobian product.
Thinking that ∂f/∂r in polar coordinates is the same as ∂f/∂x. It is not; the polar partial derivatives mix the Cartesian ones via cos θ and sin θ.
⚠️ Chain rule problems are among the most common exam questions. Practise both the scalar version (∇f · γ'(t)) and the full Jacobian product.
⚠️ Be ready to compute a directional derivative given a gradient and a direction vector. Remember to normalise the direction.
⚠️ Know the standard parameterisations (line segment, ellipse, cycloid) and be able to compute velocities from them.
⚠️ Coordinate-change problems (Cartesian to polar) are a frequent source of exam questions. Memorise the formulas for ∂f/∂r and ∂f/∂θ.
True or false: D(f ∘ g) = Dg · Df. (False; the order is Df(g(x₀)) · Dg(x₀))
Fill in the blank: the directional derivative of f in the direction v is ∇f · ____ , where v must be a ____ vector.
True or false: the gradient is parallel to level curves. (False; it is perpendicular to them)
What is the velocity vector of the cycloid at parameter t? ((1 − cos t, sin t))
True or false: changing from Cartesian to polar coordinates changes the value of f at a point. (False; only the description of the point changes, not the function value)
Q: Let f(x, y) = x²y and γ(t) = (cos t, sin t). Use the chain rule to find d/dt [f(γ(t))].
A: ∇f = (2xy, x²). At γ(t): ∇f(cos t, sin t) = (2 cos t sin t, cos² t). γ'(t) = (−sin t, cos t). So d/dt [f(γ(t))] = (2 cos t sin t)(−sin t) + (cos² t)(cos t) = −2 cos t sin²t + cos³t.
Q: Find the directional derivative of f(x, y, z) = x² + yz at the point (1, 2, 3) in the direction of the vector (1, 1, 1).
A: ∇f = (2x, z, y). At (1, 2, 3): ∇f = (2, 3, 2). The unit vector in the direction (1, 1, 1) is (1/√3, 1/√3, 1/√3). The directional derivative is (2 + 3 + 2)/√3 = 7/√3 = 7√3/3.
Q: Write the formula for ∂f/∂θ when converting from Cartesian to polar coordinates.
A: ∂f/∂θ = −r sin θ · ∂f/∂x + r cos θ · ∂f/∂y.
Q: Parameterise the ellipse x²/9 + y²/4 = 1 and find the velocity vector.
A: x = 3 cos α, y = 2 sin α. Velocity = (−3 sin α, 2 cos α).
Q: A cylinder has radius r = 3 cm and height h = 10 cm. The radius is increasing at 0.1 cm/s and the height is decreasing at 0.2 cm/s. What is dV/dt?
A: dV/dt = 2πrh(dr/dt) + πr²(dh/dt) = 2π(3)(10)(0.1) + π(9)(−0.2) = 6π − 1.8π = 4.2π ≈ 13.19 cm³/s.
The chain rule and directional derivatives are prerequisites for Lagrange multipliers and constrained optimisation, which you will meet when studying extrema of functions subject to constraints.
Parameterised curves lead directly into line integrals and arc length, which appear in vector calculus and are essential for topics like work, circulation, and flux.
The coordinate-change machinery here (Cartesian to polar) extends to cylindrical and spherical coordinates, which are central to evaluating multiple integrals in later chapters.
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