f(x, y) = xy, where x(t₀) = 4, y(t₀) = 5, dx/dt|_{t₀} = 3, and dy/dt|_{t₀} = 2. What is df/dt at t₀?
A6 — the product of the two rates of change
B8 — only the contribution from the y-branch
C15 — only the contribution from the x-branch
D23 — the sum of both partial-derivative contributions
By the multivariable chain rule, df/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) = y·3 + x·2 = 5·3 + 4·2 = 15 + 8 = 23. Options 1, 2, and 3 each represent the error of tracking only one route from t to f (or multiplying instead of summing). The key insight is that every path from the independent variable t to f contributes its own term, and all contributions are summed.
Question 2 Multiple Choice
If f: ℝ² → ℝ is differentiable and x: ℝ → ℝ² is a differentiable path, the derivative of the composition f(x(t)) is best described as:
AThe product of f′(x) and x′(t) treated as two scalars
BThe dot product of the gradient ∇f evaluated at x(t) with the velocity vector x′(t)
CThe sum of all second-order partial derivatives of f along x(t)
D∂f/∂t, computed by directly differentiating f with respect to t
In the Jacobian (matrix) formulation, the derivative of a composition is the product of the Jacobians. For scalar-valued f, the Jacobian is the gradient ∇f (a row vector), and the Jacobian of x is x′(t) (a column vector of component rates). Their product is the dot product ∇f · x′(t). Written out for two components, this is exactly (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt). Option 4 is wrong because f has no direct dependence on t — t enters only through x and y.
Question 3 True / False
If f(x, y) = x + y, x(t) = t, and y(t) = 0 for all t, then df/dt = ∂f/∂x.
TTrue
FFalse
Answer: True
Applying the chain rule: df/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) = 1·1 + 1·0 = 1. Since ∂f/∂x = 1, the statement is true. When y is constant, dy/dt = 0 and the y-branch contributes nothing, so the multivariable chain rule collapses to the single-variable rule in x. This is a useful sanity check: the multivariable formula should reduce to the familiar single-variable case when all but one intermediate variable is frozen.
Question 4 True / False
The partial derivatives ∂f/∂x and ∂f/∂y in the chain rule formula df/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) are evaluated at the value of t, not at the point (x(t), y(t)).
TTrue
FFalse
Answer: False
The partial derivatives are properties of f evaluated at a point in its domain, which is ℝ². They must be evaluated at (x(t), y(t)) — the current location of the path in the plane — not at the scalar t. This is a common notational confusion: t is the parameter of the path, but ∂f/∂x and ∂f/∂y measure how f varies in the (x, y)-plane at the specific point the path occupies at time t.
Question 5 Short Answer
In your own words, explain why the multivariable chain rule sums the partial-derivative contributions rather than, say, multiplying them.
Think about your answer, then reveal below.
Model answer: Each intermediate variable (x, y) provides an independent channel through which a change in t can affect f. A small change dt causes dx = (dx/dt)dt change in x, which then produces a (∂f/∂x)dx change in f via the x-channel; simultaneously it causes a (∂f/∂y)dy change via the y-channel. Because both channels act at the same time and their effects on f add together (not multiply), the total change in f is the sum of all individual contributions.
The additive structure comes from f being approximately linear near any point (the definition of differentiability): df ≈ (∂f/∂x)dx + (∂f/∂y)dy. The total differential is a sum because small changes in independent directions add. This is unlike multiplying because the two paths (t→x→f) and (t→y→f) are parallel routes to the same output, not sequential steps — you add parallel contributions, you multiply sequential ones.