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MathidiaMathematics Academy
Multivariable calculus
Lesson 34 of 75
Challenge

Multivariable chain rule

Objective. Differentiate composite multivariable mappings.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

The multivariable chain rule extends single-variable differentiation to composite functions where intermediate variables themselves depend on one or more other variables. If z = f(x, y) with x = x(t) and y = y(t), then dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt), summing contributions through every path of dependency. When x and y instead depend on two variables s and t, partial versions ∂z/∂s and ∂z/∂t are formed the same way, each summing over all intermediate variables. This rule is essential for implicit differentiation and for converting between coordinate systems. For z = x²y with x = t² and y = sin t, dz/dt = 2xy(2t) + x²(cos t), which after substitution gives an explicit function of t alone.

Worked example

Use the multivariable chain rule to find dz/dt for z = x² + y² where x = 3t and y = t².

3
Coordinate grid with the line 3
  1. 1Compute the partial derivatives: ∂z/∂x = 2x and ∂z/∂y = 2y.
  2. 2Compute the derivatives of the inner functions: dx/dt = 3 and dy/dt = 2t.
  3. 3Apply the chain rule: dz/dt = 2x(3) + 2y(2t) = 6x + 4ty.
  4. 4Substitute x = 3t and y = t²: dz/dt = 6(3t) + 4t(t²) = 18t + 4t³.

Answer: dz/dt = 18t + 4t³

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Common mistake. Students often omit one branch of the dependency tree, differentiating through x but forgetting the separate contribution through y; every intermediate variable the function depends on must contribute its own term.

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