The Multivariable Chain Rule · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

One term per path

Mathematics · Calculus & Analysis · ages 19-20
Name ______________________   Date ____________

  1. Nina says every route on the dependency diagram adds one product term. Is Nina right?

    Circle one:   True   False

  2. For z = f(x, y) with x(t), y(t), what is dz/dt?

    • (dz/dx)(dx/dt) + (dz/dy)(dy/dt)
    • (dz/dx) + (dz/dy)
    • (dz/dx)(dz/dy)
    • (dx/dt) + (dy/dt)

  3. For z = f(x, y) with x(t), y(t), what is dz/dt?

    • (dz/dx) + (dz/dy)
    • (dz/dx)(dx/dt) + (dz/dy)(dy/dt)
    • (dx/dt) + (dy/dt)

  4. Let z = x squared + y squared with x = t and y = 2t. What is dz/dt at t = 1?

    Answer: ______________

  5. For z = f(x, y) with x(s, t), y(s, t), what is dz/ds?

    • (dz/dx)(dx/ds) + (dz/dy)(dy/ds)
    • (dz/dx)(dx/dt) + (dz/dy)(dy/dt)
    • (dz/dx) + (dz/dy)
    • (dx/ds)(dy/ds)

  6. Let z = 3x + 4y with x = t^2 and y = t. What is dz/dt at t = 2?

    Answer: ______________

  7. Let z = x^2 + y^2 with x = t and y = 2t. What is dz/dt at t = 1?

    Answer: ______________

  8. For z = f(x, y) with x(s, t), y(s, t), what is dz/ds?

    • (dz/dx)(dx/ds) + (dz/dy)(dy/ds)
    • (dz/dx)(dx/dt) + (dz/dy)(dy/dt)
    • (dz/dx) + (dz/dy)

  9. Leo writes dz/ds = (dz/dx)(dx/dt) + (dz/dy)(dy/dt) when x and y depend on s and t. Is Leo right?

    Circle one:   True   False

  10. Let z = x squared + y with x = t and y = t squared. What is dz/dt at t = 3?

    • 6
    • 18
    • 12
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Answer key

For grown-ups. Fold this page away before handing over the rest.

One term per path W1-mt_CHSHdNxPmJ-s1

  1. True · Nina is right: terms and paths match one to one.
  2. (dz/dx)(dx/dt) + (dz/dy)(dy/dt) · Two paths give two product terms, one per intermediate variable.
  3. (dz/dx)(dx/dt) + (dz/dy)(dy/dt) · Two paths give two product terms, one per intermediate variable.
  4. 10 · z = 5t squared, so dz/dt = 10t, which is 10 at t = 1.
  5. (dz/dx)(dx/ds) + (dz/dy)(dy/ds) · Hold t fixed and follow both paths down to s.
  6. 16 · dz/dt = 3(2t) + 4 = 6t + 4, which is 16 at t = 2.
  7. 10 · z = 5t^2, so dz/dt = 10t, which is 10 at t = 1.
  8. (dz/dx)(dx/ds) + (dz/dy)(dy/ds) · Hold t fixed and follow both paths down to s.
  9. False · Those paths end at t, not s. Paths to s use dx/ds and dy/ds.
  10. 12 · dz/dt = 2t + 2t = 4t, which is 12 at t = 3.
Worksheet · LightMySky