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Math · Multivariable Calculus

Chapter 4: Partial Derivatives

The Multivariable Chain Rule

Every path the change can travel gets added up.

Lesson
3
Time
About 24 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

Suppose z depends on x and y, and both x and y depend on t. Then z depends on t through two separate routes.

The rule

dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt). One term per route, and the terms are added.

The tree diagram

Draw z branching to x and y, then each branching to t. Multiply along each branch and add the branches.

Why they add

Changing t nudges x and y at once, and each nudge moves z. Both effects happen, so both count.

Two parameters

If x and y depend on s and t, the same pattern gives ∂z/∂s and ∂z/∂t separately, one route at a time.

Implicit differentiation

For F(x, y) = 0 the chain rule gives dy/dx = −Fₓ/Fᵧ. The old technique falls out as a special case.

Every path of influence is added

If z depends on x and y, and both depend on t, then dz/dt sums the contribution through each. The multivariable chain rule adds one term per route from the input to the output.

A dependency tree organises it

Draw arrows from the final variable back to the independent one. Each complete path contributes a product of derivatives, and the rule is the sum over all paths.

It gives implicit differentiation

Applying the chain rule to F(x, y) = 0 produces dy/dx as a ratio of partial derivatives. The technique from single-variable calculus turns out to be a special case of this rule.

And multivariable related rates

A quantity depending on several changing measurements has a rate combining all of them. That is how error propagation and sensitivity analysis are computed in the sciences.

Step 2: Try It Yourself

Tap and try it out.

One vector is the pair of partials and the other is the pair of rates. The chain rule is their dot product.
  • Vector a(3, 2) · length 3.61
  • Vector b(4, 1) · length 4.12

Step 3: Watch an Example

One step at a time.

Watch Tomas Follow Both Routes

Tomas has z with ∂z/∂x = 4 and ∂z/∂y = 3, while dx/dt = 2 and dy/dt = 5.

  1. Step 1

    He takes the x route first: 4 × 2 = 8.

Step 4: Your Turn

Practice makes it stick.

The Heated Plate

Problem 1 of 2

∂T/∂x = 2 and ∂T/∂y = −1, with dx/dt = 3 and dy/dt = 4. What is dT/dt?

The Frozen Route

Problem 2 of 2

∂z/∂x = 5, ∂z/∂y = 9, dx/dt = 0 and dy/dt = 2. What is dz/dt?

Follow the Routes

1 of 8

∂z/∂x = 1, ∂z/∂y = 1, dx/dt = 6, dy/dt = 7. What is dz/dt?

2 of 8

∂z/∂x = 3, ∂z/∂y = 0, dx/dt = 4, dy/dt = 100. What is dz/dt?

3 of 8

How many terms does the chain rule have when z depends on three variables?

4 of 8

∂z/∂x = −2, ∂z/∂y = 4, dx/dt = 5, dy/dt = 3. What is dz/dt?

5 of 8

Fₓ = 6 and Fᵧ = 3 for F(x, y) = 0. What is dy/dx?

6 of 8

Are the chain rule terms added or multiplied together? Enter 1 for added, 2 for multiplied.

7 of 8

Order the steps of a chain rule computation.

  1. 1Multiply the derivatives along each route
  2. 2Add the routes together
  3. 3Substitute the values at the point
  4. 4Draw the tree of dependencies

8 of 8

∂z/∂x = 7 and dx/dt = 0, with no y dependence. What is dz/dt?

Step 5: Quick Check

Show what you know.

Question 1 of 2

∂z/∂x = 2, ∂z/∂y = 6, dx/dt = 3, dy/dt = 1. What is dz/dt?

Question 2 of 2

Why are the chain rule terms added?

What You Learned

  • When z depends on x and y, and both depend on t, change travels along two routes.
  • Multiply the derivatives along each route, then add the routes.
  • Implicit differentiation is the same rule in disguise: dy/dx = −Fₓ/Fᵧ.