The Multivariable Chain Rule

Every path the change can travel gets added up.

12 problems · about 24 minutes · F-IF.B.6, F-BF.A.1

What this lesson teaches

The student applies the chain rule when variables depend on one or more parameters.

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.

Straightforward practice. Get the method working first.

5 problems

∂z/∂x = 2, ∂z/∂y = 6, dx/dt = 3, dy/dt = 1. What is dz/dt?
Answer 12
Why 12.
Why are the chain rule terms added?
Answer Both variables change at once, so both effects on z happen together.
Why Every route contributes.
∂z/∂x = 1, ∂z/∂y = 1, dx/dt = 6, dy/dt = 7. What is dz/dt?
Answer 13
Why 6 + 7.
∂z/∂x = 3, ∂z/∂y = 0, dx/dt = 4, dy/dt = 100. What is dz/dt?
Answer 12
Why The y route is switched off.
How many terms does the chain rule have when z depends on three variables?
Answer 3
Why One per route.

The same ideas with more to keep track of.

3 problems

∂z/∂x = −2, ∂z/∂y = 4, dx/dt = 5, dy/dt = 3. What is dz/dt?
Answer 2
Why −10 + 12.
Fₓ = 6 and Fᵧ = 3 for F(x, y) = 0. What is dy/dx?
Answer -2
Why Negative Fₓ over Fᵧ.
Are the chain rule terms added or multiplied together? Enter 1 for added, 2 for multiplied.
Answer 1
Why Multiply along a route, add across routes.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Order the steps of a chain rule computation.
Answer 1. Draw the tree of dependencies 2. Multiply the derivatives along each route 3. Add the routes together 4. Substitute the values at the point
Why The tree comes first.
∂z/∂x = 7 and dx/dt = 0, with no y dependence. What is dz/dt?
Answer 0
Why Nothing is changing.
The Heated Plate: ∂T/∂x = 2 and ∂T/∂y = −1, with dx/dt = 3 and dy/dt = 4. What is dT/dt?
Answer 2
Why 2.
The Frozen Route: ∂z/∂x = 5, ∂z/∂y = 9, dx/dt = 0 and dy/dt = 2. What is dz/dt?
Answer 18
Why 18.