Multivariable Chain Rule

The multivariable chain rule finds the derivative of a function whose inputs depend on other variables. If z = f(x, y) with x = x(t) and y = y(t), then dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt). Each path from z down to t in a tree diagram gives one product, and you add the products.

Before this lesson: Chain Rule, Partial Derivatives and Vector Functions and Space Curves

Why the chain rule has more than one term

Suppose z = f(x, y) and both x and y change with time t. A small step in t moves x a little and moves y a little. Each of those moves changes z, so the total change in z is the sum of two effects.

The x effect is (rate z changes with x) × (rate x changes with t), and the y effect works the same way. That is why the multivariable chain rule is a sum of products, one product for each input.

Multivariable chain rule formulas

Case 1: one independent variable. If z = f(x, y), x = x(t) and y = y(t):

\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt}

Case 2: two independent variables. If z = f(x, y), x = x(s, t) and y = y(s, t):

\dfrac{\partial z}{\partial s} = \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial s} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial s}, \qquad \dfrac{\partial z}{\partial t} = \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial t} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial t}

• Tree diagram rule: draw the function at the top, its inputs below it, and their inputs below those. For each path from the top down to the variable you want, multiply the derivatives on the branches. Then add the products.
• More variables: w = f(x, y, z) has three branches at the top, so its chain rule has three terms.
• Implicit differentiation: if F(x, y) = 0 defines y as a function of x, then \dfrac{dy}{dx} = -\dfrac{F_x}{F_y} wherever F_y \ne 0.
• Use d for a derivative when there is one independent variable (dz/dt) and ∂ when there are several (∂z/∂s).

Multivariable chain rule examples

Example 1: One parameter

Let z = x²y, with x = t² and y = 3t + 1. Find dz/dt at t = 1.

\dfrac{\partial z}{\partial x} = 2xy, \quad \dfrac{\partial z}{\partial y} = x^2

\dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = 3

\dfrac{dz}{dt} = 2xy(2t) + x^2(3)

x = 1^2 = 1, \quad y = 3(1) + 1 = 4

\dfrac{dz}{dt} = 2(1)(4)(2) + (1)^2(3)

= 16 + 3

= 19

Answer: dz/dt = 19 at t = 1. Check: z = t⁴(3t + 1) = 3t⁵ + t⁴, so dz/dt = 15t⁴ + 4t³ = 19 at t = 1.

Example 2: Two independent variables

Let z = e^(xy), with x = s + t and y = s − t. Find ∂z/∂s and ∂z/∂t in terms of s and t.

\dfrac{\partial z}{\partial x} = ye^{xy}, \quad \dfrac{\partial z}{\partial y} = xe^{xy}

\dfrac{\partial z}{\partial s} = ye^{xy}(1) + xe^{xy}(1)

= (x + y)e^{xy}

= 2se^{s^2 - t^2}

\dfrac{\partial z}{\partial t} = ye^{xy}(1) + xe^{xy}(-1)

= (y - x)e^{xy}

= -2te^{s^2 - t^2}

Answer: ∂z/∂s = 2s e^(s² − t²) and ∂z/∂t = −2t e^(s² − t²).

Example 3: Three intermediate variables

Let w = x² + yz, with x = cos t, y = sin t and z = t. Find dw/dt at t = π/2.

\dfrac{dw}{dt} = 2x\dfrac{dx}{dt} + z\dfrac{dy}{dt} + y\dfrac{dz}{dt}

= 2x(-\sin t) + z(\cos t) + y(1)

x = 0, \quad y = 1, \quad z = \dfrac{\pi}{2}

\dfrac{dw}{dt} = 2(0)(-1) + \dfrac{\pi}{2}(0) + 1(1)

= 1

Answer: dw/dt = 1 at t = π/2.

Example 4: Implicit differentiation

Find dy/dx for the curve x³ + y³ = 6xy at the point (3, 3).

3^3 + 3^3 = 54 = 6(3)(3)

F(x, y) = x^3 + y^3 - 6xy

F_x = 3x^2 - 6y

F_x(3, 3) = 27 - 18 = 9

F_y = 3y^2 - 6x

F_y(3, 3) = 27 - 18 = 9

\dfrac{dy}{dx} = -\dfrac{9}{9}

= -1

Answer: dy/dx = −1 at (3, 3). This matches what implicit differentiation from Calculus 1 gives, with less algebra.

Example 5: Two cars moving apart (word problem)

Car A is 30 miles north of an intersection, driving north at 50 mph. Car B is 40 miles east of it, driving east at 60 mph. How fast is the distance between the cars changing?

D = \sqrt{x^2 + y^2}

D = \sqrt{40^2 + 30^2} = 50

\dfrac{\partial D}{\partial x} = \dfrac{x}{D}, \quad \dfrac{\partial D}{\partial y} = \dfrac{y}{D}

\dfrac{dD}{dt} = \dfrac{x}{D}\dfrac{dx}{dt} + \dfrac{y}{D}\dfrac{dy}{dt}

= \dfrac{40}{50}(60) + \dfrac{30}{50}(50)

= 48 + 30

= 78

Answer: the distance between the cars is growing at 78 mph.

Common mistakes

Related methods

Substitute first, then differentiate

For a single-variable answer, you can also substitute x(t) and y(t) into z and use the one-variable chain rule. This is a good way to check your work, as in Example 1. The multivariable chain rule is easier when the formulas are messy or when you only know rates at one moment, as in Example 5.

Connection to related rates and the gradient

Example 5 is a related rates problem solved with partial derivatives. In the next lesson, the chain rule for z = f(x(t), y(t)) is written as a dot product of the gradient with the velocity vector, which leads to directional derivatives.

Practice problems

1. Let z = x² + y², x = 2t and y = t². Find dz/dt.
2. Let z = xy, x = eᵗ and y = cos t. Find dz/dt.
3. Let z = x/y, x = 3t and y = t + 1. Find dz/dt at t = 1.
4. Let z = x²y, x = s + t and y = st. Find ∂z/∂s and ∂z/∂t at (s, t) = (1, 2).
5. Let w = xyz, x = t, y = t² and z = t³. Find dw/dt.
6. Let z = sin(x + y), x = s² and y = st. Find ∂z/∂t.
7. Use the chain rule to find dy/dx for x² + xy + y² = 7 at (1, 2).
8. Use the chain rule to find dy/dx for xy² + x³ = 10 at (1, 3).
9. Let w = f(x, y, z), where x, y and z are functions of u and v. Write the chain rule for ∂w/∂u. How many terms does it have?
10. A rectangle is 12 cm long and 5 cm wide. Its length grows 2 cm/s and its width shrinks 1 cm/s. How fast are its diagonal and its area changing?

\dfrac{dz}{dt} = 2x(2) + 2y(2t)

= 2(2t)(2) + 2(t^2)(2t)

= 8t + 4t^3

\dfrac{dz}{dt} = y(e^t) + x(-\sin t)

= e^t\cos t - e^t\sin t

\dfrac{\partial z}{\partial x} = \dfrac{1}{y}, \quad \dfrac{\partial z}{\partial y} = -\dfrac{x}{y^2}

x = 3, \quad y = 2

\dfrac{dz}{dt} = \dfrac{1}{2}(3) - \dfrac{3}{4}(1)

= 1.5 - 0.75

= 0.75

x = 3, \quad y = 2

\dfrac{\partial z}{\partial s} = 2xy(1) + x^2(t)

= 2(3)(2) + 3^2(2)

= 12 + 18

= 30

\dfrac{\partial z}{\partial t} = 2xy(1) + x^2(s)

= 2(3)(2) + 3^2(1)

= 12 + 9

= 21

\dfrac{dw}{dt} = yz(1) + xz(2t) + xy(3t^2)

= t^5 + 2t^5 + 3t^5

= 6t^5

\dfrac{\partial z}{\partial t} = \cos(x + y)(0) + \cos(x + y)(s)

= s\cos(s^2 + st)

F_x = 2x + y = 4

F_y = x + 2y = 5

\dfrac{dy}{dx} = -\dfrac{4}{5}

F_x = y^2 + 3x^2 = 9 + 3 = 12

F_y = 2xy = 6

\dfrac{dy}{dx} = -\dfrac{12}{6} = -2

\dfrac{\partial w}{\partial u} = \dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial u} + \dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial u} + \dfrac{\partial w}{\partial z}\dfrac{\partial z}{\partial u}

D = \sqrt{12^2 + 5^2} = 13

\dfrac{dD}{dt} = \dfrac{12}{13}(2) + \dfrac{5}{13}(-1)

= \dfrac{24}{13} - \dfrac{5}{13}

= \dfrac{19}{13}

\approx 1.46

\dfrac{dA}{dt} = w\dfrac{dl}{dt} + l\dfrac{dw}{dt}

= 5(2) + 12(-1)

= 10 - 12

= -2

Multivariable chain rule FAQ

What is the multivariable chain rule?

It is the rule for differentiating a function whose inputs depend on other variables. You multiply derivatives along each path from the function to the variable and add the results.

How do you use a tree diagram for the chain rule?

Put the function at the top, its inputs on the next row and their variables on the bottom row. Label each branch with a derivative. Every path to the variable you want gives one product.

How is it different from the single-variable chain rule?

The single-variable chain rule has one path, so it is one product, dy/dx = (dy/du)(du/dx). With several inputs there are several paths, so the answer is a sum of products.

How does the chain rule give dy/dx for an implicit curve?

If F(x, y) = 0, differentiating with the chain rule gives F_x + F_y(dy/dx) = 0. Solving gives dy/dx = −F_x/F_y.

Summary

• For z = f(x, y) with x(t) and y(t): dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
• With x(s, t) and y(s, t), find ∂z/∂s and ∂z/∂t the same way, using partial derivatives on the bottom branches.
• Tree diagram: multiply along each path, then add the paths.
• Evaluate x, y and t together when you need a number.
• Implicit curves: dy/dx = −F_x/F_y.

Previous lesson: Tangent Planes

Next lesson: Directional Derivatives and the Gradient

Back to the Calculus 3 Index