Multivariable Chain Rule- Special Cases Tutorial

What Is the Multivariable Chain Rule (And Why You Need It)

The chain rule tells you how to find derivatives when functions are composed—meaning one function feeds into another. In single-variable calculus, this is straightforward. In multivariable calculus, things get messier because your functions can depend on multiple variables, and those variables can each depend on other variables.

You need the multivariable chain rule when:

• You have a function f(x,y) where x and y both depend on t
• You're differentiating a composition of vector-valued functions
• You're working with implicit relationships between variables

This tutorial covers the special cases you'll encounter most often. Master these and the general case becomes obvious.

The Three Cases You Must Know

Every multivariable chain rule problem falls into one of these patterns:

• Case 1: One independent variable, one intermediate variable (essentially single-variable chain rule)
• Case 2: One independent variable, multiple intermediate variables
• Case 3: Multiple independent variables, multiple intermediate variables

Case 1: Single Path (The Basic Chain Rule)

z = f(x) and x = g(t)

This is the single-variable chain rule you already know:

d/dt [f(g(t))] = f'(g(t)) · g'(t)

Nothing new here. Skip ahead if this bores you.

Case 2: One t, Multiple Intermediate Variables

z = f(x, y) where x = g(t) and y = h(t)

This is where multivariable calculus starts. You have a function depending on x and y, and both x and y depend on t. You need to account for how z changes through both paths.

The formula:

dz/dt = (∂f/∂x) · (dx/dt) + (∂f/∂y) · (dy/dt)

The partial derivatives (∂f/∂x and ∂f/∂y) are evaluated at the point (x, y). The ordinary derivatives (dx/dt and dy/dt) capture how x and y change with t.

Case 3: Multiple Independent Variables

z = f(x, y) where x = g(u, v) and y = h(u, v)

Now you have two independent variables, u and v. Each path to z branches through x and y, which each branch through u and v.

∂z/∂u = (∂f/∂x)(∂x/∂u) + (∂f/∂y)(∂y/∂u)

∂z/∂v = (∂f/∂x)(∂x/∂v) + (∂f/∂y)(∂y/∂v)

You get one equation per independent variable. That's the pattern—add a term for each intermediate variable, multiplied by the derivative of that intermediate variable with respect to your target variable.

Worked Example: Case 2 in Action

Let z = x²y where x = t² and y = t + 1. Find dz/dt.

Step 1: Compute the pieces you need.

• ∂z/∂x = 2xy
• ∂z/∂y = x²
• dx/dt = 2t
• dy/dt = 1

Step 2: Plug into the formula.

dz/dt = (2xy)(2t) + (x²)(1)

Step 3: Substitute x = t² and y = t + 1.

dz/dt = 2(t²)(t+1)(2t) + (t²)²

dz/dt = 4t³(t+1) + t⁴

dz/dt = 4t⁴ + 4t³ + t⁴ = 5t⁴ + 4t³

Done. The key is don't substitute too early. Keep the partial derivatives symbolic, apply the formula, then substitute at the end. It keeps the algebra cleaner.

Worked Example: Implicit Differentiation via Chain Rule

Find dy/dx if x³ + y³ = 6xy.

Step 1: Differentiate both sides with respect to x, treating y as a function of x.

d/dx(x³) + d/dx(y³) = d/dx(6xy)

Step 2: Apply the chain rule to y³.

3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)

Step 3: Collect dy/dx terms on one side.

3y²(dy/dx) - 6x(dy/dx) = 6y - 3x²

Step 4: Solve for dy/dx.

dy/dx(3y² - 6x) = 6y - 3x²

dy/dx = (6y - 3x²) / (3y² - 6x)

dy/dx = (2y - x²) / (y² - 2x)

This is the chain rule in disguise—you're treating y as an intermediate variable between x and whatever function defines y implicitly.

The General Formula (For Reference)

If you have z = f(x₁, x₂, ..., xₙ) where each xᵢ = gᵢ(t₁, t₂, ..., tₘ), then:

∂z/∂tⱼ = Σᵢ (∂f/∂xᵢ)(∂xᵢ/∂tⱼ)

For each output variable tⱼ, you sum over all intermediate variables xᵢ. Each term is the partial of f with respect to xᵢ, times the partial of xᵢ with respect to tⱼ.

That's it. Every multivariable chain rule problem is just this summation applied to a specific setup.

Common Mistakes That Cost You Points

• Using ∂ when you need d: If x is a function of a single variable t, use ordinary derivative dx/dt. Only use partial derivatives (∂) when the intermediate variable depends on multiple independent variables.
• Forgetting to evaluate partials: ∂f/∂x gives you a function of (x, y). You can't leave it as just "∂f/∂x"—you need to substitute your expressions for x and y.
• Dropping terms: Each intermediate variable contributes one term. If you have three intermediate variables, you have three terms. No shortcuts.
• Mixing up notation: The variable in the denominator of your final derivative tells you what you're differentiating with respect to. dz/dt ≠ ∂z/∂t. The first assumes x is fixed; the second assumes all other variables (besides z and t) vary.

Practice Problems to Try

1. If z = eˣʸ, x = cos(t), y = sin(t), find dz/dt.

2. If w = x² + y² + z² where x = r cos(θ), y = r sin(θ), z = r, find ∂w/∂r and ∂w/∂θ.

3. Find dy/dx if sin(xy) = x² + y.

Check your answers by verifying the chain rule structure matches the dependency between variables.

Quick Reference Table

fₓ means ∂f/∂x, gᵤ means ∂g/∂u, and so on.