The Multivariable Chain Rule

Quick answer

When z = f(x, y) and both x and y depend on t, the rate dz/dt adds one term for each way t reaches z: dz/dt = f_x dx/dt + f_y dy/dt. The rule comes straight from linear approximation, dividing the total differential by dt. With two parameters, as when x and y depend on s and t, each partial derivative of z gets the same kind of sum, and a tree diagram keeps track of the routes. The same rule gives implicit differentiation in one line: dy/dx = −F_x / F_y for a curve F(x, y) = 0.

What you'll learn

Differentiate a function of several variables along a path
Use a tree diagram for functions of two parameters
Differentiate an implicit curve with −F_x/F_y
Solve related-rates problems with several changing quantities

Following a path through a function

A hiker walks along a trail, and the temperature at each point of the map is $T(x, y)$ . As time passes, both coordinates change, so the temperature the hiker feels changes through two routes at once. The chain rule adds them:

\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}

Each term is a rate of $f$ in one direction times how fast the path moves in that direction.

Why the terms add

The linear approximation says that small changes in the inputs change $z$ by

\Delta z \approx f_x\,\Delta x + f_y\,\Delta y

Divide by the small change $\Delta t$ that caused them and let it shrink: $\tfrac{\Delta x}{\Delta t} \to \tfrac{dx}{dt}$ and $\tfrac{\Delta y}{\Delta t} \to \tfrac{dy}{dt}$ , and the approximation becomes exact. Near any point the function is almost linear, and for a linear function the effects of the two inputs add, which is why the chain rule is a sum.

Worked examples

Common mistakes

Practice problems

For $w = xy$ with $x = t^2$ and $y = 3t$ , find $\tfrac{dw}{dt}$ at $t = 2$ .

Answer

$36$

Full solution

$\tfrac{dw}{dt} = y(2t) + x(3) = 3t \cdot 2t + 3t^2 = 9t^2$ , which is $36$ at $t = 2$ . Substituting first, $w = 3t^3$ , agrees.
For $z = x^2 + y^2$ with $x = e^t$ and $y = e^{-t}$ , find $\tfrac{dz}{dt}$ at $t = 0$ .

Answer

$0$

Full solution

$\tfrac{dz}{dt} = 2x e^t + 2y(-e^{-t}) = 2e^{2t} - 2e^{-2t}$ , which is $0$ at $t = 0$ .
For $z = xy$ with $x = st$ and $y = s + t$ , find $\tfrac{\partial z}{\partial s}$ at $(s, t) = (1, 2)$ .

Answer

$8$

Full solution

$\tfrac{\partial z}{\partial s} = y \cdot t + x \cdot 1$ . At $(1, 2)$ : $x = 2$ and $y = 3$ , so $3 \cdot 2 + 2 = 8$ .
Find $\tfrac{dy}{dx}$ on the curve $x^2 + xy + y^2 = 7$ at $(1, 2)$ .

Answer

$-\tfrac{4}{5}$

Full solution

$F_x = 2x + y = 4$ and $F_y = x + 2y = 5$ , so $\tfrac{dy}{dx} = -\tfrac{4}{5}$ .
Use $-\tfrac{F_x}{F_y}$ to find the slope of the circle $x^2 + y^2 = 25$ .

Answer

$-\tfrac{x}{y}$

Full solution

$F_x = 2x$ and $F_y = 2y$ . At $(3, 4)$ the slope is $-\tfrac{3}{4}$ , perpendicular to the radius, whose slope is $\tfrac{4}{3}$ .
A rectangle is $20$ cm long, growing at $2$ cm/s, and $10$ cm wide, shrinking at $1$ cm/s. How fast is its area changing?

Answer

$0$ cm²/s

Full solution

$A = lw$ , so $\tfrac{dA}{dt} = w\tfrac{dl}{dt} + l\tfrac{dw}{dt} = 10(2) + 20(-1) = 0$ . The two effects cancel at this instant.
Draw the tree diagram for $w = f(x, y, z)$ with $x$ , $y$ , $z$ each depending on $t$ , and write $\tfrac{dw}{dt}$ .

Answer

$\tfrac{dw}{dt} = f_x\tfrac{dx}{dt} + f_y\tfrac{dy}{dt} + f_z\tfrac{dz}{dt}$

Full solution

Three branches lead from $w$ to $x$ , $y$ and $z$ , and each continues to $t$ . Three paths give three products.
For $T = x^2 + y^2$ on the path $x = \cos t$ , $y = 2\sin t$ , where on the path does $T$ increase fastest?

Answer

At $t = \tfrac{\pi}{4}$ and $t = \tfrac{5\pi}{4}$

Full solution

$\tfrac{dT}{dt} = 3\sin 2t$ is largest, equal to $3$ , where $\sin 2t = 1$ .
For $z = f(x, y)$ with $x = r\cos\theta$ and $y = r\sin\theta$ , write $\tfrac{\partial z}{\partial r}$ .

Answer

$f_x\cos\theta + f_y\sin\theta$

Full solution

$\tfrac{\partial x}{\partial r} = \cos\theta$ and $\tfrac{\partial y}{\partial r} = \sin\theta$ . This is the rate of change of $f$ moving straight out from the origin.
A student computes $\tfrac{dz}{dt}$ for $z = x^2y$ with $x = t$ and $y = t^2$ as $2x \cdot 1 = 2t$ . What went wrong?

Hint

How many routes lead from $t$ to $z$ ?

Answer

The route through $y$ is missing. $\tfrac{dz}{dt} = 2xy \cdot 1 + x^2 \cdot 2t = 4t^3$ .

Full solution

The student also dropped the factor $y$ from $f_x = 2xy$ . With both routes, $2t \cdot t^2 + t^2 \cdot 2t = 4t^3$ , which matches differentiating $z = t^4$ directly.

Frequently asked questions

What is the multivariable chain rule?

If z = f(x, y) with x = x(t) and y = y(t), then dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt): one term for each intermediate variable.

Why does the chain rule add terms?

A small change in t moves both x and y, and the linear approximation dz ≈ f_x dx + f_y dy adds their effects. Dividing by dt gives the rule.

What is a tree diagram?

A picture with z at the top, the intermediate variables below it and the independent variables at the bottom. Each path from z down to a variable is one product in the sum.

How does the chain rule give implicit differentiation?

Differentiating F(x, y) = 0 with y = y(x) gives F_x + F_y dy/dx = 0, so dy/dx = −F_x / F_y wherever F_y is not zero.

What if x and y depend on two variables s and t?

Then ∂z/∂s = f_x x_s + f_y y_s and ∂z/∂t = f_x x_t + f_y y_t, the same rule with partial derivatives throughout.

What to learn next

Directional Derivatives and the GradientFind the rate of change of f in any direction as a dot product with the gradient, see why the gradient points uphill, and use it as a normal to level curves and surfaces.

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