Multivariable Chain Rule | Tony's Intertwingled Memex

Core Definition

Let $z = f(x, y)$ be a differentiable function of $x$ and $y$ .

Case 1: Single Independent Parameter (Trajectory Paths)
If $x = x(t)$ and $y = y(t)$ are differentiable functions of a single parameter $t$ , then $z$ is an indirect function of $t$ , and the total derivative is:
$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ In vector notation, this is the dot product of the gradient and velocity: $\frac{dz}{dt} = \nabla z \cdot \vec{v}(t) = \left\langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right\rangle \cdot \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle$
Case 2: Multiple Independent Parameters (Surface Parameterizations)
If $x = g(u, v)$ and $y = h(u, v)$ are differentiable functions of two independent variables $u$ and $v$ , then the partial rates of change are:
$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$ $\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$
Implicit Differentiation Formulas
- For a level curve defined implicitly by $F(x, y) = 0$ where $F_y \neq 0$ : $\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
- For a level surface defined implicitly by $F(x, y, z) = 0$ where $F_z \neq 0$ : $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$

Concept Explanation

The Dependency Tree Mental Model

When a system output depends on intermediate variables that simultaneously depend on one or more independent driving inputs, changes ripple across multiple paths at once. To determine the rate of change with respect to an input, use a tree diagram:

Top Level: The dependent output variable ( $z$ ).
Intermediate Level: The variables directly fed into the function ( $x, y$ ).
Bottom Level: The independent parameters driving the system ( $t$ or $u, v$ ).

             z                          z
           /   \                      /   \
          x     y                    x     y
          |     |                   / \   / \
          t     t                  u   v u   v

   Case 1: One Parameter           Case 2: Two Parameters

To compute the derivative of the top variable with respect to a bottom variable:

Multiply the derivatives along each connected branch from top to bottom.
Sum the resulting products across all parallel pathways leading to the same independent variable.

Total Derivative vs. Partial Derivative

An ordinary derivative $\frac{dz}{dt}$ represents the total rate of change experienced as the parameter $t$ advances, accounting for all internal pathways changing concurrently.
A partial derivative $\frac{\partial z}{\partial x}$ isolates the rate of change purely along the $x$ -direction, artificially freezing $y$ as a constant.
When an object moves along a trajectory $\vec{r}(t) = \langle x(t), y(t) \rangle$ , its sensory experience combines the spatial slopes of the landscape ( $\\nabla z$ ) with its travel velocity ( $\vec{r}\x27(t)$ ).

Derivation of Implicit Differentiation

Consider an implicit equation $F(x, y) = 0$ defining $y$ as a differentiable function of $x$ . Differentiating both sides with respect to $x$ using the multivariable chain rule (where $x$ depends on $x$ with $\frac{dx}{dx} = 1$ , and $y$ depends on $x$ ):

\frac{d}{dx}[F(x, y)] = \frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0

F_x (1) + F_y \frac{dy}{dx} = 0 \implies F_y \frac{dy}{dx} = -F_x \implies \frac{dy}{dx} = -\frac{F_x}{F_y}

The negative sign arises directly from moving the partial

F_x

to the opposite side of the equation when isolating

\frac{dy}{dx}

Engineering Applications

Civil & Structural Engineering: Dynamic stress intensity factors on fatigue-loaded bridge gusset plates depend on traffic load stress and physical crack length, both advancing with time (see Bridge Truss Gusset Plate Fatigue (Mechanical & Civil Engineering)).
Computer & Electrical Engineering: Processor core power dissipation depends on clock frequency and supply voltage, which dynamic voltage and frequency scaling (DVFS) algorithms adjust concurrently during computing spikes (see Processor Thermal Power Dissipation (Computer & Electrical Engineering)).
Environmental Engineering: Pollutant plumes spreading in groundwater aquifers convert spatial coordinates to radial flow paths $(r, \theta)$ to monitor municipal extraction wells (see Groundwater Contaminant Plume in Radial Aquifer (Environmental Engineering)).
Aerospace & Energy Systems: Solar array I-V non-linear diode characteristics use implicit differentiation to extract dynamic small-signal resistance without explicitly solving transcendental equations (see Photovoltaic Solar Cell Dynamic Impedance (Electrical & Energy Engineering)).

Example Problems

Example 1: Trajectory Rate of Change (Case 1)

A thermal sensor moves across a hot metal plate with temperature profile $T(x, y) = 2x^2 y + y^3$ (in $^\circ\text{C}$ ). The sensor\x27s coordinates (in meters) follow the parametric path $x(t) = 3t$ , $y(t) = t^2 + 1$ , where $t$ is in seconds. Find the rate of temperature change $\frac{dT}{dt}$ recorded by the sensor at $t = 1\text{ s}$ .

Solution

State Governing Equation: $\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}$
Compute Spatial Partials: $\frac{\partial T}{\partial x} = 4xy, \qquad \frac{\partial T}{\partial y} = 2x^2 + 3y^2$
Compute Velocity Components: $\frac{dx}{dt} = 3, \qquad \frac{dy}{dt} = 2t$
Evaluate at $t = 1$ :
- Coordinates: $x(1) = 3(1) = 3$ , $y(1) = 1^2 + 1 = 2$ .
- Partials: $\frac{\partial T}{\partial x}(3, 2) = 4(3)(2) = 24$ , $\frac{\partial T}{\partial y}(3, 2) = 2(3)^2 + 3(2)^2 = 18 + 12 = 30$ .
- Velocities: $\frac{dx}{dt} = 3$ , $\frac{dy}{dt} = 2(1) = 2$ .
Synthesize: $\frac{dT}{dt} = (24)(3) + (30)(2) = 72 + 60 = 132^\circ\text{C/s}$

Example 2: Implicit Differentiation of a Closed Surface

Find the rate of vertical elevation change $\frac{\partial z}{\partial x}$ on the ellipsoid $x^2 + 4y^2 + 9z^2 = 36$ at the point $(2, 2, \frac{4}{3})$ .

Solution

Define the Level Surface Function: $F(x, y, z) = x^2 + 4y^2 + 9z^2 - 36 = 0$
State Governing Formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$
Compute Partial Derivatives of $F$ : $F_x = \frac{\partial}{\partial x}[x^2 + 4y^2 + 9z^2 - 36] = 2x$ $F_z = \frac{\partial}{\partial z}[x^2 + 4y^2 + 9z^2 - 36] = 18z$
Assemble and Evaluate at $(2, 2, \frac{4}{3})$ : $\frac{\partial z}{\partial x} = -\frac{2x}{18z} = -\frac{x}{9z}$ $\frac{\partial z}{\partial x}\Big|_{(2, 2, 4/3)} = -\frac{2}{9\left(\frac{4}{3}\right)} = -\frac{2}{12} = -\frac{1}{6}$

Connections & References

Parent Concepts: Partial Derivatives, Single-Variable Chain Rule
Differential Linkages: Differentials and Error Propagation
Geometric Extensions: Gradient Vectors, Tangent Planes
Course Sessions: Calc-3 Session 11