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Study Guide915 words

The Multivariable Chain Rule: A Comprehensive Study Guide

The Chain Rule

The Multivariable Chain Rule: A Comprehensive Study Guide

This guide covers the extension of the single-variable Chain Rule to functions of several variables, focusing on intermediate and independent variables, the use of tree diagrams, and the generalized chain rule formula.


Learning Objectives

After studying this chapter, you should be able to:

  • State the chain rules for one or two independent variables.
  • Construct and interpret tree diagrams to visualize dependency paths.
  • Differentiate between intermediate and independent variables in a composite function.
  • Perform implicit differentiation of functions involving several variables.
  • Apply the generalized chain rule to functions with nnn independent variables and mmm intermediate variables.

Key Terms & Glossary

  • Independent Variable: The fundamental inputs at the "bottom" of the dependency chain (e.g., ttt in x(t)x(t)x(t) or u,vu, vu,v in x(u,v)x(u,v)x(u,v)).
  • Intermediate Variable: A variable that depends on others but is itself an input for a higher-level function (e.g., xxx and yyy in z=f(x,y)z = f(x,y)z=f(x,y) where xxx and yyy are functions of ttt).
  • Composite Function: A function formed by substituting one function into another.
  • Partial Derivative: The derivative of a multivariable function with respect to one variable while holding others constant.
  • Total Derivative: The rate of change of a function with respect to an independent variable, accounting for all paths through intermediate variables.

The "Big Idea"

In single-variable calculus, the Chain Rule handles functions nested like Russian dolls: f(g(x))f(g(x))f(g(x)). In multivariable calculus, the nesting becomes a web. A change in an independent variable (like time ttt) might affect multiple intermediate variables (like coordinates xxx and yyy), which in turn both affect the final output zzz. The multivariable Chain Rule is essentially a way to sum the contributions of change from every possible path through that web.


Formula / Concept Box

CaseDependencyFormula
One Independent Variablez=f(x,y)z = f(x,y)z=f(x,y), where x=g(t),y=h(t)x=g(t), y=h(t)x=g(t),y=h(t)dzdt=∂z∂xdxdt+∂z∂ydydt\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}dtdz​=∂x∂z​dtdx​+∂y∂z​dtdy​
Two Independent Variablesz=f(x,y)z = f(x,y)z=f(x,y), where x=g(u,v),y=h(u,v)x=g(u,v), y=h(u,v)x=g(u,v),y=h(u,v)∂z∂u=∂z∂x∂x∂u+∂z∂y∂y∂u\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}∂u∂z​=∂x∂z​∂u∂x​+∂y∂z​∂u∂y​
Generalized Casez=f(x1,...,xn)z = f(x_1, ..., x_n)z=f(x1​,...,xn​), xi=gi(t1,...,tm)x_i = g_i(t_1, ..., t_m)xi​=gi​(t1​,...,tm​)∂z∂tj=∑i=1n∂z∂xi∂xi∂tj\frac{\partial z}{\partial t_j} = \sum_{i=1}^{n} \frac{\partial z}{\partial x_i} \frac{\partial x_i}{\partial t_j}∂tj​∂z​=∑i=1n​∂xi​∂z​∂tj​∂xi​​

[!IMPORTANT] Note the notation difference: If the final result depends on only one variable, we use ddd (total derivative). If it depends on multiple variables, we use ∂\partial∂ (partial derivative).


Hierarchical Outline

  • I. Chain Rule for One Independent Variable
    • Dependency Path: t→(x,y)→zt \to (x, y) \to zt→(x,y)→z
    • Summation Principle: Multiply derivatives along paths; add the paths together.
  • II. Chain Rule for Two Independent Variables
    • Independence: uuu and vvv are independent; changing uuu does not change vvv.
    • Partial Differentiation: We calculate ∂z∂u\frac{\partial z}{\partial u}∂u∂z​ and ∂z∂v\frac{\partial z}{\partial v}∂v∂z​ separately.
  • III. The Generalized Chain Rule
    • Dimensionality: Works for nnn variables and mmm parameters.
    • Matrix View: Can be represented as a product of Jacobian matrices.
  • IV. Implicit Differentiation Revisited
    • Using the Chain Rule to find dydx\frac{dy}{dx}dxdy​ for F(x,y)=0F(x,y) = 0F(x,y)=0.

Visual Anchors

Dependency Tree for One Independent Variable

This diagram shows how ttt influences zzz through two different intermediate paths.

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Figure 1 — Mermaid diagram

Dependency Tree for Two Independent Variables

When we have two independent variables (u,vu, vu,v), we must follow paths to the specific variable of interest.

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Figure 2 — Mermaid diagram

Definition-Example Pairs

1. Total Derivative via Chain Rule

  • Definition: The derivative of a function with respect to an independent variable when all intermediate variables are also functions of that independent variable.
  • Example: If z=x2+y2z = x^2 + y^2z=x2+y2 (a paraboloid) and a particle moves along a path x=sin⁡(t),y=cos⁡(t)x = \sin(t), y = \cos(t)x=sin(t),y=cos(t), the total derivative dzdt\frac{dz}{dt}dtdz​ tells us how the height of the particle changes over time.

2. Generalized Chain Rule Summation

  • Definition: For a function w=f(x,y,z)w = f(x, y, z)w=f(x,y,z), the rate of change with respect to an input uuu is the sum of the partials of www with respect to each coordinate, each multiplied by how that coordinate changes with uuu.
  • Example: In a weather model where temperature TTT depends on latitude, longitude, and altitude, and each of those depends on the time of day, the Chain Rule calculates the temperature change rate for a moving weather balloon.

Worked Examples

Example 1: One Independent Variable

Problem: Calculate dzdt\frac{dz}{dt}dtdz​ given z=x2y−y2z = x^2y - y^2z=x2y−y2, where x=extsin(t)x = ext{sin}(t)x=extsin(t) and y=ety = e^ty=et.

Step 1: Find necessary partials.

  • ∂z∂x=2xy\frac{\partial z}{\partial x} = 2xy∂x∂z​=2xy
  • ∂z∂y=x2−2y\frac{\partial z}{\partial y} = x^2 - 2y∂y∂z​=x2−2y
  • dxdt=cos⁡(t)\frac{dx}{dt} = \cos(t)dtdx​=cos(t)
  • dydt=et\frac{dy}{dt} = e^tdtdy​=et

Step 2: Apply the Chain Rule formula. dzdt=(2xy)(cos⁡t)+(x2−2y)(et)\frac{dz}{dt} = (2xy)(\cos t) + (x^2 - 2y)(e^t)dtdz​=(2xy)(cost)+(x2−2y)(et)

Step 3: Substitute xxx and yyy back to get a function of ttt. dzdt=2sin⁡(t)etcos⁡(t)+(sin⁡2(t)−2et)et\frac{dz}{dt} = 2\sin(t)e^t\cos(t) + (\sin^2(t) - 2e^t)e^tdtdz​=2sin(t)etcos(t)+(sin2(t)−2et)et

Example 2: Two Independent Variables

Problem: Calculate ∂z∂u\frac{\partial z}{\partial u}∂u∂z​ for z=x2+3xyz = x^2 + 3xyz=x2+3xy, where x=u⋅vx = u \cdot vx=u⋅v and y=u+vy = u + vy=u+v.

Step 1: Identify components.

  • ∂z∂x=2x+3y\frac{\partial z}{\partial x} = 2x + 3y∂x∂z​=2x+3y
  • ∂z∂y=3x\frac{\partial z}{\partial y} = 3x∂y∂z​=3x
  • ∂x∂u=v\frac{\partial x}{\partial u} = v∂u∂x​=v
  • ∂y∂u=1\frac{\partial y}{\partial u} = 1∂u∂y​=1

Step 2: Calculate. ∂z∂u=(2x+3y)(v)+(3x)(1)\frac{\partial z}{\partial u} = (2x + 3y)(v) + (3x)(1)∂u∂z​=(2x+3y)(v)+(3x)(1) ∂z∂u=(2uv+3(u+v))v+3uv=2uv2+3uv+3v2+3uv=2uv2+6uv+3v2\frac{\partial z}{\partial u} = (2uv + 3(u+v))v + 3uv = 2uv^2 + 3uv + 3v^2 + 3uv = 2uv^2 + 6uv + 3v^2∂u∂z​=(2uv+3(u+v))v+3uv=2uv2+3uv+3v2+3uv=2uv2+6uv+3v2


Checkpoint Questions

  1. Identify: In the dependency w=f(x,y,z)w = f(x, y, z)w=f(x,y,z) where x=g(r,s)x=g(r, s)x=g(r,s), y=h(r,s)y=h(r, s)y=h(r,s), and z=k(r,s)z=k(r, s)z=k(r,s), which variables are independent and which are intermediate?
  2. Logic: Why do we add the terms in the Chain Rule formula rather than multiplying the entire expression?
  3. Visualization: Draw a tree diagram for w=f(x,y)w = f(x, y)w=f(x,y) where x=g(u,v)x = g(u, v)x=g(u,v) and y=h(u,v)y = h(u, v)y=h(u,v). How many paths lead to vvv?
  4. Application: If zzz is a function of xxx and yyy, and xxx and yyy are functions of ttt, what is the difference between ∂z∂x\frac{\partial z}{\partial x}∂x∂z​ and dzdt\frac{dz}{dt}dtdz​?
▶Click to view answers
  1. Independent: r,sr, sr,s. Intermediate: x,y,zx, y, zx,y,z.
  2. Because each intermediate variable provides an additive contribution to the total change of the output.
  3. Two paths lead to vvv (one through xxx, one through yyy).
  4. ∂z∂x\frac{\partial z}{\partial x}∂x∂z​ is the rate of change of zzz as only xxx changes; dzdt\frac{dz}{dt}dtdz​ is the total rate of change of zzz as ttt flows through both xxx and yyy.
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