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PAGE 1

15.4 The Chain Rule

if \( y = f(x) \) and \( x = g(t) \) then the Chain Rule is used to find the rate of change of \( y \) with respect to \( t \)

\[ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} \]
how \( t \) affects \( y \)
how \( y \) is affected by its own variable \( x \)
how \( t \) affects \( x \)

a good way to track variable relationships is a tree diagram

\( \frac{dy}{dt} \)
A vertical tree diagram showing variable y at the top, connected to x, which is connected to t at the bottom.
\( \frac{dy}{dx} \)
\( \frac{dx}{dt} \)
each step down is a derivative
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  • partial derivative \( \rightarrow \) isolates effect from one variable
  • multiple variables: isolate effects, then combine

example

\( z = f(x, y) = x^3 + y^2 \)
\( x = \cos(t) \quad y = \ln(t) \)

find \( \frac{dz}{dt} \)

tree:
\( \frac{dz}{dt} \)
A branching tree diagram starting from z, splitting into two paths: one through x to t, and one through y to t.

find how \( z \) is affected by \( t \)

one branch at a time

1

pretend \( y \) is not doing anything

\[ \frac{\partial z}{\partial x} \frac{dx}{dt} \]

\( \swarrow \) partial down through a split

2

pretend \( x \) is doing nothing

\[ \frac{\partial z}{\partial y} \frac{dy}{dt} \]
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Multivariable Chain Rule: Putting it Together

put them together

partial : split
regular : no split
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \]
put answer in terms of this (t)
\[ = (3x^2)(-\sin t) + (2y)(\frac{1}{t}) \]
\[ = -3 \cos^2 t \sin t + 2 \frac{\ln t}{t} \]

Quick check: what if we sub in \(x, y\) as functions of \(t\) initially?

\[ z = x^3 + y^2 \]
\[ x = \cos t \]
\[ y = \ln t \]
\[ z = (\cos t)^3 + (\ln t)^2 \]
\[ \frac{dz}{dt} = 3 (\cos t)^2 \cdot (-\sin t) + 2 (\ln t) \cdot \frac{1}{t} \]

same

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Example: Chain Rule with Multiple Independent Variables

example

\[ z = \sin(x + y) \]\[ x = u^2 + v \]\[ y = 1 - 2uv \]

find: \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \)

ultimately \(z\) is a function of \(u\) and \(v\)

\( \rightarrow z = f(u, v) \)

tree:

A partial derivative tree diagram showing variable z branching to x and y, which both branch down to u and v.
use branches w/ u at bottom
\[ \frac{\partial z}{\partial u} = \underbrace{\frac{\partial z}{\partial x} \frac{\partial x}{\partial u}}_{(1)} + \underbrace{\frac{\partial z}{\partial y} \frac{\partial y}{\partial u}}_{(2)} \]
\[ = \cos(x+y) \cdot 2u + \cos(x+y)(-2v) \]
put everything in terms of u (and possibly v)
\[ = \cos(u^2 + v + 1 - 2uv)(2u) + \cos(u^2 + v + 1 - 2uv)(-2v) \]
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\[ \frac{\partial z}{\partial v} = \underbrace{\frac{\partial z}{\partial x} \frac{\partial x}{\partial v}}_{\text{①}} + \underbrace{\frac{\partial z}{\partial y} \frac{\partial y}{\partial v}}_{\text{②}} \]
\[ = \cos(x+y) \cdot (1) + \cos(x+y) \cdot (-2u) \]
\[ = \cos(x+y)(1-2u) = \boxed{\cos(u^2+v+1-2uv)(1-2u)} \]

Implicit Differentiation

\( x^2 + 2y^2 = 4 \) \( y \) is an implicit function of \( x \) (\( y = f(x) \))

find \( \frac{dy}{dx} \)

from calc 1:

\[ \frac{d}{dx}(x^2 + 2y^2) = \frac{d}{dx}(4) \]
\[ 2x + 4y \cdot \frac{dy}{dx} = 0 \]

solve for \( \frac{dy}{dx} \) : \( \frac{dy}{dx} = -\frac{x}{2y} \)

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we can use the multivariable chain rule to find the same result

\( x^2 + 2y^2 = 4 \) \( y \) is an implicit function of \( x \) (\( y = f(x) \))

create a new function: \( F(x,y) = x^2 + 2y^2 - 4 = 0 = g(x) \)

because \( y = \text{some function of } x \)

tree: \( g(x) = F(x,y) = 0 \)

find \( \frac{dy}{dx} \) : Step down the tree

\[ \underbrace{\frac{dg}{dx}}_{\text{①}} = \frac{\partial F}{\partial x} + \underbrace{\frac{\partial F}{\partial y} \boxed{\frac{dy}{dx}}}_{\text{②}} = 0 \]

because \( g = 0 \) so \( \frac{dg}{dx} = 0 \)

solve for \( \frac{dy}{dx} \) :

\[ \frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y} = -\frac{2x}{4y} = \boxed{-\frac{x}{2y}} \]
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can do this w/ more variables ("calc 1" way gets harder)

example

\[ xy + yz + xz = 3 \]

\( z \) is an implicit function of \( x \) and \( y \) \( \longrightarrow z = f(x, y) \)

and \( x \) and \( y \) don't depend on anything else

find \( \frac{\partial z}{\partial y} \)

define \( F(x, y, z) = xy + yz + xz - 3 = 0 = g(x, y) \)

\( g(x, y) = F(x, y, z) = 0 \)

A tree diagram showing function F branching to x, y, and z. The z branch further splits into x and y.
\[ \frac{\partial z}{\partial y} : \text{collect branches w/ } y \text{ at base} \]
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\( g(x, y) = F(x, y, z) = 0 \)

\[ \frac{\partial}{\partial y}(g) = \underbrace{\frac{\partial F}{\partial y}}_{\text{①}} + \underbrace{\frac{\partial F}{\partial z} \cdot \frac{\partial z}{\partial y}}_{\text{②}} = 0 \]

Note: \( \frac{\partial z}{\partial y} \) is what we want this. The equation equals 0 because \( g \) is constant.

\[ \frac{\partial z}{\partial y} = - \frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}} = \boxed{-\frac{(x+z)}{y+x}} \]