z = f(x(t),y(t)) and , where the ordinary derivatives are evaluated at t ...

11.5 The Chain Rule
1. Chain Rule: One Independent Variable: Suppose that z = f(x, y)
has continuous first-order partial derivatives (and so f is
differentiable) and that x = g(t) y = h(t) are differentiable functions
of t. Then z is a differentiable function of t given by
z = f (x(t), y(t)) and
dz
dt = ∂f dx
∂x dt + ∂f dy
∂y dt = ∂z dx
∂x dt + ∂z dy
∂y dt ,
where the ordinary derivatives are evaluated at t and the partial
derivatives are evaluated at (x, y).
a. For this version of the chain rule, z is called the dependent
variable, x and y are called the intermediate variables, and t
is called the independent variable.
b. This derivative can be interpreted as the rate of change of z with
respect to t as (x, y) moves along the curve C given by the
parametric equations x = g(t) y = h(t).
c. This formula can be represented schematically using a "tree
diagram''.
1. Start with z at the top and move down using lines to connect
each variable above to the variables below that it depends on
directly. (Moving down: dependent to intermediate to
independent.)
2. Label each branch (line) with the derivative whose "numerator"
contains the variable at the top of the branch and whose
"denominator" contains the variable below.
3. To find the formula for dz , form the sum of the terms found
dt
by multiplying the expressions along each branch of the tree.
d. This extends to a function of more than two intermediate variables.
(In the formula, just include a term for each intermediate variable.

In the tree diagram, include a branch for each intermediate
variable.)
e. This version of the chain rule has applications to related rates,
where t is interpreted as time.
2. Chain Rule: Two Independent Variables: Suppose that z = f(x, y)
has continuous first-order partial derivatives (and so f is
differentiable) and that x = g(s, t) y = h(s, t) are differentiable
functions of s and t. Then z is a differentiable function of s and t
given by z = f (x(s,t), y(s,t)) and
∂z
∂s = ∂z ∂x
∂x ∂s + ∂z ∂y
∂y ∂s
∂z
∂t = ∂z ∂x
∂x ∂t + ∂z ∂y
∂y ∂t .
a. Note that in this case, there are two independent variables.
b. To prove hold one of the independent variables fixed and
differentiate using the one independent variable case above.
c. In the tree diagram, there will be two branches below each of
the intermediate variables, one for each independent variable.
To find each partial derivative of z, add the products of the
expressions along the branches that end in the required
independent variable.
3. Chain Rule: The General Case: Suppose that w is a differentiable
function of the variables x 1 , x 2 , . . . , x m and that each of these
variables is a differentiable function of the variables t 1 , t 2 , … , t n .
Then,
∂w
∂w
∂x1
∂w
∂x2
∂w
∂xm
= + + L +
∂t
∂x
∂t
∂x
∂t
∂x
∂t
i
for each i = 1,2,...,n.
1
i
2
i
m
i

(Note that there are as many terms in each partial of w as there are
intermediate variables and as many partial derivatives of w as
independent variables.)
A tree diagram will have as many branches below w as intermediate
variables and each of these intermediate variables will have as many
branches as it has independent variables.
4. Implicit Differentiation: Suppose that the function F(x, y) is
differentiable (for example, it has continuous first partial derivatives)
and that the equation F(x, y) = 0 implicitly defines a differentiable
function y = f(x). Then the one independent variable case of the
chain rule gives
∂F
dy
dx =− ∂x
∂F
∂y
=− F x
F y
, where ∂F
∂x ≠ 0.
5. Implicit Function Theorem: If F is defined on a disk containing
(a, b), where F(a, b) = 0, F y (a,b) ≠ 0, and F x and F y are both
continuous on the disk, then the equation F(x, y) = 0, defines y as a
function of x near the point (a, b) and the derivatives of this function
are given by the derivative given above in 4.
6. Implicit Partial Differentiation: Suppose that the function F(x, y, z)
is differentiable (for example, it has continuous first partial
derivatives) and that the equation F(x, y, z) = 0 implicitly defines a
differentiable function z = f(x, y) (for example, it has continuous first
partial derivatives). Then
∂z
∂x =−F x
F z
and
∂z
∂y =−F y
F z
, were F z = ∂F
∂z ≠ 0.
7. Implicit Function Theorem: If F is defined on a sphere containing
(a, b, c), where F(a, b, c) = 0, F y (a,b,c) ≠ 0 , and F x , F y , and F z are
both continuous inside the sphere, then the equation F(x, y, z) = 0,
defines z as a function of x and y near the point (a, b, c) and the
partial derivatives are given by the derivative given above in 6.