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21-259: Calculus in Three Dimensions

Lecture #9
Updated February 20, 2025

Relevant Textbook Sections: OpenStax 4.5-4.6, Stewart 6th: 14.5-14.6

The Chain Rule

Recall that when y = f (x) and x = g (t ) where f and g are differentiable functions, that y is also a
differentiable function of t as y = f (g (t )) and thus the Chain Rule gives

dy dy dx
= .
dt dx dt
The same is true for multivariate functions, however there may be more than one “path” to the inde-
pendent variable(s).

∂z z ∂z
The Chain Rule (Case I): Suppose that z = f (x, y) is a differen-
∂x ∂y
tiable function of x and y, where x = g (t ) and y = h(t ) are both
differentiable functions of t . Then z is a differentiable function
of t and x y
d z ∂z d x ∂z d y dx dy
= + .
d t ∂x d t ∂y d t dt dt
t t

dz
Example 1. Find if z = arctan(y/x), x = e t , and y = 1 − e −t .
dt

dw
Example 2. Find when t = 1 if w = xe y/z , x = t 2 , y = 1 − t , and z = 1 + 2t .
dt

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The Chain Rule (Case II): Suppose that z = f (x, y) is a differentiable function of x and y, where
x = g (s, t ) and y = h(s, t ) are both differentiable functions of s and t . Then z is a differentiable
function of s and t , and

∂z ∂z ∂x ∂z ∂y ∂z ∂z ∂x ∂z ∂y
= + , and = + .
∂s ∂x ∂s ∂y ∂s ∂t ∂x ∂t ∂y ∂t

∂z z ∂z ∂z z ∂z
∂x ∂y ∂x ∂y

x y x y
∂x ∂y ∂x ∂y
∂s ∂s ∂t ∂t

s t s t s t s t
∂z ∂z ∂x ∂z ∂y ∂z ∂z ∂x ∂z ∂y
= + = +
∂s ∂x ∂s ∂y ∂s ∂t ∂x ∂t ∂y ∂t

∂z ∂z ∂z
Example 3. Let z = x 2 + x y 3 where x = w v 2 + w and y = u + ve w . Find , , and when u = 2, v =
∂u ∂v ∂w
1, w = 0.

If z = f (x, y) is an implicitly-defined function by an equation of the form F (x, y, z) = 0, then we can use
the chain rule to compute the derivative with respect to x or y implicitly:

∂F ∂x ∂F ∂y ∂F ∂z
+ + = 0.
∂x ∂x ∂y ∂x ∂z ∂x

∂F ∂F ∂z ∂z
Since ∂x/∂x = 1 and ∂y/∂x = 0, we have + = 0, which can be solved for .
∂x ∂z ∂x ∂x

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Theorem: If F (x, y, z) = 0 defines z implicitly as a function of x and y, and the below partial
derivatives all exist, then
∂z Fx ∂z Fy
=− , =− .
∂x Fz ∂y Fz

dy dy Fx
This can also by used to find when F (x, y) = 0: =− .
dx dx Fy

dy 2
Example 4. Find if y 5 + x 2 y 3 = 1 + ye x .
dx

∂z
Example 5. Find if x y z = cos(x + y + z).
∂x

Directional Derivatives and the Gradient Vector

Definition: The directional derivative of f (x, y) at (x 0 , y 0 ) in the direction of a unit vector u =

〈a, b〉 is
f (x 0 + ha, y 0 + hb) − f (x 0 , y 0 )
D u f (x 0 , y 0 ) = lim .
h →0 h
If f = f (x, y, z), the directional derivative in the direction of the unit vector u = 〈a, b, c〉, then

f (x 0 + ha, y 0 + hb, z 0 + hc) − f (x 0 , y 0 , z 0 )

D u f (x 0 , y 0 , z 0 ) = lim .
h →0 h

In general, the directional derivative of f at x 0 in the direction of the unit vector u is

f (x 0 + hu) − f (x 0 )
D u f (x 0 ) = lim .
h →0 h

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Theorem: If f is a differentiable function of x and y, then f has

directional derivative in the direction of unit vector u = 〈a, b〉 and

D u f (x, y) = f x (x, y)a + f y (x, y)b.

If f is a differentiable function of x, y and z, then f has directional

derivative in the direction of unit vector u = 〈a, b, c〉 and

D u f (x, y, z) = f x (x, y, z)a + f y (x, y, z)b + f z (x, y, z)c.

Example 6. Find the directional derivative of the function f (x, y) = x 2 y 3 − 4y at the point (2, −1) in the
direction of the vector v = 2ı + 5 ȷ.

Example 7. Find the directional derivative of the function f (x, y) = x 3 − 3x y + 4y 2 in the direction of
the unit vector that makes an angle of π/6 with the positive x-axis.

p ­2
, 3, 6
®
Example 8. Find D u f at the point (1, 3, 1) if f (x, y, z) = x + y z and u = 7 7 7
.

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Definition: Let f be a differentiable function of multiple variables x 1 , . . . , x n . The gradient of f is

the vector function ∇ f given by
∂f ∂f
¿ À
∇f = ,..., .
∂x 1 ∂x n
­ ® ­ ®
For example, if f = f (x, y), ∇ f = f x , f y , and if f = f (x, y, z), then ∇ f = f x , f y , f z . An alternate
notation for ∇ f is grad f .

The operator ∇ (pronounced “grad” or “del”) is a differential operator, i.e., it is something that we can
apply to functions to produce other functions. Specifically, in three dimensions,

∂ ∂ ∂ ∂ ∂ ∂
¿ À
∇= ı+ ȷ+ k= , , ,
∂x ∂y ∂z ∂x ∂y ∂z

so ∇ f is really right scalar multiplication of ∇ by f . Note: we have that D u f = ∇ f · u.

Example 9. If f (x, y, z) = x sin y z, find ∇ f and the directional derivative of f at (1, 3, 0) in the direction
of v = ı + 2 ȷ − k.

Theorem: Let f be a differentiable function of multiple variables. The maximum value of D u f (x)
is |∇ f (x)| and it occurs when u has the same direction as ∇ f (x).

Example 10. Suppose that the temperature at a point (x, y, z) in space is given by

T (x, y, z) = 80/(1 + x 2 + 2y 2 + 3z 2 )2 ,

where T is measured in degrees Celsius and x, y, z are in meters. In which direction does the temper-
ature increase the fastest at the point (1, 1, −2)? What is the maximum rate of increase?

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The gradient gives the direction of maximum increase of a function at a point. Evaluated at x 0 ,
the vector ∇ f (x 0 ) is orthogonal to the level curves/surfaces of f that pass through the point x 0 .

Example 11. Find the equations of the tangent plane and the nor-
mal line at the point (−2, 1, −3) to the ellipsoid

x2 z2
+ y2 + = 3.
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Exercises
1. OpenStax Problems:

(a) Section 4.5: 215-220, 229, 230-234, 244-247

(b) Section 4.6: 263-287 odd, 291, 295, 297, 301, 303, 305

2. Stewart 6th Ed. Problems:

(a) Section 14.5: 1-34, 38, 39, 45-48

(b) Section 14.6: 4-17, 19-26, 28-34, 39-42

Exercise Answers
1. See OpenStax solutions.
2. See Stewart 6th Ed. solutions.

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