Partial Derivatives

Partial Derivatives
P. Sam Johnson
August 30, 2019
P. Sam Johnson Partial Derivatives August 30, 2019 1/117

Overview
The calculus of several variables is similar to single-variable calculus

applied to several variables one at a time. When we hold all but one of the
independent variables of a function constant and differentiate with respect
to that one variable, we get a partial derivative.
In this lecture, we see how partial derivatives are defined and interpreted
geometrically, and how to calculate them by applying the rules for
differentiating functions of a single variable.
The idea of differentiability for functions of several variables requires more

than the existence of the partial derivatives, but we will see that
differentiable functions of several variables behave in the same way as
differentiable single-variable functions.

Partial Derivatives of a Function of Two Variables
If (x0 , y0 ) is a point in the domain of a function f (x, y ), the vertical plane

y = y0 will cut the surface z = f (x, y ) in the curve z = f (x, y0 ).
The picture shows that the intersec-

tion of the plane y = y0 with the sur-
face z = f (x, y ), viewed from above
the first quadrant of the xy -plane.

This curve is the graph of the func-

tion z = f (x, y0 ) in the plane y =
y0 . The horizontal coordinate in this
plane is x; the vertical coordinate is
z. The y -value is held constant at
y0 , so y is not a variable.

We define the partial derivative of f with respect to x at the point (x0 , y0 )

as the ordinary derivative of f (x, y0 ) with respect to x at the point x = x0 .
To distinguish partial derivatives from ordinary derivatives we use the

symbol ∂ rather than the d previously used. The partial derivative symbol
is variously pronounced as “del”, “partial dee”, “doh”, or “dabba”.
In the definition, h represents a real number, positive or negative.

Definition 1.
The partial derivative of f(x,y) with respect to x at the point (x0 , y0 )
is

∂f f (x0 + h, y0 ) − f (x0 , y0 )
= lim ,
∂x (x0 ,y0 ) h→0 h
provided the limit exists.

An equivalent expression for the partial derivative is

d
f (x, y0 ) .
dx x=x0
The slope of the curve z = f (x, y0 ) at the point P(x0 , y0 , f (x0 , y0 )) in the
plane y = y0 is the value of the partial derivative of f with respect to x at
(x0 , y0 ).
The tangent line to the curve at P is the line in the plane y = y0 that
passes through P with this slope. The partial derivative ∂f /∂x at (x0 , y0 )
gives the rate of change of f with respect to x when y is held fixed at the
value y0 .

We use several notations for the partial derivative:

∂f ∂z ∂f ∂z
(x0 , y0 ) or fx (x0 , y0 ), , and fx , , zx , or .
∂x ∂x (x0 ,y0 )
∂x ∂x
The definition of the partial derivative of f (x, y ) with respect to y at a
point (x0 , y0 ) is similar to the definition of the partial derivative of f with
respect to x. We hold x fixed at the value x0 and take the ordinary
derivative of f (x0 , y ) with respect to y at y0 .
Definition 2.
The partial derivative of f(x,y) with respect to y at the point (x0 , y0 )
is

∂f d f (x0 , y0 + h) − f (x0 , y0 )
= f (x0 , y ) = lim ,
∂y (x0 ,y0 )
dy y =y0 h→0 h
provided the limit exists.

The intersection of the plane x = x0

with the surface z = f (x, y ), viewed
from above the first quadrant of the
xy -plane.

The slope of the curve z = f (x0 , y ) at the point P(x0 , y0 , f (x0 , y0 )) in the
vertical plane x = x0 is the partial derivative of f with respect to y at
(x0 , y0 ). The tangent line to the curve at P is the line in the plane x = x0
that passes through P with this slope. The partial derivative gives the rate
of change of f with respect to y at (x0 , y0 ) when x is held fixed at the
value x0 .
The partial derivative with respect to y is denoted the same way as the
partial derivative with respect to x:
∂f ∂f
(x0 , y0 ), fy (x0 , y0 ), , fy .
∂y ∂y

The partial derivatives of f at (a, b) are the slopes of the tangents to C1

and C2 .

The graph of f is the paraboloid z = 4 − x 2 − 2y 2 and the vertical plane

y = 1 intersects it in the parabola z = 2 − x 2 , y = 1. The slope of the
tangent line to this parabola at the point (1, 1, 1) is fx (1, 1) = −2.
Similarly, the curve C2 in which the plane x = 1 intersects the paraboloid

is the parabola z = 3 − 2y 2 , x = 1, and the slope of the tangent line at
(1, 1, 1) is fy (1, 1) = −4.

The following figures are computer-drawn counterparts to the last figures.

Part (a) shows the plane y = 1 intersecting the surface to form the curve
Ci and part (b) shows the curve Ci and the tangent line Ti , i = 1, 2.

Notice that we now have two tangent lines associated with the surface
z = f (x, y ) at the point P(x0 , y0 , f (x0 , y0 )).
The tangent lines at the point

(x0 , y0 , f (x0 , y0 )) determine a plane
that, in this picture at least, appears
to be tangent to the surface.
Is the plane they determine tangent to the surface at P? We will see that
it is for the differentiable functions, we will learn how to find the tangent
plane. First we have to learn more about partial derivatives themselves.

Calculations
The definitions of ∂f /∂x and ∂f /∂y give us two different ways of

differentiating f at a point: with respect to x in the usual way while
treating y as a constant and with respect to y in the usual way while
treating x as a constant. As the following examples show, the values of
these partial derivatives are usually different at a given point (x0 , y0 ).
Example 3.
Find the values of ∂f /∂x and ∂f /∂y at the point (4, −5) if
f (x, y ) = x 2 + 3xy + y − 1.
Solution : To find ∂f /∂x, we treat y as a constant and differentiate with
respect to x:
∂f ∂ 2
= (x + 3xy + y − 1) = 2x + 3 · 1 · y + 0 − 0 = 2x + 3y .
∂x ∂x
The value of ∂f /∂x at (4, −5) is 2(4) + 3(−5) = −7.
Solution (contd...)
To find ∂f /∂y , we treat x as a constant and differentiate with respect to

y:
∂f ∂ 2
= (x + 3xy + y − 1) = 0 + 3 · x · 1 + 1 − 0 = 3x + 1.
∂y ∂y
The value of ∂f /∂y at (4, −5) is 3(4) + 1 = 13.
Example 4.
Find ∂f /∂y as a function if f (x, y ) = y sin xy .
Solution : We treat x as a constant and f as a product of y and sin xy :

∂f ∂ ∂ ∂
= (y sin xy ) = y sin xy + (sin xy ) (y )
∂y ∂y ∂y ∂y
∂
= (y cos xy ) (xy ) + sin xy = xy cos xy + sin xy .
∂y
Example
Example 5.
2y
Find fx and fy as functions if f (x, y ) = y +cos x .
Solution : We treat f as a quotient. With y held constant, we get

∂ ∂
∂ 2y (y + cos x) ∂x (2y ) − 2y ∂x (y + cos x)
fx = ( )= 2
∂x y + cos x (y + cos x)
(y + cos x)(0) − 2y (− sin x) 2y sin x
= 2
= .
(y + cos x) (y + cos x)2
With x held constant, we get
∂ ∂
∂ 2y (y + cos x) ∂y (2y ) − 2y ∂y (y + cos x)
fy = ( )=
∂y y + cos x (y + cos x)2
(y + cos x)(2) − 2y (1) 2 cos x
= 2
= .
(y + cos x) (y + cos x)2
Solution
Implicit differentiation works for partial derivatives the way it works for
ordinary derivatives, as the next example illustrates.
Example 6.
Find ∂z/∂x if the equation yz − ln z = x + y defines z as a function of the
two independent variables x and y and the partial derivative exists.
We differentiate both sides of the equation with respect to x, holding y

constant and treating z as a differentiable function of x:
∂ ∂ ∂x ∂y
(yz) − ln z = +
∂x ∂x ∂x ∂x
1 ∂z
y− =1
z ∂x
∂z z
= .
∂x yz − 1
Example
Example 7.
The plane x = 1 intersects the paraboloid z = x 2 + y 2 in a parabola. Find
the slope of the tangent to the parabola at (1, 2, 5).
The tangent to the curve of intersec-

tion of the plane x = 1 and surface
z = x 2 + y 2 at the point (1, 2, 5).

Solution
The slope is the value of the partial derivative ∂z/∂y at (1, 2):

∂z ∂ 2 2

= (x + y ) = 2y |(1,2) = 2(2) = 4.
∂y (1,2)
∂y (1,2)
As a check, we can treat the parabola as the graph of the single-variable

function z = (1)2 + y 2 = 1 + y 2 in the plane x = 1 and ask for the slope
at y = 2. The slope, calculated now as an ordinary derivative, is

dz d 2

= (1 + y ) = 2y |y =2 = 4.
dy y =2
dy y =2

Functions of More Than Two Variables
The definitions of the partial derivatives of functions of more than two

independent variables are like the definitions for functions of two variables.
They are ordinary derivatives with respect to one variable, taken while the
other independent variables are held constant.
Example 8.
If x, y , and z are independent variables and
f (x, y , z) = x sin(y + 3z),
then
∂f ∂ ∂
= [x sin(y + 3z)] = x sin(y + 3z)
∂z ∂z ∂z
∂
= x cos(y + 3z) (y + 3z) = 3x cos(y + 3z).
∂z

Example
Example 9.
If resistors of R1 ,R2 , and R3 ohms are connected in parallel to make an
R-ohm resistor, the value of R can be found from the equation
1 1 1 1
= + + .
R R1 R2 R3
Find the value of ∂R/∂R2 when R1 = 30, R2 = 45, and R3 = 90 ohms.
Resistors arranged this way are said to be connected

in parallel. Each resistor lets a portion of the current
through. Their equivalent resistance R is calculated
with the formula
1 1 1 1
= + + .
R R1 R2 R3

Solution
To find ∂R/∂R2 , we treat R1 and R3 as constants and, using implicit

differentiation, differentiate both sides of the equation with respect to R2 :

∂ 1 ∂ 1 1 1
= + +
∂R2 R ∂R2 R1 R2 R3
1 ∂R 1
− 2 =0− 2 +0
R ∂R2 R2
2
2
∂R R R
= 2 = .
∂R2 R2 R2

Solution (contd...)
When R1 = 30, R2 = 45, and R3 = 90,

1 1 1 1 3+2+1 6 1
= + + = = = ,
R 30 45 90 90 90 15
so R = 15 and
2 2
∂R 15 1 1
= = = .
∂R2 45 3 9
Thus at the given values, a small change in the resistance R2 leads to a

change in R about 1/9th as large.

Partial Derivatives and Continuity
A function f (x, y ) can have partial derivatives with respect to both

x and y at a point without the function being continuous there.
This is different from functions of a single variable, where the

existence of a derivative implies continuity.
We shall see that if the partial derivatives of f (x, y ) exist and are
continuous throughout a disk centered at (x0 , y0 ), however, then f is
continuous at (x0 , y0 ).

Partial Derivatives and Continuity
Example 10.
(
0, xy 6= 0
Let f (x, y ) =
1, xy = 0.
(a) Find the limit of f as (x, y ) approaches (0, 0) along the line y = x.
(b) Prove that f is not continuous at the origin.
(c) Show that both partial derivatives ∂f /∂x and ∂f /∂y exist at the
origin.
(
0, xy 6= 0
The graph of f (x, y ) = consists of
1, xy = 0
the linesL1 and L2 and the four open quadrants of
the xy -plane. The function has partial derivatives
at the origin but is not continuous there.

Solution
(a) Since f (x, y ) is constantly zero along the line y = x (except at the
origin), we have

lim f (x, y ) = lim 0 = 0.
(x,y )→(0,0) y =x (x,y )→(0,0)
(b) Since f (0, 0) = 1, the limit in part (a) proves that f is not continuous
at (0, 0).
(c) To find ∂f /∂x at (0, 0), we hold y fixed at y = 0. Then f (x, y ) = 1
for all x, and the graph of f is the line L1 in the figure. The slope of
this line at any x is ∂f /∂x = 0. In particular, ∂f /∂x = 0 at (0, 0).
Similarly, ∂f /∂y is the slope of line L2 at any y , so ∂f /∂y = 0 at
(0, 0).

Relation Between Differentiability and Continuity
The above example suggests that we need a stronger requirement

for differentiability in higher dimensions than the mere existence of
the partial derivatives.
We shall define differentiability for functions of two variables (which

is slightly more complicated than for single-variable functions) and
then revisit the connection to continuity.

Second-Order Partial Derivatives
When we differentiate a function f (x, y ) twice, we produce its

second-order derivatives. These derivatives are usually denoted by
∂2f ∂2f ∂2f ∂2f
or fxx , or fyy , or fyx , and or fxy .
∂x 2 ∂y 2 ∂x∂y ∂y ∂x
The defining equations are
∂2f ∂2f

∂ ∂f ∂ ∂f
2
= , = ,
∂x ∂x ∂x ∂x∂y ∂x ∂y
and so on.
Notice the order in which the mixed partial derivatives are taken:
∂2f
Differentiate first with respect to y , then with respect to x.
∂x∂y
fyx = (fy )x Means the same thing.

Example
Example 11.
If f (x, y ) = x cos y + ye x , find the second-order derivatives
∂2f ∂2f ∂2f ∂2f

, , , and .
∂x 2 ∂y ∂x ∂y 2 ∂x∂y
Solution : The first step is to calculate both first partial derivatives.

∂f ∂ ∂f ∂
= (x cos y + ye x ) = (x cos y + ye x )
∂x ∂x ∂y ∂y
x
= cos y + ye = − x sin y + e x
Now we find both partial derivatives of each first partial:
∂2f ∂2f

∂ ∂f ∂ ∂f
= = − sin y + e x = = − sin y + e x
∂y ∂x ∂y ∂x ∂x∂y ∂x ∂y
∂2f ∂2f

∂ ∂f ∂ ∂f
= = ye x . = = −x cos y .
∂x 2 ∂x ∂x ∂y 2 ∂y ∂y

The Mixed Derivative Theorem
We have noticed in the above example that the mixed second-order partial
derivatives
∂2f ∂2f
and
∂y ∂x ∂x∂y
are equal. This is not a coincidence. They must be equal whenever
f , fx , fy , fxy , and fyx are continuous, as stated in the following theorem.
Theorem 12 (The Mixed Derivative Theorem).

If f (x, y ) and its partial derivatives fx , fy , fxy , and fyx are defined
throughout an open region containing a point (a, b) and are all continuous
at (a, b), then
fxy (a, b) = fyx (a, b).

Clairaut’s Theorem
The Mixed Derivative Theorem is also known as Clairaut’s Theorem,

named after the French mathematician Alexis Clairaut who discovered it.
The Mixed Derivative Theorem says that to calculate a mixed

second-order derivative, we may differentiate in either order, provided the
continuity conditions are satisfied. This ability to proceed in different order
sometimes simplifies our calculations.
Example 13.
Find ∂ 2 w /∂x∂y if
ey
w = xy + .
y2 + 1

Solution
The symbol ∂ 2 w /∂x∂y tells us to differentiate first with respect to y and

then with respect to x. However, if we interchange the order of
differentiation and differentiate first with respect to x we get the answer
more quickly. In two steps,
∂w ∂2w
=y and = 1.
∂x ∂y ∂x
If we differentiate first with respect to y , we obtain ∂ 2 w /∂x∂y = 1 as

well. We can differentiate in either order because the conditions of “The
Mixed Derivative Theorem” hold for w at all points (x0 , y0 ).

Partial Derivatives of Still Higher Order
Although we will deal mostly with first- and second-order partial

derivatives, because these appear the most frequently in applications, there
is no theoretical limit to how many times we can differentiate a function as
long as the derivatives involved exist. Thus, we get third- and fourth-order
derivatives denoted by symbols like
∂3f
= fyyx ,
∂x∂y 2
∂4f
= fyyxx ,
∂x 2 ∂y 2
and so on. As with second-order derivatives, the order of differentiation is
immaterial as long as all the derivatives through the order in question are
continuous.

Example 14.
Find fyxyz if f (x, y , z) = 1 − 2xy 2 z + x 2 y .
Solution : We first differentiate with respect to the variable y , then x,

then y again, and finally with respect to z:
fy = −4xyz + x 2
fyx = −4yz + 2x
fyxy = −4z
fyxyz = −4.

Differentiability
The starting point for differentiability is not the difference quotient we saw
in studying single-variable functions, but rather the idea of increment.
Recall that if y = f (x) is differentiable at x = x0 , then the change in the
value of f that results from changing x from x0 to x0 + ∆x is given by an
equation of the form
∆y = f 0 (x0 )∆x + ε∆x
in which ε → 0 as ∆x → 0.
For functions of two variables, the analogous property becomes the

definition of differentiability.

Increment Theorem for Functions of Two Variables
Theorem 15 (The Increment Theorem for Functions of Two
Variables).
Suppose that the first partial derivatives of f (x, y ) are defined throughout
an open region R containing the point (x0 , y0 ) and that fx and fy are
continuous at (x0 , y0 ). Then the change
∆z = f (x0 + ∆x, y0 + ∆y ) − f (x0 , y0 )
in the value of f that results from moving from (x0 , y0 ) to another point
(x0 + ∆x, y0 + ∆y ) in R satisfies an equation of the form
∆z = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + ε1 ∆x + ε2 ∆y
in which each of ε1 , ε2 → 0 as both ∆x, ∆y → 0.

Differentiable Functions
Definition 16.
A function z = f (x, y ) is differentiable at (x0 , y0 ) if fx (x0 , y0 ) and
fy (x0 , y0 ) exist and ∆z satisfies an equation of the form
∆z = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + ε1 ∆x + ε2 ∆y
in which each of ε1 , ε2 → 0 as both ∆x, ∆y → 0. We call f

differentiable if it is differentiable at every point in its domain, and say
that its graph is a smooth surface.
Because of this definition, an immediate corollary of “Increment Theorem

for Functions of Two Variables” is that a function is differentiable at
(x0 , y0 ) if its first partial derivatives are continuous there.

Differentiable Functions
Corollary 17.
If the partial derivatives fx and fy of a function f (x, y ) are continuous
throughout an open region R, then f is differentiable at every point of R.
If z = f (x, y ) is differentiable, then the definition of differentiability

assures that ∆z = f (x0 + ∆x, y0 + ∆y ) − f (x0 , y0 ) approaches 0 as ∆x
and ∆y approach 0. This tells us that a function of two variables is
continuous at every point where it is differentiable.
Theorem 18 (Differentiability Implies Continuity).

If a function f (x, y ) is differentiable at (x0 , y0 ), then f is continuous at
(x0 , y0 ).

Remark
A function f (x, y ) must be continuous at a point (x0 , y0 ) if fx and fy are

continuous throughout an open region containing (x0 , y0 ).
Remember, however, that it is still possible for a function of two variables

to be discontinuous at a point where its first partial derivatives exist.
Existence alone of the partial derivatives at that point is not enough, but
continuity of the partial derivatives guarantees differentiability.

Recall
The Chain Rule for functions of a single variable says that when
w = f (x) is a differentiable function of x, and

x = g (t) is a differentiable function of t,
then
w is a differentiable function of t, and

dw dw dx
dw /dt can be calculated by the formula dt = dx dt .
For functions of two or more variables, the Chain Rule has several forms.
The form depends on how many variables are involved, but once this is
taken into account, it works like the Chain Rule for functions of a single
variable.

Functions of Two Variables
The Chain Rule formula for a differentiable function w = f (x, y ) when

x = x(t) and y = y (t) are both differentiable functions of t is given in the
following theorem.
Theorem 19 (Chain Rule for Functions of Two Independent

Variables).
If w = f (x, y ) is differentiable and if x = x(t),y = y (t) are differentiable
functions of t, then the composite w = f (x(t), y (t)) is a differentiable
function of t and
dw
= fx (x(t), y (t)) · x 0 (t) + fy (x(t), y (t)) · y 0 (t),
dt
or
dw ∂f dx ∂f dy
= + .
dt ∂x dt ∂y dt

Branch Diagram : Functions of Two Variables
Often we write ∂w /∂x for the

partial derivative ∂f /∂x, so
we can rewrite the Chain Rule
in the form
dw ∂w dx ∂w dy
= + .
dt ∂x dt ∂y dt

Functions of Three Variables
We can now probably predict the Chain Rule for functions of three
variables, as it only involves adding the expected third term to the
two-variable formula.
Theorem 20 (Chain Rule for Functions of Three Independent
Variables).
If w = f (x, y , z) is differentiable and x, y , and z are differentiable
functions of t, then w is a differentiable function of t and
dw ∂w dx ∂w dy ∂w dz
= + + .
dt ∂x dt ∂y dt ∂z dt

Branch Diagram : Functions of Two Variables

Functions Defined on Surfaces
If we are interested in the temperature w = f (x, y , z) at points (x, y , z) on

the earth’s surface, we might prefer to think of x, y , and z as functions of
the variables r and s that give the point’s longitudes and latitudes.
If x = g (r , s), y = h(r , s),

and z = k(r , s), we could then
express the temperature as a
function of r and s with the
composite function
w = f (g (r , s), h(r , s), k(r , s)).

Under the conditions stated below, w has partial derivatives with respect
to both r and s that can be calculated in the following way.
Theorem 21 (Chain Rule for Two Independent Variables and

Three Intermediate Variables).
Suppose that w = f (x, y , z), x = g (r , s), y = h(r , s), and z = k(r , s). If
all four functions are differentiable, then w has partial derivatives with
respect to r and s, given by the formulas
∂w ∂w ∂x ∂w ∂y ∂w ∂z
= + +
∂r ∂x ∂r ∂y ∂r ∂z ∂r
∂w ∂w ∂x ∂w ∂y ∂w ∂z
= + + .
∂s ∂x ∂s ∂y ∂s ∂z ∂s

∂w ∂w ∂x ∂w ∂y ∂w ∂z
Branch Diagram for ∂r = ∂x ∂r + ∂y ∂r + ∂z ∂r

∂w ∂w ∂x ∂w ∂y ∂w ∂z
Branch Diagram for ∂s = ∂x ∂s + ∂y ∂s + ∂z ∂s

∂w ∂w ∂x ∂w ∂y
Branch Diagram for ∂r = ∂x ∂r + ∂y ∂r
If f is a function of two variables instead of

three, each equation in the above Theorem
becomes correspondingly one term shorter.
If w = f (x, y ), x = g (r , s), and y = h(r , s),
then
∂w ∂w ∂x ∂w ∂y
= +
∂r ∂x ∂r ∂y ∂r
and
∂w ∂w ∂x ∂w ∂y
= + .
∂s ∂x ∂s ∂y ∂s

Branch Diagram
If f is a function of x alone, our equations are

even simpler.
If w = f (x, y ), x = g (r , s), and y = h(r , s),

then
∂w dw ∂x
=
∂r dx ∂r
and
∂w dw ∂x
= .
∂s dx ∂s

Theorem 22 (A Formula for Implicit Differentiation).
Suppose that F (x, y ) is differentiable and that the equation F (x, y ) = 0
defines y as a differentiable function of x. Then at any point where
Fy 6= 0,
dy Fx
=− .
dx Fy

The result in Theorem 22 is easily extended to three variables.
Suppose that the equation F (x, y , z) = 0 defines the variable z implicitly

as a function z = f (x, y ). Then for all all (x, y ) in the domain of f , we
have
F (x, y , f (x, y )) = 0.
Assuming that F and f are differentiable functions, we can use the Chain
Rule to differentiate the equation F (x, y , z) = 0 with respect to the
independent variables x and y . Hence we get
∂z Fx ∂z Fy
=− and =− . (1)
∂x Fy ∂y Fz

Implicit Function Theorem
An important result from advanced calculus, called the Implicit Function

Theorem, states the conditions for which our results in Equations (1) are
valid.
If the partial derivatives Fx , Fy , and Fz are continuous throughout an open

region R in space containing the point (x0 , y0 , z0 ), and if for some
constant c, F (x0 , y0 , z0 ) = c and Fz (x0 , y0 , z0 ) 6= 0, then the equation
F (x, y , z) = c defines z implicitly as a differentiable function of x and y
near (x0 , y0 , z0 ), and the partial derivatives of z are given by Equations (1).

Functions of Many Variables
We have seen several different forms of the Chain Rule in this section, but
each one is just a special case of one general formula. When solving
particular problems, it may help to draw the appropriate branch diagram
by placing the dependent variable on top, the intermediate variables in the
middle, and the selected independent variable at the bottom.
To find the derivative of the dependent variable with respect to the

selected independent variable, start at the dependent variable and read
down each route of the branch diagram to the independent variable,
calculating and multiplying the derivatives along each route. Then add the
products found for the different routes.

In general, suppose that w = f (x, y , . . . , v ) is a differentiable function of

the variables x, y , . . . , v (a finite set) and the x, y , . . . , v are differentiable
functions of p, q, . . . , t (another finite set).
Then w is a differentiable function of the variables p through t, and the

partial derivatives of w with respect to these variables are given by
equations of the form
∂w ∂w ∂x ∂w ∂y ∂w ∂v
= + + ... + .
∂p ∂x ∂p ∂y ∂p ∂v ∂p
The other equations are obtained by replacing p by q, · · · , t, one at a time.

One way to remember this equation is to think of the right-hand side as

the dot product of two vectors with components
complex number
z }| {
x
|{z} + iy
|{z}
real imaginary

∂w ∂w ∂w ∂x ∂y ∂v
, ,..., and , ,..., .
∂x ∂y ∂v ∂p ∂p ∂p
| {z } | {z }
Derivatives of w with Derivatives of the
respect to the intermediate variables with
intermediate variables respect to the selected
independent variable

Calculating First-Order Partial Derivatives
Exercise 23.
Find ∂f /∂x and ∂f /∂y .
1. f (x, y ) = tan−1 (y /x)
2. f (x, y ) = (x 3 + (y /2))2/3
3. f (x, y ) = (x + y )/(xy − 1)
4. f (x, y ) = e −x sin(x + y )
5. f (x, y ) = sin2 (x − 3y )
Z y
6. f (x, y ) = g (t)dt (g continuous for all t)
x
∞
X
7. f (x, y ) = (xy )n (|xy | < 1)
n=0

Solution for (1),(2),(3) and (4) in Exercise 23
∂f 1 ∂ y y y
1. ∂x = 2 · ∂x ( x ) = − x 2 [1+( y 2 = − 2
) ] x +y 2
,
1+( yx ) x
∂f 1 ∂ y 1 x
∂y = 2 · ∂y ( x ) = x[1+( yx )2 ]
= x 2 +y 2
1+( yx )
∂f 2
2. = q 2x ∂f
, ∂y = q 1
∂x
x +( y2 )
3 3
3 3 x 3 +( y2 )
(xy −1)(1)−(x+y )(y ) −y 2 −1 ∂f
3. ∂f
∂x = (xy −1)2
= (xy ,
−1)2 ∂y
= (xy −1)(1)−(x+y
(xy −1)2
)(x) −x 2 −1
= (xy −1)2
4. ∂f
∂x = −e −x sin(x + y ) + e −x cos(x + y ), ∂y = e −x cos(x + y )
∂f

Solution for (5),(6) and (7) in Exercise 23
∂f ∂
5. ∂x = 2 sin(x − 3y ) · ∂x sin(x − 3y ) =
∂
2 sin(x − 3y ) cos(x − 3y ) · ∂x (x − 3y ) = 2 sin(x − 3y ) cos(x − 3y ),
∂f ∂
∂y = 2 sin(x − 3y ) · ∂y sin(x − 3y ) =
∂
2 sin(x − 3y ) cos(x − 3y ) · ∂y (x − 3y ) = −6 sin(x − 3y ) cos(x − 3y )
∂f ∂f
6. ∂x= −g (x), ∂y = g (y )
∞
(xy )n ,
P
7. f (x, y ) =
n=0
1 ∂f 1 ∂ y
|xy | < 1 ⇒ f (x, y ) = 1−xy ⇒ ∂x = − (1−xy )2 · ∂x (1 − xy ) = (1−xy )2
∂f 1 ∂ x
and ∂y = − (1−xy )2
· ∂y (1 − xy ) = (1−xy )2

Exercises
Exercise 24.
Find fx , fy and fz .
1. f (x, y , z) = 1 + xy 2 − 2z 2
2. f (x, y , z) = (x 2 + y 2 + z 2 )−1/2
3. f (x, y , z) = sin−1 (xyz)
4. f (x, y , z) = ln(x + 2y + 3z)
5. f (x, y , z) = tanh(x + 2y + 3z)
6. f (x, y , z) = sinh(xy − z 2 )

Solution for Exercise 24
1. fx = y 2 , fy = 2xy , fz = −4z
2. fx = −x(x 2 + y 2 + z 2 )−3/2 , fy = −y (x 2 + y 2 + z 2 )−3/2 ,
fz = −z(x 2 + y 2 + z 2 )−3/2
3. fx = √ yz2 2 2 , fy = √ xz2 2 2 , fz = √ xy2 2 2
1−x y z 1−x y z 1−x y z
1 2 3
4. fx = x+2y +3z , fy = x+2y +3z , fz = x+2y +3z
5. fx = sech2 (x + 2y + 3z), fy = 2 sech2 (x + 2y + 3z),
fz = 3 sech2 (x + 2y + 3z)
6. fx = y cosh(xy − z 2 ), fy = x cosh(xy − z 2 ), fz = −2z cosh(xy − z 2 )

Exercises
Exercise 25.
Find the partial derivative of the function with respect to each variable.
1. f (t, α) = cos(2πt − α)
2. g (r , θ, z) = r (1 − cos θ) − z
3. Work done by the heart
V δv 2
W (P, V , δ, v , g ) = PV +
2g
4. Wilson lot size formula
km hq
A(c, h, k, m, q) = + cm +
q 2

∂f ∂f
1. ∂t = −2π sin(2πt − α), ∂α = sin(2πt − α)
∂g ∂g ∂g
2. ∂r = 1 − cos θ, ∂θ = r sin θ, ∂z = −1
δv 2 Vv 2 2V δv V ∂v
3. Wp = V , Wv = P + 2g , Wδ = 2g , Wv = 2g = g ,
2
Wg = − V2gδv2
4. ∂A
∂c = m, ∂A
∂h = q2 , ∂A
∂k = m ∂A
q , ∂m = k
q + c, ∂A
∂q = − km
q2
+ h
2

Exercises
Exercise 26.
Find all the second-order partial derivatives of the functions.
1. f (x, y ) = x + y + xy
2. h(x, y ) = xe y + y + 1
3. w = x 2 tan(xy )
x−y
4. w = x 2 +y

Solution for (1), (2) and (3) in Exercise 26
∂f ∂f ∂2f ∂2f ∂2f ∂2f
1. ∂x = 1 + y, ∂y = 1 + x, ∂x 2
= 0, ∂y 2
= 0, ∂y ∂x = ∂x∂y =1
∂h ∂h ∂2h ∂2h ∂2h ∂2h
2. ∂x = ey , ∂y = xe y + 1, ∂x 2
= 0, ∂y 2
= xe y , ∂y ∂x = ∂x∂y = ey
3. ∂w
∂x
= 2x tan(xy ) + x 2 sec2 (xy ) · y = 2x tan(xy ) + x 2 y sec2 (xy ),
∂w
∂y
= x 2 sec2 (xy ) · x = x 3 sec2 (xy ),
2
∂ w
∂x 2
= 2 tan(xy ) + 2x sec2 (xy ) · y + 2xy sec2 (xy ) + x 2 y (2 sec(xy ) sec(xy ) tan(xy ) · y) =
2 tan(xy ) + 4xy sec2 (xy ) + 2x 2 y 2 sec2 (xy ) tan(xy ),
∂2w
∂y 2
= x 3 (2 sec(xy ) sec(xy ) tan(xy ) · x) = 2x 4 sec2 (xy ) tan(xy )
∂2w ∂2w
∂y ∂x
= ∂x∂y = 3x 2 sec2 (xy ) + x 3 (2 sec(xy ) sec(xy ) tan(xy ) · y ) =
3x sec (xy ) + x 3 y sec2 (xy ) tan(xy ).
2 2

Solution for (4) in Exercise 26
(x 2 +y )−(x−y )(2x) 2
4. ∂w
∂x = (x 2 +y )2
= −x(x+2xy +y
2 +y )2 ,
(x 2 +y )(−1)−(x−y ) 2
∂w
∂y = (x 2 +y )2
= (x−x2 +y−x)2 ,
∂2w (x +y ) (−2x+2y )−(−x 2 +2xy +y )2(x 2 +y )(2x)
2 2 2(x 3 −3x 2 y −3xy +y 2 )
∂x 2 = 2 2 2 = (x 2 +y )3
,
[(x +y ) ]
2 2 2 −x)2(x 2 +y )·1
∂2w
∂y 2
= (x +y ) ·0−(−x = (x 2x 2 +2x
2 +y )3 ,
[(x 2 +y )2 ]2
2 2 (x 2 +y )2 (2x+1)−(−x 2 +2xy +y )2(x 2 +y )·1 3 2 −2xy −y
∂ w ∂ w
∂y ∂x = ∂x∂y = = 2x +3x
[(x 2 +y )2 ]2 (x 2 +y )3

Mixed Partial Derivatives
Exercise 27.
1. Verify that wxy = wyx .
(a) w = e x + x ln y + y ln x
(b) w = x sin y + y sin x + xy
2. Which order of differentiation will calculate fxy faster: x first or y
first? try to answer without writing anything down.
(a) f (x, y ) = x sin y + e y
(b) f (x, y ) = y + x 2 y + 4y 3 − ln(y 2 + 1)
(c) f (x, y ) = x ln xy

∂w x y ∂w x ∂2w 1 1 ∂2w 1 1
1. (a) ∂x = e + ln y + x , ∂y = y + ln x, ∂y ∂x = y + x , and ∂x∂y = y + x
∂w ∂w
(b) ∂x = sin y + y cos x + y , ∂y = x cos y + sin x + x,
2
∂ w ∂2w
∂y ∂x = cos y + cos x + 1, and ∂x∂y = cos y + cos x + 1
2. (a) x first
(b) x first
(c) y first

Exercises
Exercise 28.
The fifth-order partial derivative ∂ 5 f /∂x 2 ∂y 3 is zero for each of the
following functions. To show this as quickly as possible, which variable
would you differentiate with respect to first: x or y ? Try to answer
without writing anything down.
1. f (x, y ) = y 2 + y (sin x − x 4 )
2 /2
2. f (x, y ) = xe y

1. y first three times

2. x first twice

Using the Partial Derivative Definition
Exercise 29.
Use the limit definition of partial derivative to compute the partial
derivatives of the functions at the specified points.
∂f ∂f
1. f (x, y ) = 1 − x + y − 3x 2 y , ∂x and ∂y at (1, 2)
√ ∂f ∂f
2. f (x, y ) = 2x + 3y − 1, ∂x and ∂y at (−2, 3)

f (1+h,2)−f (1,2) [1−(1+h)+2−6(1+h)2 ]−(2−6)
1. fx (1, 2) = lim h = lim h =
h→0 h→0
2 )+6
lim −h−6(1+2h+h
2
h lim −13h−6h
h = lim (−13 − 6h) = −13,
h→0 h→0 h→0
fy (1, 2) = lim f (1,2+h)−f h
(1,2)
= lim [1−1+(2+h)−3(2+h)]−(2−6)
h =
h→0 h→0
lim (2−6−2h)−(2−6)
h = lim (−2) = −2
h→0 h→0
√ √
f (−2+h,3)−f (−2,3) 2(−2+h)+9−1− −4+9−1
2. fx (−2, 3) = lim h = lim h
√ h→0 √ √ h→0
2h+4−2 2h+4−2 √2h+4+2 2
= lim h = lim h 2h+4+2
= lim √2h+4+2 = 12 ,
h→0 h→0 h→0
fy (−2, 3) = lim f (−2,3+h)−f h
(−2,3)
=
√ h→0
√ √
−4+3(3+h)−1− −4+9−1 3h+4−2
= lim h = lim h =
h→0
√ √ h→0
3h+4−2 √3h+4+2 3
lim h 3h+4+2
= lim √2h+4+2 = 34
h→0 h→0

Exercises
Exercise 30.
1. Let f (x, y ) = 2x + 3y − 4. Find the slope of the line tangent to this
surface at the point (2, −1) and lying in the a. plane x = 2 b. plane
y = −1.
2. Let f (x, y ) = x 2 + y 3 . Find the slope of the line tangent to this
surface at the point (−1, 1) and lying in the a. plane x = −1 b.
plane y = 1.
3. Three variables : Let w = f (x, y , z) be a function of three
independent variables and write the formal definition of the partial
derivative ∂f /∂z at (x0 , y0 , z0 ). Use this definition to find ∂f /∂z at
(1, 2, 3) for f (x, y , z) = x 2 yz 2 .
4. Three variables : Let w = f (x, y , z) be a function of three
independent variables and write the formal definition of the partial
derivative ∂f /∂y at (x0 , y0 , z0 ). Use this definition to find ∂f /∂y at
(−1, 0, 3) for f (x, y , z) = −2xy 2 + yz 2 .
1. (a) In the plane x = 2 ⇒ fy (x, y ) = 3 ⇒ fy (2, −1) = 3 ⇒ m = 3
(b) In the plane y = −1 ⇒ fx (x, y ) = 2 ⇒ fy (2, −1) = 2 ⇒ m = 2
2. (a) In the plane
x = −1 ⇒ fy (x, y ) = 3y 2 ⇒ fy (−1, 1) = 3(1)2 = 3 ⇒ m = 3
(b) In the plane
y = 1 ⇒ fx (x, y ) = 2x ⇒ fy (−1, 1) = 2(−1) = −2 ⇒ m = −2
f (x0 ,y0 ,z0 +h)−f (x0 ,y0 ,z0 )
3. fz (x0 , y0 , z0 ) = lim h ;
h→0
2
fz (1, 2, 3) = lim f (1,2,3+h)−f
h
(1,2,3)
= lim 2(3+h)h −2(9) =
h→0 h→0
12h+2h2
lim h = lim (12 + 2h) = 12
h→0 h→0
f (x0 ,y0 +h,z0 )−f (x0 ,y0 ,z0 )
4. fy (x0 , y0 , z0 ) = lim h ;
h→0
2
fy (−1, 0, 3) = lim f (−1,h,3)−f h
(−1,0,3)
= lim (2h +5h)−0
h =
h→0 h→0
lim (2h + 9) = 9
h→0

Differentiating Implicitly
Exercise 31.
1. Find the value of ∂z/∂x at the point (1, 1, 1) if the equation
xy + z 3 x − 2yz = 0
defines z as a function of the two independent variables x and y and

the partial derivative exists.
2. Find the value of ∂x/∂z at the point (1, −1, −3) if the equation
xz + y ln x − x 2 + 4 = 0
defines x as a function of the two independent variables y and z and

the partial derivative exists.

∂z ∂z ∂z
1. y + 3z 2 ∂x x + z 3 − 2y ∂x = 0 ⇒ (3xz 2 − 2y ) ∂x = −y − z 3 ⇒ at

∂z ∂z
(1, 1, 1) we have (3 − 2) ∂x = −1 − 1 or ∂x = −2
y ∂x y
2. ∂x ∂x ∂x

∂z z + x + x ∂z − 2x ∂z = 0 ⇒ (z + x − 2x) ∂z = −x ⇒ at
(1, −1, −3) we have (−3 − 1 − 2) ∂x ∂x
∂z = −1 or ∂z = 6
1

Exercises
Exercise 32.
The following exercises are about the triangle shown here.
1. Express A implicitly as a function of a, b, and c and calculate ∂A/∂a

and ∂A/∂b.
2. Express a implicitly as a function of A, b, and B and calculate ∂a/∂A
and ∂a/∂B.

1. a2 = b 2 + c 2 − 2bc cos A ⇒ 2a = (2bc sin A) ∂A ∂A

∂a ⇒ ∂a =
a
bc sin A ; also
∂A
0 = 2b − 2c cos A + (2bc sin A) ∂b ⇒ 2c cos A − 2b =
(2bc sin A) ∂A ∂A c cos A−b
∂b ⇒ ∂b ⇒= bc sin A
∂a
a b (sin A) ∂A −a cos A ∂a
2. sin A = sin B ⇒ sin A = 0 ⇒ (sin A) ∂x − a cos A = 0⇒
∂a a cos A
∂A = ; also
1
sin
∂a
A
∂a
sin A ∂B = b(− csc B cot B) ⇒ ∂B = −b csc B cot B sin A

Exercises
Exercise 33.
1. Express vx in terms of u and y if the equations x = v ln u and
y = u ln v define u and v as functions of the independent variables x
and y , and if vx exists.
(Hint : Differentiate both equations with respect to x and solve for vx
by eliminating ux .)
2. Two dependent variables: Find ∂x/∂u and ∂y /∂u if the equations
u = x 2 − y 2 and v = x 2 − y define x and y as functions of the
independent variables u and v , and the partial derivatives exist. Then
let s = x 2 + y 2 and find ∂s/∂u.

1. Differentiating each equation implicitly gives 1 = vx ln u + vu ux and 0 = ux ln v + u

v
vx
or v 
1 v
(ln u)vx + ux = 1 u
0 ln v

u

ln v

u ⇒ vx = v = (ln u)(ln v )−1
vx + (ln v )ux = 0
 ln u u
u
v
v
ln v
2. Differentiating each equation implicitly

gives

1 = (2x)xu − (2y )yu and 0 = (2x)xu − yu or
) 1
−2y
(2x)xu − (2y )yu = 1 0

−1

−1 1
⇒ xu = =
(−2x+4xy )
= 2x−4xy and
(2x)xu − yu = 0 2x −2y

2x −1

2x 1

2x 0

−2x
yu = −2x+4xy = −2x+4xy = 2x = 1−2y 1 ∂s
; next s = x 2 + y 2 ⇒ ∂u ∂x
= 2x ∂u + 2y ∂y
2x−4xy ∂u
1 1 1 2y 1+2y
2x 2x−4xy + 2y 1−2y = 1−2y + 1−2y = 1−2y

Exercises
Exercise 34.
(
y 3, y ≥0
1. Let f (x, y ) = 2
−y , y < 0.
Find fx , fy , fxy , and fyx , and state the domain for each partial
derivative. (√
x, x ≥ 0
2. Let f (x, y ) = 2
x , x < 0.
Find fx , fy , fxy , and fyx , and state the domain for each partial
derivative.

(
0 if y ≥ 0
1. fx (x, y ) = ⇒ fx (x, y ) = 0 for all points (x, y ); at y = 0,
0 if y < 0
f (x,0+h)−f (x,0) f (x,h)−0 f (x,h)
fy (x, 0) = lim h
= lim h
= lim h
= 0 because
h→0 h→0 h→0
f (x,h) h3
lim h
= lim h
= 0 and
h→0− h→0− (
f (x,h) h2 3y 2 ify ≥ 0
lim h
= lim = 0 ⇒ fy (x, y ) = ; fyx (x, y ) = fxy (x, y ) = 0 for all
h→0+ h→0+ h −2y ify < 0
points (x, y ).
f (0+h,y )−f (0,y ) f (h,y )−0 f (h,y )
2. At x = 0, fx (0, y ) = lim h
= lim h
= lim h
which does not exist
h→0 h→0 h→0
f (h,y ) 2
because lim h
= lim hh = 0 and
h→0− h→0− (
√ 1
f (h,y ) √ ifx > 0
lim h
= = lim √1 = +∞ ⇒ fx (x, y ) = 2 x
lim hh ;
h→0+ h→0 + h→0 + h 2x ifx < 0
(
0 ifx ≥ 0
fy (x, y ) = ⇒ fy (x, y ) = 0 for all points (x, y ); fyx (x, y ) = 0 for all points
0 ifx < 0
(x, y ), while fxy (x, y ) = 0 for all points (x, y ) such that x 6= 0.

Exercises
Exercise 35.
Show that each function satisfies a Laplace equation.
1. f (x, y , z) = 2z 3 − 3(x 2 + y 2 )z
2. f (x, y ) = tan−1 x
y
3. f (x, y , z) = e 3x+4y cos 5z

∂f ∂f ∂f 2 2 2 ∂2f
1. ∂x = −6xz, ∂y = −6yz, ∂z = 6z − 3(x + y ), ∂x 2 = −6z,
∂2f ∂2f ∂2f ∂2f ∂2f
∂y 2
= −6z, ∂z 2
= 12z ⇒ ∂x 2 + ∂y 2 + ∂z 2 = −6z − 6z + 12z =0
∂f 1/y y −x/y 2 −x
2. ∂x = 2 = , ∂f
y 2 +x 2 ∂y
= 2 = y 2 +x 2
,
1+ yx 1+ yx
2 2 )·0−y ·2x (y 2 +x 2 )·0−(−x)·2y
∂2f ∂2f
∂x 2
= (y +x(y 2 +x 2 )2
= (y−2xy
2 +x 2 )2 , ∂y 2 = (y 2 +x 2 )2
= (y 22xy
+x 2 )2
⇒
∂2f ∂2f −2xy 2xy
∂x 2
+ ∂y 2 = (y 2 +x 2 )2 + (y 2 +x 2 )2 = 0
∂f 3x+4y cos 5z, ∂f = 4e 3x+4y cos 5z, ∂f = −5e 3x+4y sin 5z;
3. ∂x = 3e ∂y ∂z
2
∂ f
= 9e 3x+4y cos 5z, ∂ 2 f = 16e 3x+4y cos 5z,
∂x 2 ∂y 2
∂2f 3x+4y ∂2f ∂2f ∂2f
∂z 2
= −25e cos 5z ⇒ ∂x 2 + ∂y 2 + ∂z 2 =
9e 3x+4y cos 5z + 16e 3x+4y cos 5z − 25e 3x+4y cos 5z = 0

Exercises
Exercise 36.
Show that the functions are all solutions of the wave equation
∂2w 2
2∂ w
= c ,
∂t 2 ∂x 2
where w is the wave height, x is the distance variable, t is the time
variable, and c is the velocity with which the waves are propagated.
1. w = cos(2x + 2ct)
2. w = ln(2x + 2ct)
3. w = f (u), where f is a differentiable function of u, and u = a(x + ct), where a is a
constant.
4. Does a function f (x, y ) with continuous first partial derivatives throughout an open region
R have to be continuous on R? Give reasons for your answer.
5. If a function f (x, y ) has continuous second partial derivatives throughout an open region
R, must the first-order partial derivatives of f be continuous on R? Give reasons for your
answer.

Solution for (1), (2) and (3) in Exercise 36
∂w ∂w
1. ∂x = −2 sin(2x + 2ct), ∂t = −2c sin(2x + 2ct);
2
∂ w
∂x 2
= −4 cos(2x + 2ct),
∂2w 2 2
∂t 2
= −4c 2 cos(2x + 2ct) ⇒ ∂∂tw2 = c 2 [−4 cos(2x + 2ct)] = c 2 ∂∂xw2
∂w 1 ∂w c ∂2w −1
2. ∂x = x+ct , ∂t = x+ct ; ∂x 2h = (x+ct)2,
i
∂2w −c 2 ∂2w −1 2
∂t 2
= (x+ct)2 ⇒ ∂t 2 = c 2 (x+ct)2 = c 2 ∂∂xw2
∂w ∂f ∂u ∂f ∂2w ∂2f 2 2 ∂2f
3. ∂t = ∂u ∂t = ∂u (ac) ⇒ ∂t 2 =(ac) ∂u
2 (ac) = a c ∂u 2 ;
∂w ∂f ∂u ∂f ∂2w ∂2f 2 ∂2f ∂2w
∂x = ∂u ∂x = ∂u · a ⇒ ∂x 2 = a ∂u 2 · a = a ∂u 2 ⇒ ∂t 2 =

a2 c 2 ∂2f ∂2f
= c 2 a2 ∂u
2
= c 2 ∂∂xw2
∂u 2 2

Solution for (4) and (5) in Exercise 36
1. If the first partial derivatives are continuous throughout an open

region R, then by Theorem 3 in this section of the text,
f (x, y ) = f (x0 , y0 ) + fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + 1 ∆x + 2 ∆y ,
where 1 , 2 → 0 as ∆x, ∆y → 0. Then as (x, y ) → (x0 , y0 ), ∆x → 0
and ∆y → 0 ⇒ lim f (x, y ) = f (x0 , y0 ) ⇒ f is continuous at
(x,y )→(x0 ,y0 )
every point (x0 , y0 ) in R.
2. Yes, since fxx , fyy , fxy , and fyx are all continuous on R, use the same
reasoning as in Exercise 76 with
fx (x, y ) = fx (x0 , y0 ) + fxx (x0 , y0 )∆x + fxy (x0 , y0 )∆y + 1 ∆x + 2 ∆y
and
fy (x, y ) = fy (x0 , y0 ) + fyx (x0 , y0 )∆x + fyy (x0 , y0 )∆y + 1 ∆x + 2 ∆y .
Then lim fx (x, y ) = fx (x0 , y0 ) and
(x,y )→(x0 ,y0 )
lim fy (x, y ) = fy (x0 , y0 ).
(x,y )→(x0 ,y0 )

The heat equation
Exercise 37.
An important partial differential equation that describes the distribution of
heat in a region at time t can be represented by the one-dimensional heat
equation
∂f ∂2f
= .
∂t ∂x 2
Show that u(x, t) = sin(αx) · e −βt satisfies the heat equation for
constants α and β.

To find α and β so that ut = uxx ⇒ ut = −β sin(αx)e −βt and

ux = α cos(αx)e −βt ⇒ uxx = −α2 sin(αx)e −βt ; then
ut = uxx ⇒ −β sin(αx)e −βt = −a2 sin(ax)e −βt , thus ut = uxx only if
β = α2

Exercises
Exercise 38.
xy 2
(
x 2 +y 4
, (x, y ) 6= (0, 0)
1. Let f (x, y ) =
0, (x, y ) = (0, 0).
Show that fx (0, 0) and fy (0, 0) exist, but f is not differentiable at
(0, 0).
(Hint:Show that f is not continuous at (0, 0).)
(
0, x 2 < y < 2x 2
2. Let f (x, y ) =
1, otherwise.
Show that fx (0, 0) and fy (0, 0) exist, but f is not differentiable at
(0, 0).

h02 −0
f (0+h,0)−f (0,0) h2 +04 0
1. fx (0, 0) = lim h
= lim h
= lim = 0;
h→0 h→0 h→0 h
0h2 −0
f (0,0+h)−f (0,0) 02 +h4 0
fy (0, 0) = lim h
= lim = h
= lim = 0;
h→0 h→0 h→0 h
2 2 4
(ky )y
lim f (x, y ) = lim 2 2 4 = lim 2 ky4 4 = lim 2
k
= k
k 2 +1
⇒ different limits
(x,y )→(0,0) y →0 (ky ) +y y →0 k y +y y →0 k +1
alongx=ky 2
for different values of k ⇒ lim f (x, y ) does not exist ⇒ f (x, y ) is not continuous at
(x,y )→(0,0)
(0, 0) ⇒ by Theorem 18 (Differentiability Implies Continuity), f (x, y ) is not differentiable
at (0, 0).
f (0+h,0)−f (0,0) f (h,0)−1 1−1
2. fx (0, 0) = lim h
= lim h
= lim h
= 0;
h→0 h→0 h→0
f (0,0+h)−f (0,0) f (0,h)−1
fy (0, 0) = lim h
= lim h
= lim 1−1 = 0;
h→0 h→0 h→0 h
lim f (x, y ) = lim 0 = 0 but lim f (x, y ) = lim 1 = 1 ⇒ lim f (x, y )
(x,y )→(0,0) y →0 (x,y )→(0,0) y →0 (x,y )→(0,0)
alongy =x 2 alongy =1.5x 2
does not exist ⇒ f (x, y ) is not continuous at (0, 0) ⇒ by Theorem 18 (Differentiability
Implies Continuity), f (x, y ) is not differentiable at (0, 0).

Extra Notes on Differentiability
An informal definition of differentiability of a function f (x, y ) at a point

(a, b) is that the function f (x, y ) is well-approximated by a linear function
L(x, y ) near (a, b). The function L(x, y ) is called the local linearization
of f (x, y ) near (a, b). We now focus on the precise meaning of the phrase
“well-approximated.”
By looking at examples, we shall see that local linearity requires the

existence of partial derivatives, but they do not tell the whole story.
In particular, existence of partial derivatives at a point is not sufficient to

guarantee local linearity at that point.

As an illustration, take a sheet of paper, crumple it into a ball and smooth

it out again. Wherever there is a crease it would be difficult to approximate
the surface by a plane — these are points of nondifferentiability of the
function giving the height of the paper above the floor. Yet the sheet of
paper models a graph which is continuous — there are no breaks.
As in the case of one-variable calculus, continuity does not imply

differentiability. But differentiability does require continuity: there cannot
be linear approximations to a surface at points where there are abrupt
changes in height.
For a function of two variables, as for a function of one variable, we define

differentiability at a point in terms of the error in a linear
approximation as we move from the point to a nearby point.
If the point is (a,√

b) and a nearby point is (a + h, b + k), the distance
between them is h2 + k 2 .

Definition 39.
A function f (x, y ) is differentiable at the point (a, b) if there is a linear
function L(x, y ) = f (a, b) + m(x − a) + n(y − b) such that if the error
E (x, y ) is defined by
f (x, y ) = L(x, y ) + E (x, y ),

E (a+h,b+k)
and if h = x − a, k = y − b, then the relative error √
h2 +k 2
satisfies
E (a + h, b + k)
lim √ = 0.
(h,k)→(0,0) h2 + k 2
The function f is differentiable if it is differentiable at each point of its
domain. The function L(x, y ) is called the local linearization of f (x, y )
near (a, b).

A function f (x, y ) is differentiable at the point (a, b) if there is a linear

function L(x, y ) = f (a, b) + m(x − a) + n(y − b) such that if the error
E (x, y ) is defined by f (x, y ) = L(x, y ) + E (x, y ), and if
h = x − a, k = y − b, then the relative error E (a+h,b+k)√
h2 +k 2
satisfies
E (a + h, b + k)
lim √ = 0.
(h,k)→(0,0) h2 + k 2

In the next example, we show that this definition of differentiability is

consistent with our previous notion — that is, that m = fx and n = fy and
that the graph of L(x, y ) is the tangent plane.
Theorem 40.
Show that if f is a differentiable function with local linearization
L(x, y ) = f (a, b) + m(x − a) + n(y − b),
then m = fx (a, b) and n = fy (a, b).
Proof : Since f is differentiable, we know that the relative error in L(x, y )

tends to 0 as we get close to (a, b).

Solution (contd...)
Suppose h > 0 and k = 0. Then we know that
E (a + h, b) E (a + h, b)
0 = lim √ = lim
h→0 h2 + k 2 h→0 h
f (a + h, b) − L(a + h, b)
= lim
h→0 h
f (a + h, b) − f (a, b) − mh
= lim
h→0 h
f (a + h, b) − f (a, b)
= lim −m
h→0 h
= fx (a, b) − m.
A similar result holds if h < 0, so we have m = fx (a, b). The result

n = fy (a, b) is found in a similar manner.

The previous result says that if a function is differentiable at a point, it

has partial derivatives there.
Therefore, if any of the partial derivatives fail to exist, then the function
cannot be differentiable.
This is what happens in the following example of a cone.

Example 41.
Consider the function p
f (x, y ) = x 2 + y 2.
Is f differentiable at the origin?

p
The function f (x, y ) = x 2 + y 2 is not locally linear at (0, 0).
Zooming in around (0, 0) does not make the graph look like a plane.
p
If we zoom in on the graph of the function f (x, y ) = x 2 + y 2 at the
origin, as shown in the figure, the sharp point remains; the graph never
flattens out to look like a plane. Near its vertex, the graph does not look
like it is well approximated (in any reasonable sense) by any plane.

Solution
Judging from the graph of f , we would not expect f to be differentiable at

(0, 0). Let us check this by trying to compute the partial derivatives of f
at (0, 0):
√
f (h, 0) − f (0, 0) h2 + 0 − 0 |h|
fx (0, 0) = lim = lim = lim .
h→0 h h→0 h h→0 h
Since |h|/h = ±1, depending on whether h approaches 0 from the left or

right, this limit does not exist and so neither does the partial derivative
fx (0, 0).
Thus, f cannot be differentiable at the origin. If it were, both of the

partial derivatives, fx (0, 0) and fy (0, 0), would exist.

Alternate Solution
Alternatively, we could show directly that there is no linear approximation

near (0, 0) that satisfies the small relative error criterion for differentiability.
Any plane passing through the point (0, 0, 0) has the form
L(x, y ) = mx + ny
for some constants m and n.
If E (x, y ) = f (x, y ) − L(x, y ), then

p
E (x, y ) = x 2 + y 2 − mx − ny .
Then for f to be differentiable at the origin, we would need to show that

√
h2 + k 2 − mh − nk
lim √ = 0.
(h,k)→(0,0) h2 + k 2

Alternate Solution (contd...)
Taking k = 0 gives
|h| − mh h
lim = 1 − m lim .
h→0 |h| h→0 |h|
This limit exists only if m = 0 for the same reason as before. But then the
value of the limit is 1 and not 0 as required. Thus, we again conclude f is
not differentiable.
In the above example the partial derivatives fx and fy did not exist at the
origin and this was sufficient to establish nondifferentiability there. We
might expect that if both partial derivatives do exist, then f is
differentiable. But the next example shows that this not necessarily true:
the existence of both partial derivatives at a point is not sufficient to
guarantee differentiability.

Example
Example 42.
Consider the function
f (x, y ) = x 1/3 y 1/3 .
Show that the partial derivatives fx (0, 0) and fy (0, 0) exist, but that f is
not differentiable at (0, 0).

Solution
We have f (0, 0) = 0 and we compute the partial derivatives using the

definition:
f (h, 0) − f (0, 0)
lim fx (0, 0) = lim = lim 0 − 0 = 0,
h→0 h→0 h h→0
and similarly
fy (0, 0) = 0.
So, if there did exist a linear approximation near the origin, it would have
to be L(x, y ) = 0. But we can show that this choice of L(x, y ) does not
result in the small relative error that is required for differentiability.
In fact, since E (x, y ) = f (x, y ) − L(x, y ) = f (x, y ), we need to look at the

limit
h1/3 k 1/3
lim √ .
(h,k)→(0,0) h2 + k 2

Solution (contd...)
If this limit exists, we get the same value no matter how h and k approach
0. Suppose we take k = h > 0. Then the limit becomes
h1/3 h1/3 h2/3 1

lim √ = lim √ = lim √ .
h→0 h2 + k 2 h→0 h 2 h→0 h1/3 2
But this limit does not exist, since small values for h will make the fraction
arbitrarily large. So the only possible candidate for a linear approximation
at the origin does not have a sufficiently small relative error.
Thus, this function is not differentiable at the origin, even though the
partial derivatives fx (0, 0) and fy (0, 0) exist. The figure confirms that near
the origin the graph of z = f (x, y ) is not well approximated by any plane.
In the above example, the function f was continuous at the point

where it was not differentiable.

Summary
If a function is differentiable at a point, then both partial derivatives

exist there.
Having both partial derivatives at a point does not guarantee that a
function is differentiable there.

Continuity and Differentiability
We know that differentiable functions of one variable are continuous.

Similarly, it can be shown that if a function of two variables is
differentiable at a point, then the function is continuous there. The
following example shows that even if the partial derivatives of a function
exist at a point, the function is not necessarily continuous at that point if
it is not differentiable there.
Example 43.
Suppose that f is the function of two variables defined by
(
xy
2 2, (x, y ) 6= (0, 0)
f (x, y ) = x +y
0, (x, y ) = (0, 0).
One can show that f (x, y ) is not continuous at the origin. Show that the
partial derivatives fx (0, 0) and fy (0, 0) exist. Could f be differentiable at
(0, 0)?

Solution
From the definition of the partial derivative we see that
f (h, 0) − f (0, 0) 1 0 0
fx (0, 0) = lim = lim . 2 2
= lim = 0,
h→0 h h→0 h h + 0 h→0 h
and similarly fy (0, 0) = 0.
So, the partial derivatives fx (0, 0) and fy (0, 0) exist.
However, f cannot be differentiable at the origin since it is not continuous

there.

Summary
If a function is differentiable at a point, then it is continuous there.

Having both partial derivatives at a point does not guarantee that a
function is continuous there.

How Do We Know If a Function Is Differentiable?
Question : Can we use partial derivatives to tell us if a function is

differentiable?
As we have seen from the last two examples, it is not enough that the
partial derivatives exist.
However, the following condition does guarantee differentiability:
Theorem 44 (Condition for Differentiability).

If the partial derivatives, fx and fy , of a function f exist and are
continuous on a small disk centered at the point (a, b), then f is
differentiable at (a, b).

Condition for Differentiability
We will not prove this fact, although it provides a criterion for

differentiability which is often simpler to use than the definition.
It turns out that the requirement of continuous partial derivatives is more

stringent than that of differentiability, so there exist differentiable
functions which do not have continuous partial derivatives.
However, most functions we encounter will have continuous partial

derivatives.
The class of functions with continuous partial derivatives is given the name
C 1.

Example
Example 45.
Show that the function f (x, y ) = ln(x 2 + y 2 ) is differentiable everywhere
in its domain.
Solution : The domain of f is all of 2-space except for the origin. We

shall show that f has continuous partial derivatives everywhere in its
domain (that is, the function f is in C 1 ). The partial derivatives are
2x 2y
fx = and fy = .
x2 + y2 x2 + y2
Since each of fx and fy is the quotient of continuous functions, the partial
derivatives are continuous everywhere except the origin (where the
denominators are zero). Thus, f is differentiable everywhere in its domain.

The three-dimensional Laplace equation
Exercise 46.
∂ f ∂ f ∂ f2 2 2
The three-dimensional Laplace equation ∂x 2 + ∂y 2 + ∂z 2 = 0 is
satisfied by steady-state temperature distributions T = f (x, y , z) in space,

by gravitational potentials, and by electrostatic potentials. The
∂2f ∂2f
two-dimensional Laplace equation ∂x 2 + ∂y 2 = 0, obtained by dropping
the ∂ 2 f /∂z 2 term from the previous equation, describes potentials and
steady-state temperature distributions in a plane (see the accompanying
figure). The plane (a) may be treated as a thin slice of the solid (b)
perpendicular to the z-axis.
∂2f ∂2f
(a). ∂x 2 + ∂y 2 = 0
∂2f ∂2f ∂2f
(b). ∂x 2
+ ∂y 2
+ ∂z 2
=0

The Wave Equation
Exercise 47.
If we stand on an ocean shore and take a snapshot of the waves, the
picture shows a regular pattern of peaks and valleys in an instant of time.
We see periodic vertical motion in space, with respect to distance. If we
stand in the water, we can feel the rise and fall of the water as the waves
go by. We see periodic vertical motion in time. In physics, this beautiful
symmetry is expressed by the one-dimensional wave equation
∂2w 2
2∂ w
= c ,
∂t 2 ∂x 2
where w is the wave height, x is the distance variable, t is the time
variable, and c is the velocity with which the waves are propagated.
In our example, x is the distance across the ocean’s surface, but in other
applications, x might be the distance along a vibrating string, distance
through air (sound waves), or distance through space (light waves). The
number c varies with the medium and type of wave.
Summary
Most functions built up from elementary functions have continuous partial

derivatives, except perhaps at a few obvious points.
Thus, in practice, we can often identify functions as being C 1 without

explicitly computing the partial derivatives.

References
1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus, 11th

Edition, Pearson Publishers.
2. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translated
by George Yankovsky).
3. S.C. Malik and Savitha Arora, Mathematical Analysis, New Age
Publishers.
4. R. G. Bartle, D. R. Sherbert, Introduction to Real Analysis, Wiley
Publishers.

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Partial Derivatives

August 30, 2019

P. Sam Johnson Partial Derivatives August 30, 2019 1/117

The calculus of several variables is similar to single-variable calculus

The idea of differentiability for functions of several variables requires more

P. Sam Johnson Partial Derivatives August 30, 2019 2/117

If (x0 , y0 ) is a point in the domain of a function f (x, y ), the vertical plane

The picture shows that the intersec-

P. Sam Johnson Partial Derivatives August 30, 2019 3/117

This curve is the graph of the func-

P. Sam Johnson Partial Derivatives August 30, 2019 4/117

We define the partial derivative of f with respect to x at the point (x0 , y0 )

To distinguish partial derivatives from ordinary derivatives we use the

In the definition, h represents a real number, positive or negative.

provided the limit exists.

An equivalent expression for the partial derivative is

P. Sam Johnson Partial Derivatives August 30, 2019 6/117

We use several notations for the partial derivative:

provided the limit exists.

The intersection of the plane x = x0

P. Sam Johnson Partial Derivatives August 30, 2019 8/117

P. Sam Johnson Partial Derivatives August 30, 2019 9/117

The partial derivatives of f at (a, b) are the slopes of the tangents to C1

P. Sam Johnson Partial Derivatives August 30, 2019 10/117

The graph of f is the paraboloid z = 4 − x 2 − 2y 2 and the vertical plane

Similarly, the curve C2 in which the plane x = 1 intersects the paraboloid

P. Sam Johnson Partial Derivatives August 30, 2019 11/117

The following figures are computer-drawn counterparts to the last figures.

P. Sam Johnson Partial Derivatives August 30, 2019 12/117

The tangent lines at the point

P. Sam Johnson Partial Derivatives August 30, 2019 13/117

The definitions of ∂f /∂x and ∂f /∂y give us two different ways of

To find ∂f /∂y , we treat x as a constant and differentiate with respect to

Solution : We treat x as a constant and f as a product of y and sin xy :

Solution : We treat f as a quotient. With y held constant, we get

We differentiate both sides of the equation with respect to x, holding y

The tangent to the curve of intersec-

P. Sam Johnson Partial Derivatives August 30, 2019 18/117

As a check, we can treat the parabola as the graph of the single-variable

P. Sam Johnson Partial Derivatives August 30, 2019 19/117

The definitions of the partial derivatives of functions of more than two

f (x, y , z) = x sin(y + 3z),

P. Sam Johnson Partial Derivatives August 30, 2019 20/117

Resistors arranged this way are said to be connected

P. Sam Johnson Partial Derivatives August 30, 2019 21/117

To find ∂R/∂R2 , we treat R1 and R3 as constants and, using implicit

P. Sam Johnson Partial Derivatives August 30, 2019 22/117

When R1 = 30, R2 = 45, and R3 = 90,

Thus at the given values, a small change in the resistance R2 leads to a

P. Sam Johnson Partial Derivatives August 30, 2019 23/117

A function f (x, y ) can have partial derivatives with respect to both

This is different from functions of a single variable, where the

P. Sam Johnson Partial Derivatives August 30, 2019 24/117

P. Sam Johnson Partial Derivatives August 30, 2019 25/117

P. Sam Johnson Partial Derivatives August 30, 2019 26/117

The above example suggests that we need a stronger requirement

We shall define differentiability for functions of two variables (which

P. Sam Johnson Partial Derivatives August 30, 2019 27/117

When we differentiate a function f (x, y ) twice, we produce its