Partial diferential good

Intermediate Mathematics
Introduction to Partial
Differentiation
R Horan & M Lavelle
The aim of this document is to provide a short, self
assessment programme for students who wish to acquire
a basic understanding of partial differentiation.
Copyright c

2004 rhoran@plymouth.ac.uk , mlavelle@plymouth.ac.uk
Last Revision Date: May 25, 2005 Version 1.0

Table of Contents
1. Partial Differentiation (Introduction)
2. The Rules of Partial Differentiation
3. Higher Order Partial Derivatives
4. Quiz on Partial Derivatives
Solutions to Exercises
Solutions to Quizzes
The full range of these packages and some instructions,
should they be required, can be obtained from our web
page Mathematics Support Materials.

Section 1: Partial Differentiation (Introduction) 3
1. Partial Differentiation (Introduction)
In the package on introductory differentiation, rates of change
of functions were shown to be measured by the derivative. Many
applications require functions with more than one variable: the ideal
gas law, for example, is
pV = kT
where p is the pressure, V the volume, T the absolute temperature of
the gas, and k is a constant. Rearranging this equation as
p =
kT
V
shows that p is a function of T and V . If one of the variables, say T,
is kept fixed and V changes, then the derivative of p with respect to
V measures the rate of change of pressure with respect to volume. In
this case, it is called the partial derivative of p with respect to V and
written as
∂p
∂V
.

Example 1 If p =
kT
V
, find the partial derivatives of p:
(a) with respect to T, (b) with respect to V .
Solution
(a) This part of the example proceeds as follows:
p =
kT
V
,
∴
∂p
∂T
=
k
V
,
where V is treated as a constant for this calculation.
(b) For this part, T is treated as a constant. Thus
p = kT
1
V
= kTV −1
,
∴
∂p
∂V
= −kTV −2
= −
kT
V 2
.

The symbol ∂ is used whenever a function with more than one variable
is being differentiated but the techniques of partial differentiation are
exactly the same as for (ordinary) differentiation.
Example 2 Find
∂z
∂x
and
∂z
∂y
for the function z = x2
y3
.
Solution
z = x2
y3
∴
∂z
∂x
= 2xy3
,
and
∂z
∂y
= x2
3y2
,
= 3x2
y2
.
For the first part y3
is treated as
a constant and the derivative of
x2
with respect to x is 2x.
For the second part x2
is treated
as a constant and the derivative
of y3
with respect to y is 3y2
.
Exercise 1. Find
∂z
∂x
and
∂z
∂y
for each of the following functions.
(Click on the green letters for solutions.)
(a) z = x2
y4
, (b) z = (x4
+ x2
)y3
, (c) z = y
1
2 sin(x).

Section 2: The Rules of Partial Differentiation 6
2. The Rules of Partial Differentiation
Since partial differentiation is essentially the same as ordinary differ-
entiation, the product, quotient and chain rules may be applied.
Example 3 Find
∂z
∂x
(a) z = xy cos(xy) , (b) z =
x − y
x + y
, (c) z = (3x + y)2
.
Solution
(a) Here z = uv, where u = xy and v = cos(xy) so the product rule
applies (see the package on the Product and Quotient Rules).
u = xy and v = cos(xy)
∴
∂u
∂x
= y and
∂v
∂x
= −y sin(xy) .
Thus
∂z
∂x
=
∂u
∂x
v + u
∂v
∂x
= y cos(xy) − xy2
sin(xy) .

(b) Here z = u/v, where u = x − y and v = x + y so the quotient rule
applies (see the package on the Product and Quotient Rules).
u = x − y and v = x + y
∴
∂u
∂x
= 1 and
∂v
∂x
= 1 .
Thus
∂z
∂x
=
v
∂u
∂x
− u
∂v
∂x
v2
=
(x + y) − (x − y)
(x + y)2
=
2y
(x + y)2
.
(c) In this case z = (3x + y)2
so z = u2
where u = 3x + y, and the
chain rule applies (see the package on the Chain Rule).
z = u2
and u = 3x + y
∴
∂z
∂u
= 2u and
∂u
∂x
= 3 .
Thus ∂z
∂x
=
∂z
∂u
∂u
∂x
= 2(3x + y)3 = 6(3x + y) .

Exercise 2. Find
∂z
∂x
and
∂z
∂y
(Click on the green letters for solutions.)
(a) z = (x2
+3x) sin(y), (b) z =
cos(x)
y5
, (c) z = ln(xy),
(d) z = sin(x) cos(xy), (e) z = e(x2
+y2
)
, (f) z = sin(x2
+ y).
Quiz If z = cos(xy), which of the following statements is true?
(a)
∂z
∂x
=
∂z
∂y
, (b)
∂z
∂x
=
1
x
∂z
∂y
,
(c)
1
y
∂z
∂x
=
∂z
∂y
, (d)
1
y
∂z
∂x
=
1
x
∂z
∂y
.

Section 3: Higher Order Partial Derivatives 9
3. Higher Order Partial Derivatives
Derivatives of order two and higher were introduced in the package on
Maxima and Minima. Finding higher order derivatives of functions
of more than one variable is similar to ordinary differentiation.
Example 4 Find
∂2
z
∂x2
if z = e(x3
+y2
)
.
Solution First differentiate z with respect to x, keeping y constant,
then differentiate this function with respect to x, again keeping y
constant.
z = e(x3
+y2
)
∴
∂z
∂x
= 3x2
e(x3
+y2
)
using the chain rule
∂2
z
∂x2
=
∂(3x2
)
∂x
e(x3
+y2
)
+ 3x2 ∂(e(x3
+y2
)
)
∂x
using the product rule
∂2
z
∂x2
= 6xe(x3
+y2
)
+ 3x2
(3x2
e(x3
+y2
)
)
= (9x4
+ 6x)e(x3
+y2
)

In addition to both
∂2
z
∂x2
and
∂2
z
∂y2
, when there are two variables there
is also the possibility of a mixed second order derivative.
Example 5 Find
∂2
z
∂x∂y
if z = e(x3
+y2
)
.
Solution The symbol
∂2
z
∂x∂y
is interpreted as
∂
∂x

∂z
∂y

; in words,
first differentiate z with respect to y, keeping x constant, then differ-
entiate this function with respect to x, keeping y constant. (It is this
differentiation, first with respect to x and then with respect to y, that
leads to the name of mixed derivative.)
First with x constant
∂z
∂y
= 2ye(x3
+y2
)
(using the chain rule.)
Second with y constant
∂2
z
∂x∂y
=
∂
∂x

2ye(x3
+y2
)

= 3x2
2ye(x3
+y2
)
= 6x2
ye(x3
+y2
)
.

The obvious question now to be answered is: what happens if the
order of differentiation is reversed?
Example 6 Find
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

if z = e(x3
+y2
)
.
Solution
First with y constant
∂z
∂x
= 3x2
e(x3
+y2
)
(using the chain rule).
Second with x constant
∂2
z
∂y∂x
=
∂
∂y

3x2
e(x3
+y2
)

= 2y3x2
e(x3
+y2
)
= 6x2
ye(x3
+y2
)
=
∂2
z
∂x∂y
.
As a general rule, when calculating mixed derivatives the order of
differentiation may be reversed without affecting the final result.

Exercise 3. Confirm the statement on the previous page by finding
both
∂2
z
∂x∂y
and
∂2
z
∂y∂x
for each of the following functions, whose first
order partial derivatives have already been found in exercise 2. (Click
on the green letters for solutions.)
(a) z = (x2
+3x) sin(y), (b) z =
cos(x)
y5
, (c) z = ln(xy),
(d) z = sin(x) cos(xy), (e) z = e(x2
+y2
)
, (f) z = sin(x2
+ y).
Notation For first and second order partial derivatives there is a
compact notation. Thus
∂f
∂x
can be written as fx and
∂f
∂y
as fy.
Similarly
∂2
f
∂x2
is written fxx while
∂2
f
∂x∂y
is written fxy.
Quiz If z = e−y
sin(x), which of the following is zxx + zyy?
(a) e−y
sin(x), (b) 0, (c) −e−y
sin(x), (d) −e−y
cos(x).

Section 4: Quiz on Partial Derivatives 13
4. Quiz on Partial Derivatives
Choose the correct option for each of the following.
Begin Quiz
1. If z = x2
+ 3xy + y3
then
∂z
∂x
is
(a) 2x + 3y + 3y2
, (b) 2x + 3x + 3y2
,
(c) 2x + 3x , (d) 2x + 3y .
2. If w = 1/r, where r2
= x2
+ y2
+ z2
, then xwx + ywy + zwz is
(a) −1/r , (b) 1/r , (c) −1/r2
, (d) 1/r2
.
3. If u =
r
x
y
then uxx is
(a) −
1
4
p
y3x3
, (b) −
1
4
p
yx3
, (c) −
1
8
p
y3x3
, (d) −
1
8
p
yx3
.
End Quiz

Solutions to Exercises 14
Solutions to Exercises
Exercise 1(a) To calculate the partial derivative
∂z
∂x
of the function
z = x2
y4
, the factor y4
is treated as a constant:
∂z
∂x
=
∂
∂x
x2
y4

=
∂
∂x
x2

× y4
= 2x(2−1)
× y4
= 2xy4
.
Similarly, to find the partial derivative
∂z
∂y
, the factor x2
is treated
as a constant:
∂z
∂y
=
∂
∂y
x2
y4

= x2
×
∂
∂y
y4

= x2
× 4y(4−1)
= 4x2
y3
.
Click on the green square to return

Exercise 1(b) To calculate
∂z
∂x
for the function z = (x4
+ x2
)y3
, the
factor y3
is treated as a constant:
∂z
∂x
=
∂
∂x
(x4
+ x2
)y3

=
∂
∂x
x4
+ x2

× y3
= (4x3
+ 2x)y3
.
To find the partial derivative
∂z
∂y
the factor (x4
+ x2
) is treated as a
constant:
∂z
∂y
=
∂
∂y
(x4
+ x2
)y3

= (x4
+ x2
) ×
∂
∂y
y3
= 3(x4
+ x2
)y2
.

Exercise 1(c) If z = y
1
2 sin(x) then to calculate
∂z
∂x
the y
1
2 factor is
kept constant:
∂z
∂x
=
∂
∂x

y
1
2 sin(x)

= y
1
2 ×
∂
∂x
(sin(x)) = y
1
2 cos(x) .
Similarly, to evaluate the partial derivative
∂z
∂y
the factor sin(x) is
treated as a constant:
∂z
∂y
=
∂
∂y

y
1
2 sin(x)

=
∂
∂y
y
1
2 × sin(x) =
1
2
y− 1
2 sin(x) .

Exercise 2(a) The function z = (x2
+ 3x) sin(y) can be written as
z = uv , where u = (x2
+ 3x) and v = sin(y) . The partial derivatives
of u and v with respect to the variable x are
∂u
∂x
= 2x + 3 ,
∂v
∂x
= 0 ,
while the partial derivatives with respect to y are
∂u
∂y
= 0 ,
∂v
∂y
= cos(y) .
Applying the product rule
∂z
∂x
=
∂u
∂x
v + u
∂v
∂x
= (2x + 3) sin(y) .
∂z
∂y
=
∂u
∂y
v + u
∂v
∂y
= (x2
+ 3x) cos(y) .

Exercise 2(b)
The function z =
cos(x)
y5
can be written as z = cos(x)y−5
.
Treating the factor y−5
as a constant and differentiating with respect
to x:
∂z
∂x
= − sin(x)y−5
= −
sin(x)
y5
.
Treating cos(x) as a constant and differentiating with respect to y:
∂v
∂y
= cos(x)(−5y−6
) = −5
cos(x)
y6
.

Exercise 2(c) The function z = ln(xy) can be rewritten as (see the
package on logarithms)
z = ln(xy) = ln(x) + ln(y) .
Thus the partial derivative of z with respect to x is
∂z
∂x
=
∂
∂x
(ln(x) + ln(y)) =
∂
∂x
ln(x) =
1
x
.
Similarly the partial derivative of z with respect to y is
∂z
∂y
=
∂
∂y
(ln(x) + ln(y)) =
∂
∂y
ln(y) =
1
y
.

Exercise 2(d) To calculate the partial derivatives of the function
z = sin(x) cos(xy) the product rule has to be applied
∂z
∂x
= cos(xy)
∂
∂x
sin(x) + sin(x)
∂
∂x
cos(xy) ,
∂z
∂y
= cos(xy)
∂
∂y
sin(x) + sin(x)
∂
∂y
cos(xy) .
Using the chain rule with u = xy for the partial derivatives of cos(xy)
∂
∂x
cos(xy) =
∂ cos(u)
∂u
∂u
∂x
= − sin(u)y = −y sin(xy) ,
∂
∂y
cos(xy) =
∂ cos(u)
∂u
∂u
∂y
= − sin(u)x = −x sin(xy) .
Thus the partial derivatives of z = sin(x) cos(xy) are
∂z
∂x
= cos(xy) cos(x) − y sin(x) sin(xy) ,
∂z
∂y
= −x sin(x) sin(xy) .

Exercise 2(e) To calculate the partial derivatives of z = e(x2
+y2
)
the
chain rule has to be applied with u = (x2
+ y2
):
∂z
∂x
=
∂
∂u
(eu
)
∂u
∂x
= eu ∂u
∂x
,
∂z
∂y
=
∂
∂u
(eu
)
∂u
∂y
= eu ∂u
∂y
.
The partial derivatives of u = (x2
+ y2
) are
∂u
∂x
=
∂(x2
)
∂x
= 2x ,
∂u
∂y
=
∂(y2
)
∂y
= 2y .
Therefore the partial derivatives of the function z = e(x2
+y2
)
are
∂z
∂x
= eu ∂u
∂x
= 2x e(x2
+y2
)
,
∂z
∂x
= eu ∂u
∂x
= 2y e(x2
+y2
)
.

Exercise 2(f) Applying the chain rule with u = x2
+ y the partial
derivatives of the function z = sin(x2
+ y) can be written as
∂z
∂x
=
∂
∂u
(sin(u))
∂u
∂x
= cos(u)
∂u
∂x
,
∂z
∂y
=
∂
∂u
(sin(u))
∂u
∂y
= cos(u)
∂u
∂y
.
The partial derivatives of u = x2
+ y are
∂u
∂x
=
∂x2
∂x
= 2x ,
∂u
∂y
=
∂y
∂y
= 1 .
Thus the partial derivatives of the function z = sin(x2
+ y) are
∂z
∂x
= cos(u)
∂u
∂x
= 2x cos(x2
+ y) ,
∂z
∂y
= cos(u)
∂u
∂y
= cos(x2
+ y) .

Exercise 3(a)
From exercise 2(a), the first order partial derivatives of
z = (x2
+ 3x) sin(y) are
∂z
∂x
= (2x + 3) sin(y) ,
∂z
∂y
= (x2
+ 3x) cos(y) .
The mixed second order derivatives are
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x
(x2
+ 3x) cos(y)

= (2x + 3) cos(y) ,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y
((2x + 3) sin(y)) = (2x + 3) cos(y) .

Exercise 3(b)
From exercise 2(b), the first order partial derivatives of z =
cos(x)
y5
are
∂z
∂x
= −
sin(x)
y5
,
∂z
∂y
= − 5
cos(x)
y6
,
so the mixed second order derivatives are
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x

−5
cos(x)
y6

= 5
sin(x)
y6
,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y

−
sin(x)
y5

= 5
sin(x)
y6
.

Exercise 3(c)
From exercise 2(c), the first order partial derivatives of z = ln(xy)
are
∂z
∂x
=
1
x
,
∂z
∂y
=
1
y
.
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x

1
y

= 0 ,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y

1
x

= 0 .

Exercise 3(d) From exercise 2(d), the first order partial derivatives
of z = sin(x) cos(xy) are
∂z
∂x
= cos(x) cos(xy) − y sin(x) sin(xy) ,
∂z
∂y
= − x sin(x) sin(xy) .
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x
(−x sin(x) sin(xy))
= − sin(x) sin(xy) − x cos(x) sin(xy) − xy sin(x) cos(xy) ,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y
(cos(x) cos(xy) − y sin(x) sin(xy))
= −x cos(x) sin(xy) − sin(x) sin(xy) − xy sin(x) cos(xy) .
N.B. In the solution above a product of three functions has been
differentiated. This can be done by using two applications of the
product rule.

Exercise 3(e) From exercise 2(e), the first order partial derivatives
of z = e(x2
+y2
)
are
∂z
∂x
= 2xe(x2
+y2
)
,
∂z
∂y
= 2ye(x2
+y2
)
.
The mixed second order derivatives are thus
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x

2ye(x2
+y2
)

= 4xye(x2
+y2
)
,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y

2xe(x2
+y2
)

= 4yxe(x2
+y2
)
.

Exercise 3(f) From exercise 2(f), the first order partial derivatives
of z = sin(x2
+ y) are
∂z
∂x
= 2x cos(x2
+ y) ,
∂z
∂y
= cos(x2
+ y) .
The mixed second order derivatives are thus
∂2
z
∂x∂y
=
∂
∂x

∂z
∂y

=
∂
∂x
cos(x2
+ y)

= −2x sin(x2
+ y) ,
∂2
z
∂y∂x
=
∂
∂y

∂z
∂x

=
∂
∂y
2x cos(x2
+ y)

= −2x sin(x2
+ y) .

Solutions to Quizzes 29
Solutions to Quizzes
Solution to Quiz:
To determine which of the options is correct, the partial derivatives
of z = cos(xy) must be calculated. From the calculations of exercise
2(d) the partial derivatives of z = cos(xy) are
∂
∂x
cos(xy) =
∂ cos(u)
∂u
∂u
∂x
= − sin(u)y = −y sin(xy) ,
∂
∂y
cos(xy) =
∂ cos(u)
∂u
∂u
∂y
= − sin(u)x = −x sin(xy) .
Therefore
1
y
∂
∂x
cos(xy) = − sin(xy) =
1
x
∂
∂y
cos(xy) .
The other choices, if checked, will be found to be false. End Quiz

Solutions to Quizzes 30
Solution to Quiz:
The first order derivatives of z = e−y
sin(x) are
zx = e−y
cos(x) , zy = − e−y
sin(x) ,
where e−y
is kept constant for the first differentiation and sin(x) for
the second. Continuing in this way, the second order derivatives zxx
and zyy are given by the expressions
zxx =
∂
∂x

∂z
∂x

=
∂
∂x
e−y
cos(x)

= −e−y
sin(x) ,
zyy =
∂
∂y

∂z
∂y

=
∂
∂y
−e−y
sin(x)

= e−y
sin(x) .
Adding these two equations together gives
zxx + zyy = 0 .
End Quiz

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