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.
2.
3.
4.
Partial Differentiation (Introduction)
The Rules of Partial Differentiation
Higher Order Partial Derivatives
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
kT
p=
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
Section 1: Partial Differentiation (Introduction)
kT
, find the partial derivatives of p:
V
(a) with respect to T ,
(b) with respect to V .
Example 1 If p =
Solution
(a) This part of the example proceeds as follows:
kT
,
V
∂p
k
∴
=
,
∂T
V
where V is treated as a constant for this calculation.
(b) For this part, T is treated as a constant. Thus
1
p = kT
= kT V −1 ,
V
∂p
kT
∴
= −kT V −2 = − 2 .
∂V
V
p
=
4
Section 1: Partial Differentiation (Introduction)
5
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.
∂z
∂z
Example 2 Find
and
for the function z = x2 y 3 .
∂x
∂y
Solution
z = x2 y 3
For the first part y 3 is treated as
∂z
a constant and the derivative of
∴
= 2xy 3 ,
∂x
x2 with respect to x is 2x.
∂z
For the second part x2 is treated
and
= x2 3y 2 ,
∂y
as a constant and the derivative
2 2
of y 3 with respect to y is 3y 2 .
= 3x y .
∂z
∂z
and
for each of the following functions.
∂x
∂y
(Click on the green letters for solutions.)
1
(a) z = x2 y 4 ,
(b) z = (x4 + x2 )y 3 , (c) z = y 2 sin(x).
Exercise 1. Find
Section 2: The Rules of Partial Differentiation
6
2. The Rules of Partial Differentiation
Since partial differentiation is essentially the same as ordinary differentiation, the product, quotient and chain rules may be applied.
∂z
Example 3 Find
for each of the following functions.
∂x
x−y
(a) z = xy cos(xy) , (b) z =
, (c) z = (3x + y)2 .
x+y
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
∂u
∴
∂x
Thus
= xy
and
= y
and
v
∂v
∂x
= cos(xy)
= −y sin(xy) .
∂z ∂u
∂v
=
v+u
= y cos(xy) − xy 2 sin(xy) .
∂x ∂x
∂x
Section 2: The Rules of Partial Differentiation
7
(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
∂u
∴
∂x
Thus
∂z
∂x
= x−y
and
= 1
and
v
∂v
∂x
= x+y
= 1.
∂u
∂v
−u
∂x
∂x
v2
(x + y) − (x − y)
2y
=
.
(x + y)2
(x + y)2
v
=
=
(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
= 3.
∂u
∂x
Thus
∂z
∂z ∂u
=
= 2(3x + y)3 = 6(3x + y) .
∂x
∂u ∂x
Section 2: The Rules of Partial Differentiation
8
∂z
∂z
and
for each of the following functions.
∂x
∂y
(Click on the green letters for solutions.)
Exercise 2. Find
(a) z = (x2 +3x) sin(y),
(d) z = sin(x) cos(xy),
(b) z =
cos(x)
,
y5
2
(e) z = e(x
+y 2 )
,
(c) z = ln(xy),
(f) z = sin(x2 + y).
Quiz If z = cos(xy), which of the following statements is true?
(a)
∂z
∂z
=
,
∂x
∂y
(b)
∂z
1 ∂z
=
,
∂x
x ∂y
(c)
1 ∂z
∂z
=
,
y ∂x
∂y
(d)
1 ∂z
1 ∂z
=
.
y ∂x
x ∂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.
∂2z
3
2
Example 4 Find
if z = e(x +y ) .
∂x2
Solution First differentiate z with respect to x, keeping y constant,
then differentiate this function with respect to x, again keeping y
3
2
constant.
z = e(x +y )
3
2
∂z
∴
= 3x2 e(x +y ) using the chain rule
∂x
(x3 +y 2 )
∂2z
∂(3x2 ) (x3 +y2 )
)
2 ∂(e
=
e
+ 3x
using the product rule
∂x2
∂x
∂x
2
3
2
3
2
∂ z
= 6xe(x +y ) + 3x2 (3x2 e(x +y ) )
2
∂x
3
2
= (9x4 + 6x)e(x +y )
Section 3: Higher Order Partial Derivatives
10
∂2z
∂2z
and
, when there are two variables there
∂x2
∂y 2
is also the possibility of a mixed second order derivative.
∂2z
3
2
Example 5 Find
if z = e(x +y ) .
∂x∂y
∂2z
∂ ∂z
Solution The symbol
is interpreted as
; in words,
∂x∂y
∂x ∂y
first differentiate z with respect to y, keeping x constant, then differentiate 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.)
3
2
∂z
First with x constant
= 2ye(x +y ) (using the chain rule.)
∂y
∂2z
∂ (x3 +y2 ) Second with y constant
=
2ye
∂x∂y
∂x
In addition to both
3
= 3x2 2ye(x
3
= 6x2 ye(x
+y 2 )
+y 2 )
.
Section 3: Higher Order Partial Derivatives
11
The obvious question now to be answered is: what happens if the
order of differentiation is reversed?
∂2z
∂ ∂z
3
2
Example 6 Find
=
if z = e(x +y ) .
∂y∂x
∂y ∂x
Solution
3
2
∂z
First with y constant
= 3x2 e(x +y ) (using the chain rule).
∂x
∂2z
∂ 2 (x3 +y2 ) Second with x constant
=
3x e
∂y∂x
∂y
3
= 2y3x2 e(x
+y 2 )
∂2z
.
∂x∂y
As a general rule, when calculating mixed derivatives the order of
differentiation may be reversed without affecting the final result.
3
= 6x2 ye(x
+y 2 )
=
Section 3: Higher Order Partial Derivatives
12
Exercise 3. Confirm the statement on the previous page by finding
∂2z
∂2z
both
and
for each of the following functions, whose first
∂x∂y
∂y∂x
order partial derivatives have already been found in exercise 2. (Click
on the green letters for solutions.)
cos(x)
(a) z = (x2 +3x) sin(y),
(b) z =
,
(c) z = ln(xy),
y5
2
2
(d) z = sin(x) cos(xy),
(e) z = e(x +y ) , (f) z = sin(x2 + y).
Notation For first and second order partial derivatives there is a
∂f
∂f
compact notation. Thus
can be written as fx and
as fy .
∂x
∂y
∂2f
∂2f
is written fxx while
is written fxy .
Similarly
2
∂x
∂x∂y
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 + y 3 then
(a) 2x + 3y + 3y 2 ,
(c) 2x + 3x ,
∂z
is
∂x
(b) 2x + 3x + 3y 2 ,
(d) 2x + 3y .
2. If w = 1/r, where r2 = x2 + y 2 + z 2 , then xwx + ywy + zwz is
(a) −1/r ,
(b) 1/r ,
r
x
3. If u =
then uxx is
y
(c) −1/r2 ,
(d) 1/r2 .
1
1
1
1
(a) − p
, (b) − p
, (c) − p
, (d) − p
.
3
3
3
3
3
4 y x
4 yx
8 y x
8 yx3
End Quiz Score:
Correct
Solutions to Exercises
14
Solutions to Exercises
Exercise 1(a) To calculate the partial derivative
∂z
of the function
∂x
z = x2 y 4 , the factor y 4 is treated as a constant:
∂z
∂
∂
=
x2 y 4 =
x2 × y 4 = 2x(2−1) × y 4 = 2xy 4 .
∂x
∂x
∂x
∂z
Similarly, to find the partial derivative
, the factor x2 is treated
∂y
as a constant:
∂
∂z
∂
x2 y 4 = x2 ×
y 4 = x2 × 4y (4−1) = 4x2 y 3 .
=
∂y
∂y
∂y
Click on the green square to return
Solutions to Exercises
15
∂z
for the function z = (x4 + x2 )y 3 , the
∂x
factor y 3 is treated as a constant:
∂z
∂
∂
=
(x4 + x2 )y 3 =
x4 + x2 × y 3 = (4x3 + 2x)y 3 .
∂x
∂x
∂x
∂z
To find the partial derivative
the factor (x4 + x2 ) is treated as a
∂y
constant:
∂z
∂
∂ 3
=
(x4 + x2 )y 3 = (x4 + x2 ) ×
y = 3(x4 + x2 )y 2 .
∂y
∂y
∂y
Click on the green square to return
Exercise 1(b) To calculate
Solutions to Exercises
16
∂z
1
1
Exercise 1(c) If z = y 2 sin(x) then to calculate
the y 2 factor is
∂x
kept constant:
1
1
∂
∂z
∂ 1
y 2 sin(x) = y 2 ×
=
(sin(x)) = y 2 cos(x) .
∂x
∂x
∂x
∂z
Similarly, to evaluate the partial derivative
the factor sin(x) is
∂y
treated as a constant:
∂ 1
1 1
∂z
∂ 1
y 2 sin(x) =
=
y 2 × sin(x) = y − 2 sin(x) .
∂y
∂y
∂y
2
Click on the green square to return
Solutions to Exercises
17
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
∂v
= 2x + 3 ,
= 0,
∂x
∂x
while the partial derivatives with respect to y are
∂u
∂v
= 0,
= cos(y) .
∂y
∂y
Applying the product rule
∂z ∂u
∂v
=
= (2x + 3) sin(y) .
v+u
∂x ∂x
∂x
∂v
∂z ∂u
=
v+u
= (x2 + 3x) cos(y) .
∂y ∂y
∂y
Click on the green square to return
Solutions to Exercises
18
Exercise 2(b)
cos(x)
can be written as z = cos(x)y −5 .
y5
Treating the factor y −5 as a constant and differentiating with respect
to x:
The function z =
∂z
sin(x)
= − sin(x)y −5 = − 5 .
∂x
y
Treating cos(x) as a constant and differentiating with respect to y:
∂v
cos(x)
.
= cos(x)(−5y −6 ) = −5
∂y
y6
Click on the green square to return
Solutions to Exercises
19
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
∂
∂
1
=
(ln(x) + ln(y)) =
ln(x) = .
∂x
∂x
∂x
x
Similarly the partial derivative of z with respect to y is
∂z
∂
1
∂
=
(ln(x) + ln(y)) =
ln(y) = .
∂y
∂y
∂y
y
Click on the green square to return
Solutions to Exercises
20
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
∂z
∂y
∂
∂
sin(x) + sin(x)
cos(xy) ,
∂x
∂x
∂
∂
= cos(xy)
sin(x) + sin(x)
cos(xy) .
∂y
∂y
= cos(xy)
Using the chain rule with u = xy for the partial derivatives of cos(xy)
∂
cos(xy) =
∂x
∂
cos(xy) =
∂y
∂ cos(u)
∂u
∂ cos(u)
∂u
∂u
= − sin(u)y = −y sin(xy) ,
∂x
∂u
= − sin(u)x = −x sin(xy) .
∂y
Thus the partial derivatives of z = sin(x) cos(xy) are
∂z
= cos(xy) cos(x) − y sin(x) sin(xy) ,
∂x
Click on the green square to return
∂z
= −x sin(x) sin(xy) .
∂y
Solutions to Exercises
21
2
Exercise 2(e) To calculate the partial derivatives of z = e(x
chain rule has to be applied with u = (x2 + y 2 ):
∂z
∂x
∂z
∂y
=
=
∂ u
(e )
∂u
∂ u
(e )
∂u
+y 2 )
the
∂u
∂u
= eu
,
∂x
∂x
∂u
∂u
= eu
.
∂y
∂y
The partial derivatives of u = (x2 + y 2 ) are
∂u
∂(x2 )
=
= 2x ,
∂x
∂x
∂u
∂(y 2 )
=
= 2y .
∂y
∂y
2
Therefore the partial derivatives of the function z = e(x
2
2
∂z
∂u
= eu
= 2x e(x +y ) ,
∂x
∂x
2
2
∂z
u ∂u
= e
= 2y e(x +y ) .
∂x
∂x
Click on the green square to return
+y 2 )
are
Solutions to Exercises
22
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
∂z
∂y
=
=
∂
(sin(u))
∂u
∂
(sin(u))
∂u
∂u
∂u
,
= cos(u)
∂x
∂x
∂u
∂u
= cos(u)
.
∂y
∂y
The partial derivatives of u = x2 + y are
∂u
∂x2
= 2x ,
=
∂x
∂x
∂u
∂y
= 1.
=
∂y
∂y
Thus the partial derivatives of the function z = sin(x2 + y) are
∂z
∂u
= 2x cos(x2 + y) ,
= cos(u)
∂x
∂x
∂z
∂u
= cos(u)
= cos(x2 + y) .
∂y
∂y
Click on the green square to return
Solutions to Exercises
23
Exercise 3(a)
From exercise 2(a), the first order partial derivatives of
z = (x2 + 3x) sin(y) are
∂z
∂z
= (2x + 3) sin(y) ,
= (x2 + 3x) cos(y) .
∂x
∂y
The mixed second order derivatives are
∂2z
∂ ∂z
∂
=
=
(x2 + 3x) cos(y) = (2x + 3) cos(y) ,
∂x∂y
∂x ∂y
∂x
∂2z
∂ ∂z
∂
=
=
((2x + 3) sin(y)) = (2x + 3) cos(y) .
∂y∂x
∂y ∂x
∂y
Click on the green square to return
Solutions to Exercises
24
Exercise 3(b)
From exercise 2(b), the first order partial derivatives of z =
are
∂z
sin(x)
∂z
cos(x)
=−
,
= −5
,
∂x
y5
∂y
y6
so the mixed second order derivatives are
∂
sin(x)
∂2z
∂ ∂z
cos(x)
=
=
=5 6 ,
−5
∂x∂y
∂x ∂y
∂x
y6
y
2
∂ z
∂ ∂z
∂
sin(x)
sin(x)
=
=
− 5
=5 6 .
∂y∂x
∂y ∂x
∂y
y
y
cos(x)
y5
Click on the green square to return
Solutions to Exercises
25
Exercise 3(c)
From exercise 2(c), the first order partial derivatives of z = ln(xy)
are
∂z 1
∂z 1
= ,
= .
∂x x
∂y y
The mixed second order derivatives are
∂ 1
∂2z
∂ ∂z
=
=
= 0,
∂x∂y
∂x ∂y
∂x y
∂2z
∂ ∂z
∂ 1
=
=
= 0.
∂y∂x
∂y ∂x
∂y x
Click on the green square to return
Solutions to Exercises
26
Exercise 3(d) From exercise 2(d), the first order partial derivatives
of z = sin(x) cos(xy) are
∂z
∂z
= cos(x) cos(xy) − y sin(x) sin(xy) ,
= − x sin(x) sin(xy) .
∂x
∂y
The mixed second order derivatives are
∂
∂2z
∂ ∂z
=
=
(−x sin(x) sin(xy))
∂x∂y
∂x ∂y
∂x
= − sin(x) sin(xy) − x cos(x) sin(xy) − xy sin(x) cos(xy) ,
∂2z
∂ ∂z
∂
=
=
(cos(x) cos(xy) − y sin(x) sin(xy))
∂y∂x
∂y ∂x
∂y
= −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.
Click on the green square to return
Solutions to Exercises
27
Exercise 3(e) From exercise 2(e), the first order partial derivatives
2
2
of z = e(x +y ) are
2
2
2
2
∂z
∂z
= 2xe(x +y ) ,
= 2ye(x +y ) .
∂x
∂y
The mixed second order derivatives are thus
2
2
∂2z
∂ ∂z
∂ (x2 +y2 ) =
=
2ye
= 4xye(x +y ) ,
∂x∂y
∂x ∂y
∂x
2
2
2
2
∂2z
∂ ∂z
∂ 2xe(x +y ) = 4yxe(x +y ) .
=
=
∂y∂x
∂y ∂x
∂y
Click on the green square to return
Solutions to Exercises
28
Exercise 3(f ) From exercise 2(f ), the first order partial derivatives
of z = sin(x2 + y) are
∂z
∂z
= 2x cos(x2 + y) ,
= cos(x2 + y) .
∂x
∂y
The mixed second order derivatives are thus
∂
∂2z
∂ ∂z
=
=
cos(x2 + y) = −2x sin(x2 + y) ,
∂x∂y
∂x ∂y
∂x
∂2z
∂ ∂z
∂
=
=
2x cos(x2 + y) = −2x sin(x2 + y) .
∂y∂x
∂y ∂x
∂y
Click on the green square to return
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
∂
∂ cos(u) ∂u
cos(xy) =
= − sin(u)y = −y sin(xy) ,
∂x
∂u
∂x
∂
∂ cos(u) ∂u
= − sin(u)x = −x sin(xy) .
cos(xy) =
∂y
∂u
∂y
Therefore
1 ∂
1 ∂
cos(xy) = − sin(xy) =
cos(xy) .
y ∂x
x ∂y
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
∂
∂
∂z
zxx =
=
e−y cos(x) = −e−y sin(x) ,
∂x ∂x
∂x
∂
∂ ∂z
zyy =
=
−e−y sin(x) = e−y sin(x) .
∂y ∂y
∂y
Adding these two equations together gives
zxx + zyy = 0 .
End Quiz