4 Simultaneous Equations: Two Equations with Two Unknowns
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54
Mathematics for Physical Chemistry
We substitute this into the first equation to obtain an
equation in one unknown
x 2 − 4x + 2x 2 − x = 0,
2
3x − 5x = 0.
Since we can factor x out of this equation, we have two
solutions:
x =0
and
5
.
3
We substitute each of these into the first equation. The first
solution set is
x = 0, y = 2.
x=
5.4.3 Consistency and Independence in
Simultaneous Equations
There are two common difficulties that can arise with pairs
of simultaneous inhomogeneous equations: these are (1)
that the equations might be inconsistent and (2) that the
equations might not be independent. If two equations are
inconsistent, there is no solution that can satisfy both of
them. If the equations are not independent, they express the
same information, so that there is really only one equation,
which can be solved for one variable in terms of the other but
cannot be solved to give numerical values for both variables.
Example 5.16.
inconsistent:
2x + 3y = 15,
4x + 6y = 45.
The second solution set is
x=
Show that the pair of equations is
5
1
, y= .
3
3
In this example there are two solution sets, since the first
equation is quadratic in x.
Exercise 5.15. Solve the simultaneous equations by the
method of substitution:
2
x − 2x y − x = 0,
x + y = 0.
5.4.2 The Method of Elimination
This method consists of the process of subtracting one
equation from another to obtain a simpler equation. If
necessary, we can multiply each equation by a constant
before subtracting.
Example 5.15. Solve the following pair of equations:
x + y = 3,
2x + y = 0.
We subtract the first equation from the second to obtain
x = −3.
This is substituted into either of the original equations to
obtain
y = 6.
Exercise 5.16. Solve the set of equations
3x + 2y = 40,
2x − y = 10.
We attempt a solution by elimination. We multiply the
first equation by 2 and subtract the second from the first,
obtaining
0 = −15,
which is obviously not correct. The equations are
inconsistent.
Example 5.17.
independent:
Show that the equations are not
3x + 4y = 7,
6x + 8y = 14.
We attempt a solution by elimination, multiplying the
first equation by 2. We have just one independent equation
instead of two, so that we could solve for x in terms of y or for
y in terms of x, but not for numerical values of either x or y.
We can understand consistency and independence in
simultaneous equations by looking at the graphs of the
equations. Each of the graphs represents y as a linear
function of x.
Figure 5.1 shows the lines representing two consistent
and independent equations. The lines cross at the point
whose coordinates represent the solution, consisting of a
value of x and a value of y that satisfy the equations.
Figure 5.2 shows the lines representing two inconsistent
equations. Since the lines do not cross, there is no solution to
this pair of inconsistent equations. A single line represents
two equations that are not independent. Any point on the
line satisfies both equations. The line represent y as a
function of x.
5.4.4 Homogeneous Linear Equations
A pair of homogeneous linear equations can be written in
the form:
CHAPTER | 5 Problem Solving and the Solution of Algebraic Equations
55
y
Both of these functions are represented by straight lines
with zero intercept. There are two possibilities: Either the
lines cross at the origin or they coincide everywhere. In
other words, either x = 0, y = 0 is the solution, or else the
equations are linearly dependent (are the same equation).
The solution x = 0, y = 0 is called a trivial solution. The
two equations must be linearly dependent in order for a
nontrivial solution to exist, so that there is really only one
equation. A nontrivial solution consists of specifying y as a
function of x, not in finding constant values for both x and y.
(−3, 6)
0
Example 5.18. Determine whether the set of equations
has a nontrivial solution, and find the solution if it exists:
x
y = 3 −x
7x + 15y = 0,
y = −2x
101x + 195y = 0.
From the first equation
y=−
FIGURE 5.1 Graphical representation of the two consistent and linearly
independent equations of Example 5.15.
From the second equation
y=−
15 2
x
−
3
3
101x
= −0.51785x.
195
These two equations are represented by lines that cross at
the origin, so there is no nontrivial solution.
y
y=
7x
= −0.46667x.
15
Exercise 5.17. Determine whether the set of equations has
a nontrivial solution, and find the solution if it exists
5x + 12y = 0,
15x + 36y = 0.
5.4.5 Using Mathematica to Solve
Simultaneous Equations
x
y=5−
2x
3
The Solve statement can also be used to solve simultaneous
equations as well as single equations. The equations are
typed inside curly brackets with commas between them,
and the variables are listed inside curly brackets.
FIGURE 5.2 Graphical representation of the two inconsistent equations
of Example 5.16.
Example 5.19. Solve the equations
ax + by = c,
a11 x + a12 y = 0,
(5.18a)
a21 x + a22 y = 0.
(5.18b)
−a11 x
,
a12
−a21 x
y =
.
a22
We type the input entry
In[1]:=Solve[{a ∗ x + b ∗ y == c,g ∗ x + h ∗ y ==
k},{x,y}]
We solve both of these equations for y in terms of x:
y =
gx + hy = k.
(5.19)
(5.20)
The output is
Out[1] =
x →−
−ch + bk
−cg + ak
,y → −
−bg + ah
bg − ah
56
Mathematics for Physical Chemistry
Blank spaces could be used instead of the asterisks to
denote multiplication. Braces (curly brackets) are used to
notify Mathematica that we have a list of two equations to be
solved. The two variables to be solved for must be included
inside braces. If numerical values for the coefficients are
specified, Mathematica will give the numerical solution set.
Exercise 5.18. Use Mathematica to solve the simultaneous equations
2x + 3y = 13,
x − 4y = −10.
The Eliminate statement is used to eliminate one or
more of the variables in a set of simultaneous equations.
For example, to obtain a single equation in x from the set of
equations above, you would type the input entry (note the
double equal signs):
Eliminate[{a x + b y == c, g x + h y == k},y]
and would receive the output:
Out[1] = b k − b g x + a h x == c h
we solve this equation for x by typing
Solve[%,x]
and receive the output
Out[2]=
x→
−ch + bk
bg − ah
Find the pH of a solution formed from 0.075 mol of
NH3 and enough water to make 1.00 l of solution. The
ionization that occurs is
−
NH+
4 + OH .
NH3 + H2 O
The equilibrium expression in terms of molar
concentrations is
−
◦
◦
([NH+
4 ]/c )([OH ]/c )
x(H2 O)([NH3 ]/c◦ )
−
◦
◦
([NH+
4 ]/c )([OH ]/c )
≈
,
◦
([NH3 ]/c )
Kb =
where x(H2 O) represents the mole fraction of water,
which is customarily used instead of its molar
concentration. Since the mole fraction of the solvent
in a dilute solution is nearly equal to unity, we can
use the approximate version of the equation. The base
ionization constant of NH3 , denoted by K b , equals
1.80 × 10−5 at at 25 ◦ C.
5. The acid ionization constant of chloroacetic acid is
equal to 1.40 × 10−3 at 25 ◦ C. Assume that activity
coefficients are equal to unity and find the hydrogenion concentration at the following stoichiometric
molarities:
(a) 0.100 mol l−1 .
(b) 0.0100 mol l−1 .
6. Find the real roots of the following equations by
graphing:
(a) x 3 − x 2 + x − 1 = 0.
(b) e−x − 0.5x = 0.
(c) sin (x)/x − 0.75 = 0.
7. Make a properly labeled graph of the function y(x) =
ln (x) + cos (x) for values of x from 0 to 2.
PROBLEMS
1. Solve the quadratic equations:
(a) x 2 − 3x + 2 = 0.
(b) x 2 − 1 = 0.
(c) x 2 + 2x + 2 = 0.
2. Solve the following equations by factoring:
(a) x 3 + x 2 − x − 1 = 0.
(b) x 4 − 1 = 0.
3. Rewrite the factored quadratic equation (x − x1 )(x −
x2 ) = 0 in the form x 2 − (x1 + x2 )x + x12 x = 0.
Apply the quadratic formula to this version and show
that the roots are x = x1 and x = x2 .
4. The pH is defined for our present purposes as
pH = − log10 ([H+ ]/c◦ ).
(a) Use Excel.
(b) Use Mathematica.
8. When expressed in terms of “reduced variables,” the
van der Waals equation of state is
Pr +
3
Vr2
Vr −
1
3
=
8Tr
.
3
(a) Using Excel, construct a graph containing three
curves of Pr as a function of Vr : for the range
0.4 < Vr < 2:one for Tr = 0.6, one for Tr = 1,
and one for Tr = 1.4.
(b) Repeat part a using Mathematica.
9. Using a graphical method, find the two positive roots
of the following equation.
e x − 3.000x = 0.
CHAPTER | 5 Problem Solving and the Solution of Algebraic Equations
57
10. The following data were taken for the thermal
decomposition of N2 O3 at a constant temperature:
where c is the gross acid concentration. This equation
is based on the assumption that the concentration
of unionized acid is approximately equal to the
★
✥ gross acid concentration. Consider a solution of
t/s
0
184 426 867 1877
HCN (hydrocyanic acid) with stoichiometric acid
−1
concentration equal to 1.00 × 10−5 mol l−1 . K a =
[N2 O3 ]/mol l
2.33 2.08 1.67 1.36 0.72
✧
✦ 4 × 10−10 for HCN. At this temperature, K w =
1.00 × 10−14 .
Using Excel, make three graphs: one with
(a) Calculate [H+ ]/c◦ using this equation.
ln ([N2 O3 ]) as a function of t, one with 1/[N2 O3 ] as
2
(b) Calculate [H+ ]/c◦ using Eq. (5.9).
a function of t, and one with 1/[N2 O3 ] as a function
of t. Determine which graph is most nearly linear. If
the first graph is most nearly linear, the reaction is first
order; if the second graph is most nearly linear, the
reaction is second order, and if the third graph is most
nearly linear, the reaction is third order.
11. Write an Excel worksheet that will convert a list of
distance measurements in meters to miles, feet, and
inches. If the length in meters is typed into a cell in
column A, let the corresponding length in miles appear
on the same line in column B, the length in feet in
column C, and the length in inches in column C.
12. The van der Waals equation of state is
P+
n2a
V2
(V − nb) = n RT ,
where a and b are temperature-independent parameters that have different values for each gas. For
carbon dioxide, a = 0.3640 Pa m6 mol−2 and b =
4.267 × 10−5 m3 mol−1 .
(a) Write this equation as a cubic equation in V.
(b) Use the NSolve statement in Mathematica to
find the volume of 1.000 mol of carbon dioxide
at P = 1.000 bar (100000 Pa) and T = 298.15 K.
Notice that two of the three roots are complex and
must be ignored. Compare your result with the
prediction of the ideal gas equation of state.
(c) Use the FindRoot statement in Mathematica to find
the real root in part b.
(d) Repeat part b for P = 10.000 bar (1.0000 ×
106 Pa) and T = 298.15 K. Compare your result
with the prediction of the ideal gas equation of
state.
13. An approximate equation for the ionization of a weak
acid, including consideration of the hydrogen ions
from water is2
[H+ ]/c◦ =
K a c/c◦ + K w ,
14. Find the smallest positive root of the equation.
sinh(x) − x 2 − x = 0.
15. Solve the cubic equation by trial and error, factoring,
or by using Mathematica or Excel:
x 3 + x 2 − 4x − 4 = 0.
16. Find the real root of the equation
x 2 − e−x = 0.
17. Find the root of the equation
x − 2.00 sin (x) = 0.
18. Find two positive roots of the equation
ln (x) − 0.200x = 0.
19. Find the real roots of the equation
x 2 − 2.00 − cos (x) = 0.
20. In the theory of black-body radiation, the following
equation
x = 5(1 − e−x )
needs to be solved to find the wavelength of maximum
spectral radiant emittance. The variable x is
x=
hc
,
λmax kB T
where λmax is the wavelength of maximum spectral
radiant emittance, h is Planck’s constant, c is the speed
of light, kB is Boltzmann’s constant, and T is the
absolute temperature. Solve the equation numerically
for a value of x. Find the value of λmax for T = 6600 K.
In what region of the electromagnetic spectrum does
this value lie?
21. Solve the simultaneous equations by hand, using the
method of substitution:
x 2 + x + 3y = 15,
2 Henry F. Holtzclaw, Jr, William R. Robinson, and Jerone D. Odom,
General Chemistry, 9th ed., p. 545, Heath, Lexington, MA, 1991.
3x + 4y = 18.
58
Mathematics for Physical Chemistry
Use Mathematica to check your result. Since the first
equation is a quadratic equation, there will be two
solution sets.
22. Stirling’s approximation for ln (N !) is
ln (N !) ≈
1
ln (2π N ) + N ln (N ) − N .
2
(a) Determine the validity of this approximation and
of the less accurate version
ln (N !) ≈ N ln (N ) − N
for several values of N up to N = 100. Use a
calculator, Excel, or Mathematica.
23. The Dieterici equation of state is
Pea/Vm RT (Vm − b) = RT ,
where P is the pressure, T is the temperature, Vm is
the molar volume, and R is the ideal gas constant.
The constant parameters a and b have different
values for different gases. For carbon dioxide, a =
0.468 Pa m6 mol−2 , b = 4.63 × 10−5 m3 mol−1 .
Without linearization, find the molar volume of carbon
dioxide if T = 298.15 K and P = 10.000 atm =
1.01325 × 106 Pa. Use Mathematica, Excel, or trial
and error.
24. Determine which, if any, of the following sets of
equations is inconsistent or linearly dependent. Draw
a graph for each set of equations, showing both
equations. Find the solution for any set that has a
unique solution.
(a) x + 3y = 4,
2x + 6y = 8.
(b) 2x + 4y = 24,
x + 2y = 8.
(c) 3x1 + 4x2 = 10,
4x1 − 2x2 = 6.
25. Solve the set of equations using Mathematica or by
hand with the method of substitution:
x 2 − 2x y + y 2 = 0,
2x + 3y = 5.
✤
✜
Chapter 6
✣
✢
Differential Calculus
Principal Facts and Ideas
6.1 THE TANGENT LINE AND THE
DERIVATIVE OF A FUNCTION
• The derivative of a function is a measure of how rapidly
the dependent variable changes with changes in the
value of the independent variable.
• The derivative of a function is defined by a
mathematical limit.
• The derivatives of many simple functions can be
obtained by applying a few simple rules, either
separately or in combination.
• A finite increment in a dependent variable, y, can
sometimes be calculated approximately by use of the
formula
dy
x.
y≈
dx
A curve representing a nonlinear function has a different
direction (different steepness) at different points on the
curve. The line tangent to the curve at a given point is a
line that coincides with the curve at that point but does not
cross it (except at an inflection point, which we discuss
later). The tangent line has the same steepness as the curve
at the given point.
Figure 6.1 shows the line that is tangent to the curve
representing a function y(x) at x = x1 . If the tangent line
is represented by the formula
• A relative minimum or maximum value of a variable
y(x) is found at a point where dy/dx = 0.
• The curvature of a function is defined by
the slope of the tangent line is equal to m. The figure includes
a horizontal line intersecting the curve and the tangent line
at x = x1 . In the figure, we have labeled another point at
2 3/2
.
• At an inflection point, the curvature changes sign.
Objectives
After studying this chapter, you should be able to
• Obtain the formula for the derivative of any simple
function without consulting a table;
• Sketch a rough graph of any fairly simple function and
identify important features on the graph;
• Estimate a finite increment of a function using its
derivative;
• Find maximum and minimum values of a function of
one variable;
• Find inflection points of a function of one variable.
Mathematics for Physical Chemistry. http://dx.doi.org/10.1016/B978-0-12-415809-2.00006-9
© 2013 Elsevier Inc. All rights reserved.
y(x
)
1 + ddxy
y
d2 y/dx 2
(6.1)
y
K =
y = mx + b
Tangent line
of slope m
y2
y1 + mΔx
mΔx
Δy
y1
Δx
x1
x2
x
FIGURE 6.1 The curve representing the function y = y(x) and its
tangent line.
59
60
Mathematics for Physical Chemistry
x = x2 . At this point the vertical distance from the
horizontal line to the tangent line is given by
m(x2 − x1 ) = m x,
y ≈ m x,
(6.2)
where we use a standard notation:
x = x2 − x1 .
(6.3)
The distance m x is not necessarily equal to the distance
from the horizontal line to the curve, which is given by
y(x2 ) − y(x1 ) = y2 − y1 =
y = y(x) such as the function represented in Figure 6.1. If
x is not too large, we can write as an approximation
y.
(6.4)
Exercise 6.1. Using graph paper plot the curve
representing y = sin (x) for values of x lying between 0
and π/2 radians. Using a ruler, draw the tangent line at
x = π/4. By drawing a right triangle on your graph and
measuring its sides, find the slope of the tangent line.
There is a case in which the definition of the tangent
line to a curve at a point x1 is more complicated than in
the case shown in Figure 6.1. In this case, the point x1 lies
between a region in which the curve is concave downward
and a region in which the curve is concave upward. Such
a point is called an inﬂection point. For such a point, we
must consider tangent lines at points that are taken closer
and closer to x1 . As we approach closer and closer to x1
from either direction the tangent line will approach more
and more closely to a line that is the tangent line at x1 .
This line does cross the curve at the point that it shares with
the curve. Figure 6.2 shows a graph of a function with an
inflection point and the tangent line at this point.
6.1.1 The Derivative
We now discuss a mathematical way to construct the slope
of the tangent line to a curve. Consider a nonlinear function
(6.5)
where m is the slope of the tangent line. We divide both
sides of Eq. (6.5) by x:
m≈
y
y(x2 ) − y(x1 )
y2 − y1
=
=
.
x
x2 − x1
x2 − x1
(6.6)
If the curve is smooth, this equation becomes a better
approximation for m as x becomes smaller, and if we
take the mathematical limit as x2 → x1 , it becomes exact
if the limit exists:
m = lim
x2 →x1
y(x2 )−y(x1 )
x2 −x1
.
(6.7)
If the limit in Eq. (6.7) exists and has the same value when
x2 approaches x1 from either side, it is called the derivative
of the function at x = x1 . The derivative is denoted by the
symbol dy/dx or the symbol y or the symbol y (1) and is
defined by
dy
y(x2 ) − y(x1 )
= y = lim
(definition).
x2 →x1
dx
x2 − x1
(6.8)
If the derivative is to be evaluated at x1 , we would write
y (x1 ) or (dy/dx)x1 .
If the limit in Eq. (6.7) does not exist at x = x1 or if
the limit has a different value when x2 approaches x1 from
the two sides, the function is not differentiable at x = x1 .
If y(x) has a discontinuity at x = x1 , the function is not
differentiable at that point. If the function has a cusp at x =
x1 , the function is not differentiable at that point. A cusp
corresponds to a “corner” or abrupt change in direction of a
curve representing a function that is continuous at the point
in question. The curve has different tangents immediately
to the right and left of a cusp.
Figure 6.3 shows the graph of a function that has a step
discontinuity at x = b and a cusp at x = a. It is not
1.6
1.4
1.2
y ═ y(x)
1
y
0.8
0.6
0.4
0.2
0
−0.5
−0.2 0
−0.4
0.5
1
1.5
x
FIGURE 6.2 The curve representing the function y(x) = x 3 −2x 2 +x +1
and its tangent line at the inflection point.
a
b
x
FIGURE 6.3 A function that is not differentiable at x = a and at y = b.
CHAPTER | 6 Differential Calculus
61
differentiable at these points, although it is differentiable
elsewhere in the region shown in the graph.
Example 6.1. Decide where the following functions are
differentiable and where they are not differentiable:
(a) y = |x|.
This is differentiable everywhere except at x = 0,
where
√there is a cusp.
(b) y = x.
This function has real values for x 0. It is differentiable
for all positive values ofx, but the limit does not exist at
x = 0 , so it is not differentiable at x = 0. The tangent to
the curve at x = 0 is vertical.
●
●
●
A positive derivative is larger when the tangent line is
steeper.
The derivative is negative where the function decreases
as x increases.
A negative derivative is more negative (has a larger
magnitude) when the tangent line is steeper (has a
negative slope of larger magnitude).
Example 6.3. Find the derivative of y = x n where n is an
integer.
Using the binomial formula,
(x +
n!
x n−1 x
(n − 1)!1!
n!
x n−2 ( x)2 + · · · ,
+
(n − 2)!2!
x)n = x n +
Exercise 6.2. Decide where the following functions are
differentiable:
1
·
(a) y = 1−x
√
(b) y = x + 2 x.
(c) y = tan (x).
d xn
dx
x)n − x n
,
x→0
x
n!
n!
x n+ (n−1)!1!
x n−1 x + (n−2)!2!
x n−2 ( x)2
= lim
(x +
6.1.2 Derivatives of Specific Functions
= lim
+ · · · − xn
x
The derivatives of particular functions can be found from
the definition of the derivative.
= lim
n!
x n−1
(n − 1)!1!
n!
x n−2 ( x) + · · · = nx n−1 .
+
(n − 2)!2!
Example 6.2. Find the derivative of the function
y = ax 2 .
Let x = x2 −x1 and let
y = y(x2 )−y(x1 ) = y2 −y1 .
y = ax22 − ax12 = a x1 +
x
2
− ax12
= 2ax1 x + ( x)2 .
y
= 2ax1 + x.
x
We now take the limit as x2 → x1 ( x → 0) The first term,
2ax1 , is not affected. The second term, x, vanishes. Thus,
if we use the symbol x instead of x1
y
= 2ax.
x
ax 2
Figure 6.4 shows a graph of the function y =
and
a graph of its derivative, dy/dx = 2ax.
The graph in Figure 6.4 exhibits some important general
characteristics of derivatives:
●
●
x→0
Where the function has a horizontal tangent line, the
derivative is equal to zero.
The derivative is positive in regions where the function
increases as x increases.
,
Exercise 6.3. The exponential function can be represented
by the following power series
ebx = 1 + bx +
= a x12 + 2x1 x + ( x)2 − ax12
d(ax 2 )
= lim
x→0
dx
x→0
1 2 2
1
1
b x + b3 x 3 + · · · + bn x n · · · ,
2!
3!
n!
where the ellipsis ( · · · ) indicates that additional terms
follow. The notation n! stands for n factorial, such that
n! = n(n − 1)(n − 2) · · · (3)(2)(1) for any positive integral
value of n and 0! = 1. Derive the expression for the
derivative of ebx from this series.
Table 6.1 gives the derivatives of several functions,
where a, b, and c represent constants. Other derivatives
are found in Appendix D.
Exercise 6.4. Draw rough graphs of several functions from
Table 6.1. Below each graph, on the same sheet of paper,
make a rough graph of the derivative of the same function.
6.2 DIFFERENTIALS
In Eq. (6.5), we wrote as an approximation
y≈m x
(6.9)
62
dy
═ 2ax
dx
y ═ ax2
Mathematics for Physical Chemistry
x
0
x
FIGURE 6.4 A graph of the function y = ax 2 and its derivative, dy/d x = 2ax.
✬
✩
so that if n and T are kept fixed, we can differentiate with
TABLE 6.1 Some Elementary Functions and Their
respect to V:
Derivatives
Function, y = y(x)
Derivative, dy/dx = y (x)
ax n
anx n−1
ae bx
abe bx
a
0
a sin (bx)
ab cos (x)
a cos (bx)
−ab sin (bx)
a ln (x)
a/x
dP
−nRT
=
dV
V2
−(1.000 mol)(0.08206 l atm mol−1 K−1 )(273.15 K)
=
(22.414l)2
= −0.0446 atm l−1 .
We can approximate an increment in P :
dP
V = ( − 0.0446 atm l−1 )( − 1.000 l)
dV
≈ 0.0446 atm.
P ≈
a tan (x)
✫
✪
We determine the accuracy of this result by calculating the
actual change:
which is equivalent to
y≈
dy
dx
x.
(6.10)
This approximation will generally be more nearly correct
when x is made smaller.
Example 6.4. Using Eq. (6.10), estimate the change in
the pressure of 1.000 mol of an ideal gas at 0 ◦ C when its
volume is changed from 22.414l to 21.414l.
An ideal gas obeys the equation
P=
nRT
,
V
P = P(21.414 l) − P(22.414 l)
= 1.0468 atm − 1.0000 atm
= 0.0468 atm.
Our estimate was wrong by about 5%. If the change in
volume had been 0.10 l, the error would have been about
0.5% . In this case it would have been easier to make the
exact calculation.
Exercise 6.5. Assume that y = 3.00x 2 − 4.00x + 10.00.
If x = 4.000 and x = 0.500, find the value of y using
Eq. (6.10). Find the correct value of y.
CHAPTER | 6 Differential Calculus
63
Since Eq. (6.10) becomes more nearly exact as x is
made smaller, we make it into an exact equation by making
x become inﬁnitesimal. That is, we make it smaller in
magnitude than any nonzero quantity one might specify. In
this limit, x becomes the differential of the independent
variable, dx:
dy = m dx =
dy
dx
dx .
6.3.3 The Derivative of a Product of Two
Functions
If y and z are both functions of x,
dz
dy
d(yz)
=y
+z
.
dx
dx
dx
(6.15)
(6.11)
The infinitesimal quantity dy is the differential of the
dependent variable. It represents the change in y that results
from the infinitesimal increment dx in x. It is proportional
to dx and to dy/dx. Since x is an independent variable, dx
is arbitrary, or subject to our choice. Since y is a dependent
variable, its differential dy is determined by dx, as specified
by Eq. (6.11) and is not under our control once we have
chosen a value for dx. Although Eq. (6.11) has the appearance of an equation in which the dx in the denominator is
canceled by the dx in the denominator of a fraction, this is
not a cancelation since dy/dx is the limit that a fraction
approaches, which is not the same thing. In numerical
calculations, differentials are not directly useful. Their use
lies in the construction of formulas, especially through the
process of integration, which we discuss in the next chapter.
6.3.4 The Derivative of the Sum of Two
Functions
If y and z are both functions of x,
dy
dz
d(y + z)
=
+
.
dx
dx
dx
(6.16)
6.3.5 The Derivative of the Difference of
Two Functions
If y and z are both functions of x,
dy
dz
d(y − z)
=
−
.
dx
dx
dx
(6.17)
6.3.6 The Derivative of the Quotient of Two
Functions
6.3 SOME USEFUL DERIVATIVE
IDENTITIES
We present some useful identities involving derivatives,
which, together with the formulas for the derivatives of
simple functions presented in Table 6.1, will enable you to
obtain the derivative of almost any function that you will
encounter in physical chemistry.
If y and z are both functions of x, and z is not zero.
d(y/z)
=
dx
z
dy
dx
−y
dz
dx
z2
.
(6.18)
An equivalent result can be obtained by considering y/z to
be a product of y and 1/z and using Eq. (6.15),
6.3.1 The Derivative of a Constant
If c is a constant,
dc
=0 .
dx
From this follows the simple but important fact:
dy
d(y + c)
=
.
dx
dx
(6.12)
(6.13)
d
dx
y
1
z
=
1 dz
1 dy
−y 2
.
z dx
z dx
(6.19)
Many people think that Eq. (6.19) is more convenient to use
than Eq. (6.18).
If we add any constant to a function, we do not change its
derivative.
6.3.7 The Derivative of a Function of a
Function (the Chain Rule)
6.3.2 The Derivative of a Function Times a
Constant
If u is a differentiable function of x, and f is a differentiable
function of u,
d f du
df
=
.
(6.20)
dx
du dx
If y is a function of x and c is a constant,
dy
d(cy)
=c
.
dx
dx
(6.14)
The function f is sometimes referred to as a composite
function. It is indirectly a function of x because x is a
function of u. Specifying a value of x specifies a value of u,
64
which specifies a value of f. This can be communicated by
the notation
f (x) = f [u(x)].
(6.21)
Here we have used the same letter for the function f whether
it is expressed as a function of u or of x.
We now illustrate how these facts about derivatives can
be used to obtain formulas for the derivatives of various
functions.
Example 6.5. Find the derivative of tan (ax) by using the
formulas for the derivatives of the sine and cosine.
d
d sin (ax)
tan (ax) =
dx
dx cos (ax)
cos (ax)a cos (ax) + sin (ax)a sin ax
=
cos2 (ax)
cos2 (ax) + sin2 (ax)
a
=a
=
cos2 (ax)
cos2 (ax)
= a sec2 (ax).
We have used several trigonometric identities.
Example 6.6. Find dP/dT if P(T ) = ke−Q/T .
Let u = −Q/T . From the chain rule,
dP
dP du
Q
=
= keu 2
dT
du dT
T
Q
= ke−Q/T
.
T2
Exercise 6.6. Find the following derivatives. All letters
stand for constants except for the dependent and
independent variables indicated:
f
Mathematics for Physical Chemistry
Actual root
of f(x) ═ 0
x2
The process is as follows:
Step 1. Guess at a value, x0 , which is not too far from the
actual root. A rough graph of the function f (x)
can help you to choose a good value for x0 .
Step 2. Find the value of f (x) and the value of d f/dx at
x = x0 .
Step 3. Using the value of f (x) and d f/dx, find the value
of x at which the tangent line to the curve at x = x0
crosses the axis. This value of x, which we call x1 ,
is our next approximation to the root. It is given
by
f (x0 )
x1 = x0 − (1)
,
(6.23)
f (x0 )
where we use the notation
6.4 NEWTON’S METHOD
This method, which is also called the Newton-Raphson
method, is an iterative procedure for obtaining a numerical
solution to an algebraic equation. An iterative procedure
is one that is repeated until the desired degree of accuracy
is attained. The procedure is illustrated in Figure 6.5. We
assume that we have an equation written in the form
where f (x) is a differentiable function.
df
dx
.
(6.24)
x=x0
or the derivative evaluated at x = x0 .
Step 4. Repeat the process until you are satisfied with the
accuracy obtained. The nth approximation is given
by
xn = xn−1 −
f (x) = 0,
x0 x
FIGURE 6.5 Graph to illustrate Newton’s method.
f (1) (x0 ) =
dy
, where y = (ax 2 + bx + c)−3/2 .
(a)
dx
d ln (P)
(b)
, where P = ke−Q/T .
dT
dy
, where y = a cos (bx 3 ).
(c)
dx
x1
(6.22)
f (xn−1 )
f (1) (xn−1 )
.
(6.25)
Example 6.7. Using Newton’s method, find the positive
root of the equation
e−x − x 2 = 0.
A graph indicates that the root is near x = 0.7. Choose
x0 = 0.7000
f (0.7000) = 0.0065853,
df
= f (1) = −e−x − 2x,
dx