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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

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

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