Exponential, Logarithmic, Hyperbolic Functions, and Their Derivatives

On True Compound Interest

Let there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some fixed rate; for the bigger the capital, the bigger the amount of interest on it in a given time.

Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call “simple interest,” or by what they call “compound interest.” For in the former case the capital remains fixed, while in the latter the interest is added to the capital, which therefore increases by successive additions.

(1) At simple interest. Consider a concrete case. Let the capital at start be \$100, and let the rate of interest be 10 percent per annum. Then the increment to the owner of the capital will be \$10 every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for 10 years, by the end of that time he will have received 10 increments of \$10 each, or \$100, making, with the original \$100, a total of \$200 in all. His property will have doubled itself in 10 years. If the rate of interest had been 5 percent, he would have had to hoard for 20 years to double his property. If it had been only 2 percent, he would have had to hoard for 50 years. It is easy to see that if the value of the yearly interest is  of the capital, he must go on hoarding for n years in order to double his property.

Or, if y be the original capital, and the yearly interest is , then, at the end of n years, his property will be

(2)  At compound interest. As before, let the owner begin with a capital of \$100, earning interest at the rate of 10 percent per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to \$110; and in the second year (still at 10%) this will earn \$11 interest. He will start the third year with \$121, and the interest on that will be \$12.1; so that he starts the fourth year with \$133.1, and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to \$259.374. In fact, we see that at the end of each year, each dollar will have earned  of a dollar, and therefore, if this is always added on, each year multiplies the capital by ; and if continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by 2.59374. Let us put this into symbols. Put y0 for the original capital;  for the fraction added on at each of the n operations; and yn for the value of the capital at the end of the n operation. Then

But this mode of reckoning compound interest once a year, is really not quite fair; for even during the first year the \$100 ought to have been growing. At the end of half a year it ought to have been at least \$105, and it certainly would have been fairer had the interest for the second half of the year been calculated on \$105. This would be equivalent to calling it 5% per half-year; with 20 operations, therefore, at each of which the capital is multiplied by . If reckoned this way, by the end of ten years the capital would have grown to \$265.33.; for

But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into 10 parts, and reckon a one percent interest for each tenth of the year. We now have 100 operations lasting over the ten years; or which works out to \$270.481.

Even this is not final. Let the ten years be divided into 1000 periods, each of  of a year; the interest being  percent for each such period; then which works out to \$271.692.

Go even more minutely, and divide the ten years into 10,000 parts, each  of a year, with interest at  of 1 percent. Then which amounts to \$271.815.

Finally, it will be seen that what we are trying to find is in reality the ultimate value of the expression , which, as we see, is greater than 2; and which, as we take n larger and larger, grows closer and closer to a particular limiting value. However big you make n, the value of this expression grows nearer and nearer to the figure a number never to be forgotten.

Let us take geometrical illustrations of these things. In the following figure, OP stands for the original value. OT is the whole time during which the value is growing. It is divided into 10 periods, in each of which there is an equal step up. Here  is a constant; and if each step up is  of the original OP, then, by 10 such steps, the height is doubled. If we had taken 20 steps, each of half the height shown, at the end the height would still be just doubled. Or n such steps, each of  of the original height OP, would suffice to double the height. This is the case of simple interest. Here is 1 growing till it becomes 2.

Fig. 14.1
Figure 1 Fig. 14.1

In the next figure, we have the corresponding illustration of the geometrical progression. Each of the successive ordinates is to be , that is, times as high as its predecessor. The steps up are not equal, because each step up is now  of the ordinate at that part of the curve. If we had literally 10 steps, with for the multiplying factor, the final total would be or 2.594 times the original 1. But if only we take n sufficiently large (and the corresponding sufficiently small), then the final value to which unity will grow will be 2.71828.

Fig. 14.2
Figure 2 Fig. 14.2

The Number e

To this mysterious number 2.7182818…, the mathematicians have assigned the letter e. This number is often called Euler’s number after the Swiss mathematician Leonhard Euler. All fifth graders know that the Greek letter π (called pi) stands for 3.141592…; but how many of them know that e means 2.71828…? Yet it is an even more important number than π!

What, then, is e?

Suppose we were to let 1 grow at simple interest till it became 2; then, if at the same nominal rate of interest, and for the same time, we were to let 1 grow at true compound interest, instead of simple, it would grow to the value of the number e.

This process of growing proportionately, at every instant, to the magnitude at that instant, some people call an exponential rate of growing. Unit exponential rate of growth is that rate which in unit time will cause 1 to grow to 2.718281. It might also be called the organic rate of growing because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.

If we take 100 percent as the unit of rate, and any fixed period as the unit of time, then the result of letting 1 grow arithmetically at unit rate, for unit time, will be 2, while the result of letting 1 grow exponentially at unit rate, for the same time, will be 2.71828… .

A little more about the number e

We have seen that we require to know what value is reached by the expression , when n becomes indefinitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of a calculator) got by assuming n = 2; n = 5; n = 10; and so on, up to n = 10,000.

It is, however, worth while to find another way of calculating this immensely important figure.

Accordingly, we will avail ourselves of the binomial theorem, and expand the expression in that well-known way.

The binomial theorem gives the rule that Putting a = 1 and , we get

Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then , , and , etc., will all be sensibly equal to n; and then the series becomes

By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms:

 1.000000
dividing by 11.000000
dividing by 20.500000
dividing by 30.166667
dividing by 40.041667
dividing by 50.008333
dividing by 60.001389
dividing by 70.000198
dividing by 80.000025
dividing by 90.000002
Total2.718281

e is incommensurable with 1, and resembles π in being an interminable non-recurrent decimal.

The Exponential Series

We shall have need of yet another series.

Let us, again making use of the binomial theorem, expand the expression , which is the same as ex when we make n indefinitely great.

But, when n is made indefinitely great, this simplifies down to the following:

This series is called the exponential series.

Derivative of the Exponential Function

The great reason why e is regarded of importance is that ex possesses a property, not possessed by any other function of x, that when you differentiate it its value remains unchanged; or, in other words, its derivative is the same as itself. This can be instantly seen by differentiating it with respect to x, thus: or which is exactly the same as the original series.

Now we might have gone to work the other way, and said: Go to; let us find a function of x, such that its derivative is the same as itself. Or, is there any expression, involving only powers of x, which is unchanged by differentiation? Accordingly; let us assume as a general expression that (in which the coefficients A, B, C, etc. will have to be determined), and differentiate it.

Now, if this new expression is really to be the same as that from which it was derived, it is clear that A must = B; that ; that ; that , etc.

The law of change is therefore that

If, now, we take A = 1 for the sake of further simplicity, we have

Differentiating it any number of times will give always the same series over again.

If, now, we take the particular case of A = 1, and evaluate the series, we shall get simply and therefore thus finally demonstrating that

Of course it follows that ey remains unchanged if differentiated with respect to y. Also eax, which is equal to (ea)x, will, when differentiated with respect to x, be aeax, because a is a constant.

Natural or Naperian Logarithms

Another reason why e is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If y is the value of ex, then x is the logarithm, to the base e, of y. Or, if then

The logarithm with base e is called the natural logarithm. The natural logarithm is so important that it has its own shorthand:

The two curves plotted in Figs. 14.3 and 14.4 represent these equations.

The points calculated are:

x−2−1−0.500.511.52
y = ex0.140.370.6111.652.714.507.39
For Fig. 14.3

y0.10.512348
x = ln y−2.30−0.6900.691.101.392.08
For Fig. 14.4

Fig. 14.3
Figure 3 Fig. 14.3
Fig. 14.4
Figure 4 Fig. 14.4

 

It will be seen that, though the calculations yield different points for plotting, yet the result is identical. The two equations really mean the same thing.

As many persons who use ordinary logarithms, which are calculated to base 10 instead of base e, are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or Also the rule of powers holds good; But as 10 is no longer the basis, one cannot multiply by 100 or 1000 by merely adding 2 or 3 to the index. One can change the natural logarithm to the ordinary logarithm1 simply by multiplying it by ; or and conversely,

Using A Calculator to Find ex and ln x

Modern scientific and graphing calculators are equipped with buttons for exponentiation with the base e and buttons for calculating natural or common logarithms. The exponential function ex is sometimes denoted by exp(x). Therefore, when using a calculator, we may need to locate the or button. Many calculators provide two distinct buttons for calculating the natural logarithm (ln x) and the common logarithm (log10 x). If your calculator has an button for the natural logarithm, the button is likely designed to return log10 x.

Derivatives of Logarithmic and Exponential Functions

Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.

Take the equation: First transform this into whence, since the derivative of ey with regard to y is the original function unchanged (see here), and, reverting from the inverse to the original function,

Now this is a very curious result. It may be written

Note that x−1 is a result that we could never have got by the Power Rule for differentiating powers. That rule is to multiply by the power, and reduce the power by 1. Thus, differentiating x3 gave us 3x2; and differentiating x2 gave 2x. But differentiating x0 does not give us x−1 or , because x0 is itself = 1, and is a constant. We shall have to come back to this curious fact that differentiating ln x gives us when we reach the chapter on integrating.

Now, try to differentiate that is, we have , since the differential of ey remains ey.

This gives hence, reverting to the original function (see the Derivative of an Inverse Function), we get2

Next try

First change to natural logarithms by multiplying by the modulus . This gives us whence

In general, because and is a constant, we have

The next thing is not quite so simple. Try this:

Taking the logarithm of both sides, we get or

Since is a constant, we get hence, reverting to the original function.

We see that, since

We shall find that whenever we have an expression such as ln y = a function of x, we always have the derivative of the function of x, so that we could have written at once, from ln y = x ln a,

In summary

Let us now attempt further examples.

Examples

Example 1.

Differentiate y with respect to x if y = eax.

Solution. Let −ax = z; then y = ez.

Or thus:

Example 2.

Differentiate y with respect to x if .

Solution. Let ; then y = ez.

Or thus:

Example 3.

Given , find .

Solution. hence

Check by writing .

.. For if and , ,

Check by writing .

Example 4.

If , find .

Solution. Let ; then y = ln z.

Example 5.

If , find .

Solution. Let ; then y = ln z.

Example 6.

If , find .

Solution. Taking the logarithm of both sides, we get Differentiating both sides, we get

Example 7.

If , find .

Solution. Taking the logarithm of both sides, we get For if , let and . Similarly, if , and

Example 8.

If , find .

Solution. Differentiating and

Example 9.

If , find .

Solution.

Example 10.

If , find .

Solution. Let z = ln x; .

Example 11.

If , find .

Solution. Differentiating and

Try now the following exercises.

Exercises I

Exercise 1.

Differentiate .

 

Answer

.

 

 

 

 

Solution

(1)

 

 

Exercise 2.

Find the derivative of the expression with respect to t.

 

Answer

.

 

 

 

 

Solution

 

 

 

Exercise 3.

If y = nt, find .

 

Answer

ln n.

 

 

 

 

Solution

 

Exercise 4.

Show that if ,.

 

Solution

 

 

 

Exercise 5.

If , find .

 

Answer

.

 

 

 

 

Solution

 

 

 

Differentiate

Exercise 6.

.

 

Answer

.

 

 

 

 

Solution

 

 

Exercise 7.

.

 

Answer

.

 

 

 

 

Solution

Let . Then , and Using the Chain Rule:

 

 

 

Exercise 8.

.

 

Answer

.

 

 

 

 

Solution

Using the Product Rule, we get

 

 

 

Exercise 9.

.

 

Answer

.

 

 

 

 

Solution

Let . Now using the Chain Rule

 

 

 

Exercise 10.

.

 

Answer

.

 

 

 

 

Solution

Using the Product Rule:

 

 

 

Exercise 11.

.

 

Answer

.

 

 

 

 

Solution

Using the Quotient Rule:

 

Exercise 12.

.

 

Answer

.

 

 

 

 

Solution

 

 

 

Exercise 13.

It was shown by Lord Kelvin that the speed of signaling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called y, then the number of signals s that can be sent per minute can be expressed by the formula where a is a constant depending on the length and the quality of the materials. Show that if these are given, s will be a maximum if .

 

Solution

To differentiate , let u = y−1. Then Now using the Product Rule, we get

 

When , Therefore, it follows from the Second Derivative Test that s is a minimum if .

 

 

Exercise 14.

Find the maximum or minimum of

 

Answer

Min.: y = 0.7 for x = 0.694.

 

 

 

 

Solution

 

To show that this specific value of x makes y a minimum, we apply the Second Derivative Test. First we differentiate to find the second derivative Since when is a minimum occurring when .

 

 

Exercise 15.

Differentiate .

 

Answer

.

 

 

 

 

Solution

 

Recall that ln(AB) = ln A + ln B and . Using these properties, we can rewrite y as Since ln e = 1, Now we can differentiate term by term

 

Exercise 16.

Differentiate .

 

Answer

.

 

 

 

 

Solution

Using the Chain Rule

 

 

 

The Logarithmic Curve

Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation y = bpx.

We can see, by putting x = 0, that b is the initial height of y.

Then when

Also, we see that p is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In the following figure, we have taken p as ; each ordinate being  as high as the preceding one.

Fig. 14.5
Figure 5 Fig. 14.5

If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant difference; so that, if we should plot out a new curve, the following figure, with values of ln y as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that whence Now, since ln p is a mere number, and may be written as ln p = a, it follows that and the equation takes the new form

Fig. 14.6
Figure 6 Fig. 14.6

The Die-away Curve

If we were to take p as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in the next figure, where each successive ordinate is  of the height of the preceding one.

The equation is still

Figure 7

but since p is less than one, ln p will be a negative quantity, and may be written −a; so that p = ea,3 and now our equation for the curve takes the form

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton’s celebrated “law of cooling”) by the equation where θ0 is the original excess of temperature of a hot body over that of its surroundings, θt the excess of temperature at the end of time t, and a is a constant—namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula, is used to express the charge of an electrified body, originally having a charge Q0, which is leaking away with a constant of decrement a; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.

In fact eat serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols,  is proportional at every moment to the value that y has at that moment. For we have only to inspect the curve, the above figure, to see that, at every part of it, the slope  is proportional to the height y; the curve becoming flatter as y grows smaller. In symbols, thus or and, differentiating, hence or, in words, the slope of the curve is downward, and proportional to y and to the constant a.

We should have got the same result if we had taken the equation in the form for then But giving us as before.

The Time-constant. In the expression for the “die-away factor” eat, the quantity a is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol T. Then the die-away factor will be written ; and it will be seen, by making t = T that the meaning of T is that this is the length of time which it takes for the original quantity (called θ0 or Q0 in the preceding instances) to die away 1/eth part—that is to 0.3678—of its original value.

As an example, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when t = 0) it is 72°C hotter than the surrounding objects, and if the time-constant of its cooling is 20 minutes (that is, if it takes 20 minutes for its excess of temperature to fall to 1/e part of 72 degrees), then we can calculate to what it will have fallen in any given time t. For instance, let t be 60 minutes. Then , and we shall have to find the value of e−3, and then multiply the original 72 degrees by this. Since e−3 is 0.0498, at the end of 60 minutes the excess of temperature will have fallen to 72°C × 0.0498 = 3.586°C.

Further Examples

Example 12.

The strength of an electric current in a conductor at a time t seconds after the application of the electromotive force producing it is given by the expression .

The time constant is .

If E = 10, R = 1, L = 0.01; then when t is very large the term  becomes 1, and ; also

Its value at any time may be written: the time-constant being 0.01. This means that it takes 0.01 sec. for the variable term to fall by 1/e = 0.3678 of its initial value .

To find the value of the current when t = 0.001s, say, , e−0.1 = 0.9048.

It follows that, after 0.001 s, the variable term is , and the actual current is .

Similarly, at the end of 0.1 s, the variable term is , the current being 9.9995.

Example 13.

The intensity I of a beam of light which has passed through a thickness l cm of some transparent medium is , where I0 is the initial intensity of the beam and K is a “constant of absorption.”

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through 10 cm of a certain transparent medium, this means that or e−10K = 0.82. Now we take the natural logarithm of both sides or "> hence K ≈ 0.02.

To find the thickness that will reduce the intensity to half its value, one must find the value of l which satisfies the equality , or 0.5 = e−0.02l. It is found by putting this equation in its logarithmic form, namely, which gives

Example 14.

The quantity Q of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity Q0 of the substance by the relation , where λ is a constant and t the time in seconds elapsed since the transformation began.

For “Radium A,” if time is expressed in seconds, experiment shows that . Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

Solution. We have 0.5 = e−0.00385t."> and

Exercises II

Exercise 17.

Draw the curve ; where b = 12, T = 8, and t is given various values from 0 to 20.

 

Answer

Use your calculator to evaluate for different values of t within the range 0 ≤ t ≤ 20. Then, connect the resulting points together.

 

 

 

 

Solution

To sketch the curve by hand, we can evaluate at various values of t (0 ≤ t ≤ 20). For example,

 

ty = 12et/8
012.000
110.590
56.423
103.438
151.840
200.985

Then plot each of these points on a set of axes, and connect these points with a smooth curve.

There are also multiple tools for plotting the curve for 0 ≤ t ≤ 20. For example, you may visit WolframAlpha.com and in the search bar simply type:
plot 12 e^ (-t/8) from t=0 to t=20.

 

 

Exercise 18.

If a hot body cools so that in 24 minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to 1 per cent. of the original excess.

 

Answer

T = 34.625; minutes.

 

 

 

 

Solution

The equation of cooling is

 

After 24 minutes

time-constant is approximately .

Now we want to find t such that

 

Exercise 19.

Plot the curve

 

Solution

When x = 0, y = 0

 

When x is a large positive number, the term e−2t becomes negligible (≈ 0). Consequently, we have

On the other hand, when x is a large negative number, the terms e−2t becomes very large positive. As a result y becomes numerically large but negative.

The curve is shown below.

 

 

Exercise 20.

The following equations give very similar curves: Draw all three curves, taking a = 10 units; b = 3 units.

 

Solution

Let’s assume b > 0.

 

When x = 0, then

is zero.

is zero because e0 = 1.

is zero because arctan 0 = 0.

When x is very large positive

is ≈ a because we can ignore b compared to x (for example, when b = 3 and , then ).

is ≈ a because is negligible for large values of x.

is ≈ a because when x is large, .

For negative values of x, these curves are not comparable because is very large when x is close to −b.

When x is large negative,

is large positive. Therefore is large negative when x is large negative.

is ≈ a because again b is negligible compared to x and thus .

is ≈ −a because when x is large negative, .

These curves are shown below.

 

 

Exercise 21.

Find the derivative of y with respect to x, if

 

Answer

a) ; (b) ; (c) .

 

 

 

 

Solution

 

(a)

Taking the natural logarithm of both sides and recalling :

Now differentiating and using the Chain Rule for the left-hand side and the Product Rule for the right-hand side

(b)

Method 1)

Taking the natural logarithm of both sides and using the property , we have

Now let’s differentiate both sides with respect to x:

Simplifying, we find

Method 2) Recall that . Therefore

Let u = x2 and applying the Chain Rule with where u = x2, we get

(c)

Method 1) Let Then using the result of part (a) and the Chain Rule, we obtain

Method 2) Take ln of both sides

Now differentiate both sides with respect to x to get Therefore

 

Exercise 22.

For “Thorium A,” the value of λ is 5; find the “mean life,” that is, the time taken by the transformation of a quantity Q of “Thorium A” equal to half the initial quantity Q0 in the expression t being in seconds.

 

Answer

0.14 seconds.

 

 

 

 

Solution

We need to find t such that

 

Taking the natural logarithm of both sides

 

Exercise 23.

A condenser of capacity , charged to a potential V0 = 20, is discharging through a resistance of 10,000 ohms. Find the potential V after (a) 0.1 seconds; (b) 0.01 seconds; assuming that the fall of potential follows the rule .

 

Answer

(a) 1.642;(b) 15.58.

 

 

 

 

Solution

Here or

 

(a) When t = 0.1

(b) When t = 0.01

 

Exercise 24.

The charge Q of an electrified insulated metal sphere is reduced from 20 to 16 units in 10 minutes. Find the coefficient μ of leakage, if ; Q0 being the initial charge and t being in seconds. Hence find the time taken by half the charge to leak away.

 

Answer

μ = 0.00037, .

 

 

 

 

Solution

Since we can find μ

 

Now we need to find t such that

 

Exercise 25.

The damping on a telephone line can be ascertained from the relation , where  is the strength, after t seconds, of a telephonic current of initial strength ; l is the length of the line in kilometres, and β is a constant. For the Franco-English submarine cable laid in 1910, β = 0.0114. Find the damping at the end of the cable (40 kilometres), and the length along which  is still 8% of the original current (limiting value of very good audition).

 

Answer

is 63.4% of , 220 kilometres.

 

 

 

 

Solution

 

(a)

Therefore, the strength becomes 63.38% of the initial strength.

(b)

 

 

Exercise 26.

The pressure p of the atmosphere at an altitude h kilometres is given by ;  being the pressure at sea-level (760 millimetres).

The pressures at 10, 20 and 50 kilometres being 199.2, 42.2, 0.32 respectively, find k in each case. Using the mean value of k, find the percentage error in each case.

 

Answer

0.1339, 0.1445, 0.1555, mean 0.1446; −10.2%, practically nil, +71.9%.

 

 

 

 

Solution

 

Calculating k when h = 10 km,

Calculating k when h = 20 km

Calculating k when h = 50 km

Average k (the mean value of k):

Calculating p using kav when h = 10 km:

Calculating p using kav when h = 20 km:

Calculating p using kav when h = 50 km:

 

 

Exercise 27.

Find the minimum or maximum of y = xx.

 

Answer

Min. for .

 

 

 

 

Solution

In Exercise S, we showed that

 

Setting

Since xx ≠ 0, at the minimum or maximum, we must have Then

To determine if this value of x makes y a maximum or a minimum, we apply the Second Derivative Test:

When , then Therefore, x = 1/e ≈ 0.369 corresponds to a minimum value of y.

The graph of y = xx is shown below.

 

 

Exercise 28.

Find the minimum or maximum of y = x1/x.

 

Answer

Max. for x = e.

 

 

 

 

Solution

 

To differentiate, we first take the natural logarithm of both sides then use the Chain Rule:

Since x1/x ≠ 0,

or

To determine whether x = e makes y a minimum or a maximum, we compare the value of y at this point with the values of y at some nearby points:

When x = e

When x = 2

When x = 3

Therefore x = e makes y = x1/x a maximum.

We can also apply the Second Derivative Test:

Using the Quotient Rule

Therefore,

When x = e, both and 1 − ln x are zero. Therefore, when x = e

It follows from the Second Derivative Test that is a maximum occurring when x = e.

The graph of y = x1/x is shown below

 

 

Exercise 29.

Find the minimum or maximum of .

 

Answer

Min. for x = ln a.

 

 

 

 

Solution

 

Using the Product Rule

To find , let and use the Chain Rule

Therefore

Since ,

or

Using the Second Derivative Test

When x = ln a

Therefore is a minimum occurring when x = ln a.

 

 

Hyperbolic Functions

Hyperbolic functions are specific combinations of exponential functions that arise frequently in various applications, leading mathematicians to give them distinctive names and thoroughly investigate their properties. Although hyperbolic functions are combined exponential functions, in some ways they resemble the trigonometric functions. As a result, the hyperbolic functions are given individual names such as hyperbolic sine, hyperbolic cosine, hyperbolic tangent, and others. Their definitions are as follows:

One can easily sketch the graphs of y = sinh x and y = cosh x by plotting the curves y = ex and y = ex, adding and subtracting the ordinates, and taking half of each. The following figure displays their graphs.

Fig. 14.8
Figure 8 Fig. 14.8

A catenary, which is the graph of the hyperbolic cosine, is a curve that describes the shape of a homogeneous chain or cord hanging under its own weight.

When x is positive large, ex is very large, while ex is extremely small; hence ex ± exex leading to: Similarly, when x is negative large, ex is negligible, making ex ± ex ≈ ±ex, and hence: The graph of y = tanh x is shown below.

Fig. 14.9
Figure 9 Fig. 14.9

Identities Between Hyperbolic Functions

The properties of the hyperbolic functions closely resemble the corresponding properties of the trigonometric functions.

It follows directly from the definitions that Now using A2B2 = (AB)(A + B), we get As such we have proved one of the fundamental identities: Now if we divide each term of the above equation by cosh2 x, we obtain Similarly, dividing each term of cosh2 x − sinh2 x = 1 by sinh2 x gives us

Example 15.

Prove sinh(x + y) = sinh x cosh y + cosh x sinh y.

Solution. Let’s simplify the right-hand side

Therefore, In a similar way, we can prove It follows from the above identity that Hence, Similarly, Therefore,

Another important set of identities, which follow at once from the definitions, is:4

Derivatives of the Hyperbolic Functions

As the hyperbolic functions are combinations of exponential functions, and we have just learned how to differentiate exponential functions, we can apply the differentiation rules learned in Chapters 5 and 6 to determine the derivatives of the hyperbolic functions.

For example, since sinh x = ½(exex), we have Similarly, we can show that .

Example 16.

Show .

Solution. Since tanh x = sinh x / cosh x, we can use the Quotient Rule:

In summary,

Inverse Hyperbolic Functions

The inverse hyperbolic sine function is denoted by arcsinh x or sinh−1 x, and is defined by We may also define the inverse hyperbolic cosine function, denoted by arccosh x or cosh−1 x, as However, note that if x = cosh y, then cosh(−y) is also equal to x. This means that the above definition assigns two values of y to each value of x. To make arccosh x a single-valued function, we agree that it produces only non-negative values of y. Furthermore, because cosh y ≥ 1, the inverse hyperbolic cosine function does not take any values of x less than 1.

Similarly, the inverse hyperbolic tangent function is defined by Since −1 < tanh y < 1, the inverse hyperbolic tangent function does not take any values of x greater than or equal to 1 or less than or equal to −1.

In summary,

We can find explicit formulas for the inverse hyperbolic functions.

Example 17.

Show that .

Solution. If y = arcsinh x, then Multiplying by y by 2ey, we obtain or This is a quadratic equation in terms of ex. Therefore, by the quadratic formula, Since ey > 0 and is greater than x, only the positive sign is acceptable. Hence, Taking the natural logarithm of each side produces Therefore,

In a similar fashion, we can derive explicit formulas for the other inverse hyperbolic functions.

Derivatives of the Inverse Hyperbolic Functions

With what we have learned so far, we can find the derivatives of the inverse hyperbolic functions.

Example 18.

If (also denoted by sinh−1 x), find .

Solution. If , then x = sinh y, and By the derivative of an inverse function (see here)

We can also find the derivative of arcsinh x by differentiating . Using the chain rule and the fact that , we get This confirms the result we obtained earlier.

Example 19.

If y = arccosh x (also denoted by cosh−1 x), find .

Solution. If , then x = cosh y, and Hence,

Example 20.

If y = arctanh x (also denoted by tanh−1 x), find . Show that

Solution. If y = arctanh x, then x = tanh y, and Thus

In summary,


The ordinary logarithm or the 10-th based logarithm is often called the common logarithm.↩︎

Alternatively, we can use the Chain Rule. Let u = x + a. Then y = ln u and ↩︎

ln p = −ap = ea.↩︎

We say that the hyperbolic sine and hyperbolic tangent functions are odd, and the hyperbolic cosine is an even function (similar to the corresponding trigonometric functions). A function is called odd if and is called even if ↩︎