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 \(\dfrac{1}{n}\) 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 \(\dfrac{y}{n}\), then, at the end of n years, his property will be \[y + n\dfrac{y}{n} = 2y.\]

(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 \(\frac{1}{10}\) of a dollar, and therefore, if this is always added on, each year multiplies the capital by \(\frac{11}{10}\); 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; \(\dfrac{1}{n}\) for the fraction added on at each of the n operations; and y_n for the value of the capital at the end of the n operation. Then \[y_n = y_0\left(1 + \frac{1}{n}\right)^n.\]

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 \(\frac{21}{20}\). If reckoned this way, by the end of ten years the capital would have grown to \$265.33.; for \[\left(1 + \frac{1}{20}\right)^{20} ={2.6533}.\]

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 \[y_n = \$100 \left( 1 + \frac{1}{100} \right)^{100};\] which works out to \$270.481.

Even this is not final. Let the ten years be divided into 1000 periods, each of \(\frac{1}{100}\) of a year; the interest being \(\frac{1}{10}\) percent for each such period; then \[y_n = \$ 100 \left( 1 + \frac{1}{1000} \right)^{1000};\] which works out to \$271.692.

Go even more minutely, and divide the ten years into 10,000 parts, each \(\frac{1}{1000}\) of a year, with interest at \(\frac{1}{100}\) of 1 percent. Then \[y_n = \$100 \left( 1 + \frac{1}{10,000} \right)^{10,000};\] 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 \(\left(1 + \dfrac{1}{n}\right)^n\), 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 \[2.71828\ldots\] 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 \(\dfrac{dy}{dx}\) is a constant; and if each step up is \(\frac{1}{10}\) 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 \(\dfrac{1}{n}\) 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.

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

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 \(\left(1 + \dfrac{1}{n}\right)^n\), 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. \[\begin{align} &\left(1 + \frac{1}{2}\right)^2 &=& 2.25. \\ &\left(1 + \frac{1}{5}\right)^5 &=& 2.488. \\ &\left(1 + \frac{1}{10}\right)^{10} &=& 2.594. \\ &\left(1 + \frac{1}{20}\right)^{20} &=& 2.653. \\ &\left(1 + \frac{1}{100}\right)^{100} &=& {2.705}. \\ &\left(1 + \frac{1}{1000}\right)^{1000} &=& {2.7169}. \\ &\left(1 + \frac{1}{10,000}\right)^{10,000} &=& {2.7181}. \end{align}\]

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 \(\left(1 + \dfrac{1}{n}\right)^n\) in that well-known way.

The binomial theorem gives the rule that \[\begin{align} (a + b)^n &= a^n + n \dfrac{a^{n-1} b}{1!} + n(n - 1) \dfrac{a^{n-2} b^2}{2!} + n(n -1)(n - 2) \dfrac{a^{n-3} b^3}{3!} + \cdots. \end{align}\] Putting a = 1 and \(b = \dfrac{1}{n}\), we get \[\begin{align} \left(1 + \dfrac{1}{n}\right)^n &= 1 + 1 + \dfrac{1}{2!} \left(\dfrac{n - 1}{n}\right) + \dfrac{1}{3!} \dfrac{(n - 1)(n - 2)}{n^2} + \dfrac{1}{4!} \dfrac{(n - 1)(n - 2)(n - 3)}{n^3} + \cdots. \end{align}\]

Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then \(n - 1\), \(n - 2\), and \(n - 3\), etc., will all be sensibly equal to n; and then the series becomes \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \cdots.}\]

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:

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 \(\left(1 + \dfrac{1}{n}\right)^{nx}\), which is the same as e^x when we make n indefinitely great. \[\begin{align} e^x &= 1^{nx} + nx \frac{1^{nx-1} \left(\dfrac{1}{n}\right)}{1!} + nx(nx - 1) \frac{1^{nx - 2} \left(\dfrac{1}{n}\right)^2}{2!} + nx(nx - 1)(nx - 2) \frac{1^{nx-3} \left(\dfrac{1}{n}\right)^3}{3!} + \cdots\\ &= 1 + x + \frac{1}{2!} \cdot \frac{n^2x^2 - nx}{n^2} + \frac{1}{3!} \cdot \frac{n^3x^3 - 3n^2x^2 + 2nx}{n^3} + \cdots \\ &= 1 + x + \frac{x^2 -\dfrac{x}{n}}{2!} + \frac{x^3 - \dfrac{3x^2}{n} + \dfrac{2x}{n^2}}{3!} + \cdots. \end{align}\]

But, when n is made indefinitely great, this simplifies down to the following: \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots.}\]

This series is called the exponential series.

Derivative of the Exponential Function

The great reason why e is regarded of importance is that e^x 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: \[\begin{align} \frac{d(e^x)}{dx} &= 0 + 1 + \frac{2x}{1 \cdot 2} + \frac{3x^2}{1 \cdot 2 \cdot 3} + \frac{4x^3}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{5x^4}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \cdots. \end{align}\] or \[\begin{align} \frac{d(e^x)}{dx}= 1 + x + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \frac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \cdots, \end{align}\] which is exactly the same as the original series.

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\dfrac{d(e^x)}{dx}=e^x}\]

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 \[y = A + Bx + Cx^2 + Dx^3 + Ex^4 + \cdots,\] (in which the coefficients A, B, C, etc. will have to be determined), and differentiate it. \[\dfrac{dy}{dx} = B + 2Cx + 3Dx^2 + 4Ex^3 + \cdots.\]

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 \(C=\dfrac{B}{2}=\dfrac{A}{1\cdot 2}\); that \(D = \dfrac{C}{3} = \dfrac{A}{1 \cdot 2 \cdot 3}\); that \(E = \dfrac{D}{4} = \dfrac{A}{1 \cdot 2 \cdot 3 \cdot 4}\), etc.

The law of change is therefore that

\[y = A\left(1 + \dfrac{x}{1} + \dfrac{x^2}{1 \cdot 2} + \dfrac{x^3}{1 \cdot 2 \cdot 3} + \dfrac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \cdots.\right).\]

If, now, we take A = 1 for the sake of further simplicity, we have \[y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1 \cdot 2} + \dfrac{x^3}{1 \cdot 2 \cdot 3} + \dfrac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \cdots.\]

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 \[\begin{align} \text{when } x &= 1,\quad & y &= 2.718281\dots; & \text{that is, } y &= e; \\ \text{when } x &= 2,\quad & y &=(2.718281\dots)^2; & \text{that is, } y &= e^2; \\ \text{when } x &= 3,\quad & y &=(2.718281\dots)^3; & \text{that is, } y &= e^3; \end{align}\] and therefore \[\text{when } x=x,\quad y=(2.718281\dots.)^x;\quad\text{that is, } y=e^x,\] thus finally demonstrating that \[e^x = 1 + \dfrac{x}{1} + \dfrac{x^2}{1\cdot2} + \dfrac{x^3}{1\cdot 2\cdot 3} + \dfrac{x^4}{1\cdot 2\cdot 3\cdot 4} + \cdots.\]

Of course it follows that e^y remains unchanged if differentiated with respect to y. Also e^ax, which is equal to (e^a)^x, will, when differentiated with respect to x, be ae^ax, 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 e^x, then x is the logarithm, to the base e, of y. Or, if \[y = e^x,\] then \[x = \log_e y.\]

The logarithm with base e is called the natural logarithm. The natural logarithm is so important that it has its own shorthand: \[\log_e y \qquad\text{is often written as}\qquad \ln y.\]

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

The points calculated are:

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 \[\ln a + \ln b = \ln (ab).\] Also the rule of powers holds good; \[n \times \ln a = \ln a^n.\] 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 \(\frac{1}{\ln 10}\approx 0.4343\); or \[\begin{align} \log_{10} x = \frac{1}{\ln 10}\ln x\approx0.4343 \times \ln x, \end{align}\] and conversely, \[\begin{align} \ln x = \frac{1}{\log_{10} e}\times \log_{10} x=\ln 10\times \log_{10}x\approx2.3026 \times \log_{10} x. \end{align}\]

Using A Calculator to Find e^x 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 e^x is sometimes denoted by exp(x). Therefore, when using a calculator, we may need to locate the \(\boxed{e^x}\) or \(\boxed{\text{exp}(x)}\) 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 \(\boxed{\text{ln}}\) button for the natural logarithm, the \(\boxed{\text{log}}\) 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: \[y = \ln x.\] First transform this into \[e^y = x,\] whence, since the derivative of e^y with regard to y is the original function unchanged (see here), \[\frac{dx}{dy} = e^y,\] and, reverting from the inverse to the original function, \[\frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ } = \frac{1}{e^y} = \frac{1}{x}.\]

Now this is a very curious result. It may be written \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\dfrac{d(\ln x)}{dx} = x^{-1}.}\]

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 \(0 \times x^{-1}\), 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 \(\dfrac{1}{x}\) when we reach the chapter on integrating.

Now, try to differentiate \[\begin{align} y = \ln(x+a), \end{align}\] that is, \[\begin{align} e^y = x+a; \end{align}\] we have \(\dfrac{d(x+a)}{dy} = e^y\), since the differential of e^y remains e^y.

This gives \[\frac{dx}{dy} = e^y = x+a;\] hence, reverting to the original function (see the Derivative of an Inverse Function), we get² \[\frac{dy}{dx} = \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{x+a}.\]

Next try \[y = \log_{10} x.\]

First change to natural logarithms by multiplying by the modulus \(\log_{10}e=\dfrac{1}{\ln 10}\approx 0.4343\). This gives us \[\begin{align} y = \frac{1}{\ln 10} \ln x; \end{align}\] whence \[\begin{align} \frac{dy}{dx} = \frac{1}{x\ln 10}. \end{align}\]

In general, because \[\log_a x=\frac{\log_e x}{\log_e a}=\frac{\ln x}{\ln a}\] and \(\dfrac{1}{\ln a}\) is a constant, we have \[\frac{d\left(\log_a x\right)}{dx}=\frac{d\left(\dfrac{1}{\ln a}\cdot \ln x \right)}{dx}=\frac{1}{\ln a}\cdot\frac{d(\ln x)}{dx}=\frac{1}{\ln a\cdot x}\]

The next thing is not quite so simple. Try this: \[y = a^x.\]

Taking the logarithm of both sides, we get \[\begin{align} \ln y &= x \ln a, \end{align}\] or \[\begin{align} x &= \frac{\ln y}{\ln a}\\ &= \frac{1}{\ln a} \times \ln y. \end{align}\]

Since \(\dfrac{1}{\ln a}\) is a constant, we get \[\frac{dx}{dy} = \frac{1}{\ln a} \times \frac{1}{y} = \frac{1}{a^x \times \ln a};\] hence, reverting to the original function. \[\frac{dy}{dx} = \frac{1}{\;\dfrac{dx}{dy}\;} = a^x \times \ln a.\]

We see that, since \[\frac{dx}{dy} \times \frac{dy}{dx} =1\quad\text{and}\quad \frac{dx}{dy} = \frac{1}{y} \times \frac{1}{\ln a},\quad \frac{1}{y} \times \frac{dy}{dx} = \ln a.\]

We shall find that whenever we have an expression such as ln y = a function of x, we always have \(\dfrac{1}{y}\, \dfrac{dy}{dx} =\) the derivative of the function of x, so that we could have written at once, from ln y = x ln a, \[\frac{1}{y}\, \frac{dy}{dx} = \ln a\quad\text{and}\quad \frac{dy}{dx} = a^x \ln a.\]

In summary

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ y=\log_a x \qquad\Rightarrow \qquad \dfrac{dy}{dx}=\dfrac{1}{x\cdot \ln a} }\]

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ y=a^x \qquad\Rightarrow \qquad \dfrac{dy}{dx}=a^x\cdot\ln a }\]

Let us now attempt further examples.

Examples

Try now the following exercises.

Exercises I

Differentiate

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 = bp^x.

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

Then when \[x=1,\quad y=bp;\qquad x=2,\quad y=bp^2;\qquad x=3,\quad y=bp^3,\quad \text{etc.}\]

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 \(\frac{6}{5}\); each ordinate being \(\frac{6}{5}\) as high as the preceding one.

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 \[\begin{align} \ln y = \ln b + x \cdot \ln p, \end{align}\] whence \[\begin{align} \ln y - \ln b = x \cdot \ln p. \end{align}\] Now, since ln p is a mere number, and may be written as ln p = a, it follows that \[\ln \frac{y}{b}=ax,\] and the equation takes the new form \[y = be^{ax}.\]

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 \(\frac{3}{4}\) of the height of the preceding one.

The equation is still \[y=bp^x;\]

but since p is less than one, ln p will be a negative quantity, and may be written −a; so that p = e^−a,³ and now our equation for the curve takes the form \[y=be^{-ax}.\]

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 \[\theta_t=\theta_0 e^{-at};\] where θ₀ is the original excess of temperature of a hot body over that of its surroundings, θ_t 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, \[Q_t=Q_0 e^{-at},\] 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 e^−at 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, \(\dfrac{dy}{dt}\) 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 \(\dfrac{dy}{dx}\) is proportional to the height y; the curve becoming flatter as y grows smaller. In symbols, thus \[y=be^{-ax}\] or \[\ln y = \ln b - ax \ln e = \ln b - ax,\] and, differentiating, \[\frac{1}{y}\, \frac{dy}{dx} = -a;\] hence \[\frac{dy}{dx} = be^{-ax} \times (-a) = -ay;\] 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 \[\begin{align} y &= bp^x; \end{align}\] for then \[\begin{align} \frac{dy}{dx}= bp^x \times \ln p. \end{align}\] But \[\ln p = -a;\] giving us \[\begin{align} \frac{dy}{dx} = y \times (-a) = -ay, \end{align}\] as before.

The Time-constant. In the expression for the “die-away factor” e^−at, 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 \(e^{-\frac{t}{T}}\); and it will be seen, by making t = T that the meaning of T \(\left(\text{or of}~\dfrac{1}{a}\right)\) 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 \(\dfrac{t}{T} = \dfrac{60}{20} = 3\), 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

Exercises II

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:

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{align} \sinh x&=\frac{e^{x}-e^{-x}}{2},\\ \cosh x&=\frac{e^{x}+e^{-x}}{2},\\ \tanh x&=\frac{\sinh x}{\cosh x}\\ \text{coth } x&=\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}\\ \text{sech }x &=\frac{1}{\cosh x}\\ \text{csch }x &=\frac{1}{\sinh x} \end{align}}\]

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

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, e^x is very large, while e^−x is extremely small; hence e^x ± e^−x ≈ e^x leading to: \[\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\approx\frac{e^{x}}{e^{x}}=1.\] Similarly, when x is negative large, e^x is negligible, making e^x ± e^−x ≈ ±e^−x, and hence: \[\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\approx\frac{-e^{x}}{e^{x}}=-1.\] The graph of y = tanh x is shown below.

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 \[\begin{align} \cosh x-\sinh x & =e^{-x}\\ \cosh x+\sinh x & =e^{x} \end{align}\] Now using A² − B² = (A − B)(A + B), we get \[\begin{align} \cosh^2 x-\sinh^2 x&=(\cosh x-\sinh x)(\cosh x+\sinh x)\\ &= e^{-x}\cdot e^x=e^{-x+x}=e^0=1 \end{align}\] As such we have proved one of the fundamental identities: \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{ \cosh^{2}x-\sinh^{2}x=1}\] Now if we divide each term of the above equation by cosh² x, we obtain \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{1-\tanh^{2}x=\text{sech}^{2}x=\dfrac{1}{\cosh^{2}x}}\] Similarly, dividing each term of cosh² x − sinh² x = 1 by sinh² x gives us \[\text{coth}^2 x-\text{csch}^{2}x=1\]

Therefore, \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\sinh(x+ y)=\sinh x\cosh y+\cosh x\sinh y.}\] In a similar way, we can prove \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\cosh(x+ y)=\cosh x\cosh y+\sinh x\sinh y.}\] It follows from the above identity that \[\cosh 2x=\cosh^2 x+\sinh^2 x=(1+\sinh^2 x)+\sinh^2 x=1+2\sinh^2 x\] Hence, \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\sinh^2 x=\dfrac{1}{2}\left(\cosh 2x-1\right).}\] Similarly, \[\cosh 2x=\cosh^2x+\sinh^2 x=\cosh^2 x+(\cosh^2 x-1)=2\cosh^2 x-1\] Therefore, \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\cosh^2 x=\frac{1}{2}\left(\cosh 2x+1\right).}\]

Another important set of identities, which follow at once from the definitions, is:⁴ \[\sinh(-x)=-\sinh x\] \[\cosh(-x)=\cosh x\] \[\tanh(-x)=-\tanh x\]

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(e^x − e^−x), we have \[\begin{align} \frac{d(\sinh x)}{dx}&=\frac{d}{dx}\left(\frac{e^x}{2}-\frac{e^{-x}}{2}\right)\\ &=\frac{1}{2}\frac{d(e^x)}{dx}-\frac{1}{2}\frac{d(e^{-x})}{dx}\\ &=\frac{e^x}{2}-\frac{-e^{-x}}{2}=\frac{e^x+e^{-x}}{2}=\cosh x. \end{align}\] Similarly, we can show that \(\dfrac{d(\cosh x)}{dx}=\sinh x\).

In summary, \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{array}{rl} \dfrac{d(\sinh x)}{dx} & =\cosh x\\[9pt] \dfrac{d(\cosh x)}{dx} & =\sinh x\\[9pt] \dfrac{d(\tanh x)}{dx} & =\text{sech}^{2}x=1-\tanh^{2}x \end{array}}\]

Inverse Hyperbolic Functions

The inverse hyperbolic sine function is denoted by arcsinh x or sinh⁻¹ x, and is defined by \[y=\text{arcsinh }x=\sinh^{-1} x\qquad \text{if}\qquad x=\sinh y.\] We may also define the inverse hyperbolic cosine function, denoted by arccosh x or cosh⁻¹ x, as \[y=\text{arccosh }x=\cosh^{-1} x\qquad \text{if}\qquad x=\cosh y.\] 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 \[y=\text{arctanh }x=\tanh^{-1} x\qquad \text{if}\qquad x=\tanh y.\] 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,

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{align}y=\text{ arcsinh } x & \Leftrightarrow y=\sinh x \\ y=\text{ arccosh } x & \Leftrightarrow x=\cosh y & & (x\geq 1,y\geq 0)\\ y=\text{ arctanh } x & \Leftrightarrow x=\tanh y & & (-1<x<1) \end{align} }\]

We can find explicit formulas for the inverse hyperbolic functions.

In a similar fashion, we can derive explicit formulas for the other inverse hyperbolic functions. \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{align} & \text{ arcsinh }x=\ln\left(x+\sqrt{x^{2}+1}\right) & & (x \text{ can be any number})\\ & \text{ arccosh }x=\ln\left(x+\sqrt{x^{2}-1}\right) & & (x\geq1)\\ & \text{ arctanh }x=\frac{1}{2}\ln\frac{1+x}{1-x} & & (-1<x<1) \end{align} }\]

Derivatives of the Inverse Hyperbolic Functions

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

In summary,

\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{align} & \frac{d\left(\text{arcsinh }x\right)}{dx}=\frac{d\left(\sinh^{-1}x\right)}{dx}=\frac{1}{\sqrt{x^{2}+1}} & & (x\in\mathbb{R})\\ & \frac{d\left(\text{arccosh }x\right)}{dx}=\frac{d\left(\cosh^{-1}x\right)}{dx}=\frac{1}{\sqrt{x^{2}-1}} & & (x>1)\\ & \frac{d\left(\text{arctanh }x\right)}{dx}=\frac{d\left(\tanh^{-1}x\right)}{dx}=\frac{1}{1-x^{2}} & & (-1<x<1) \end{align} }\]

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 \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{u}\times 1=\frac{1}{x+a}.\]↩︎

ln p = −a ⇒ p = e^−a.↩︎

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 \[f(-x)=-f(x)\qquad \text{for every } x,\] and is called even if \[f(-x)=f(x)\qquad\text{for every } x.\]↩︎