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 \(10\) per cent. 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 \(5\) per cent., he would have had to hoard for \(20\) years to double his property. If it had been only \(2\) per cent., 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 \(10\) per cent. 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 £\(12\)\(2\)s.; so that he starts the fourth year with £\(133\)\(2\)s., 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 £\(259\)\(7\)s\(6\)d. In fact, we see that at the end of each year, each pound will have earned \(\tfrac{1}{10}\) of a pound, and therefore, if this is always added on, each year multiplies the capital by \(\tfrac{11}{10}\); 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 \(y_0\) 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 \(n^{\text th}\) 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 \(\tfrac{21}{20}\). If reckoned this way, by the end of ten years the capital would have grown to £\(265\)\(6\)s\(7\)d.; for \[(1 + \tfrac{1}{20})^{20} = {2.653}.\]

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-per-cent. interest for each tenth of the year. We now have \(100\) operations lasting over the ten years; or \[y_n = £100 \left( 1 + \tfrac{1}{100} \right)^{100};\] which works out to £\(270\)\(9\)s\(7\frac{1}{2}\)d.

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 \(\frac{1}{10}\) per cent. for each such period; then \[y_n = £100 \left( 1 + \tfrac{1}{1000} \right)^{1000};\] which works out to £\(271\)\(13\)s\(10\)d.

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 \(1\) per cent. Then \[y_n = £100 \left( 1 + \tfrac{1}{10,000} \right)^{10,000};\] which amounts to £\(271\)\(16\)s\(3\frac{1}{2}\)d.

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 Fig. 36, \(OP\) 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 Fig. 37, 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 \((1 + \tfrac{1}{10})^{10}\) 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 $e$.[1]

To this mysterious number (2.7182818) etc., the mathematicians have assigned as a symbol the letter $e$ (called Euler’s number). All schoolboys know that the Greek letter \(\pi\) (called pi) stands for \(3.141592\) etc.; but how many of them know that e means \(2.71828\)? Yet it is an even more important number than \(\pi\)!

What, then, is the number 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 epsilon.

This process of growing proportionately, at every instant, to the magnitude at that instant, some people call a logarithmic rate of growing. Unit logarithmic 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 \(100\) per cent. 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 logarithmically at unit rate, for the same time, will be \(2.71828\ldots\) .

A little bit more about 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 an ordinary table of logarithms) got by assuming \(n = 2\); \(n = 5\); \(n = 10\); and so on, up to \(n = 10,000\). \[\begin{aligned}  &(1 + \tfrac{1}{2})^2 &&= 2.25. \\ &(1 + \tfrac{1}{5})^5 &&= {2.488}. \\ &(1 + \tfrac{1}{10})^{10} &&= 2.594. \\ &(1 + \tfrac{1}{20})^{20} &&= 2.653. \\ &(1 + \tfrac{1}{100})^{100} &&= {2.705}. \\ &(1 + \tfrac{1}{1000})^{1000} &&= {2.7169}. \\ &(1 + \tfrac{1}{10,000})^{10,000} &&= {2.7181}.\end{aligned}\]

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 \[\begin{aligned} (a + b)^n &= a^n + n \dfrac{a^{n-1} b}{1!} + n(n – 1) \dfrac{a^{n-2} b^2}{2!} \\ & \phantom{= a^n\ } + n(n -1)(n – 2) \dfrac{a^{n-3} b^3}{3!} + \text{etc}. \end{aligned}\] Putting $a = 1$ and $b = \dfrac{1}{n}$, we get \[\begin{aligned} \left(1 + \dfrac{1}{n}\right)^n &= 1 + 1 + \dfrac{1}{2!} \left(\dfrac{n – 1}{n}\right) + \dfrac{1}{3!} \dfrac{(n – 1)(n – 2)}{n^2} \\ &\phantom{= 1 + 1\ } + \dfrac{1}{4!} \dfrac{(n – 1)(n – 2)(n – 3)}{n^3} + \text{etc}.\end{aligned}\]

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 \[e = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \text{etc}.\ldots\]

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:

dividing by \(1\) \(1.000000\)
dividing by \(2\) \(0.500000\)
dividing by \(3\) \(0.166667\)
dividing by \(4\) \(0.041667\)
dividing by \(5\) \(0.008333\)
dividing by \(6\) \(0.001389\)
dividing by \(7\) \(0.000198\)
dividing by \(8\) \(0.000025\)
dividing by \(9\) \(0.000002\)
Total \(2.718281\)

\(e\) is incommensurable with \(1\), and resembles \(\pi\) 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. \[\begin{aligned} e^x &= 1^{nx} + nx \frac{1^{nx-1} \left(\dfrac{1}{n}\right)}{1!} + nx(nx – 1) \frac{1^{nx – 2} \left(\dfrac{1}{n}\right)^2}{2!} \\ & \phantom{= 1^{nx}\ } + nx(nx – 1)(nx – 2) \frac{1^{nx-3} \left(\dfrac{1}{n}\right)^3}{3!} + \text{etc}.\\ &= 1 + x + \frac{1}{2!} \cdot \frac{n^2x^2 – nx}{n^2} + \frac{1}{3!} \cdot \frac{n^3x^3 – 3n^2x^2 + 2nx}{n^3} + \text{etc}. \\ &= 1 + x + \frac{x^2 -\dfrac{x}{n}}{2!} + \frac{x^3 – \dfrac{3x^2}{n} + \dfrac{2x}{n^2}}{3!} + \text{etc}.\end{aligned}\]

But, when \(n\) is made indefinitely great, this simplifies down to the following: \[\bbox[#F2F2F2,5px,border:2px solid black]{e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \text{etc.}\dots}\]

This series is called the exponential series.

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 differential coefficient is the same as itself. This can be instantly seen by differentiating it with respect to \(x\), thus:

\[ \begin{aligned} \frac{d(e^x)}{dx} &= 0 + 1 + \frac{2x}{1 \cdot 2} + \frac{3x^2}{1 \cdot 2 \cdot 3} + \frac{4x^3}{1 \cdot 2 \cdot 3 \cdot 4} \\ &\phantom{= 0 + 1 + \frac{2x}{1 \cdot 2} + \frac{3x^2}{1 \cdot 2 \cdot 3}\ } + \frac{5x^4}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \text{etc}. \\  &= 1 + x + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \frac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} + \text{etc}., \end{aligned}\]

which is exactly the same as the original series.

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx} e^x=e^x}\]

Now we might have gone to work the other way, and said: Go to; let us find a function of \(x\), such that its differential coefficient 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 \[\begin{aligned} y &= A + Bx + Cx^2 + Dx^3 + Ex^4 + \text{etc}.,\end{aligned}\] (in which the coefficients $A$, $B$, $C$, etc. will have to be determined), and differentiate it.. \[\begin{aligned}\dfrac{dy}{dx} &= B + 2Cx + 3Dx^2 + 4Ex^3 + \text{etc}.\end{aligned}\]

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

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

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

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



\[x=\log_e y\]

The two curves plotted in Figs. 38 and 39  represent these equations.

The points calculated are:

For Fig. 38  \begin{array}{|c||*{5}{c|}} \hline {x} & {0} & {0.5} & {1} & {1.5} & {2} \\ \hline {y} & {1} & {1.65} & {2.71} & {4.50} & {7.39} \\ \hline \end{array} For Fig. 39  \begin{array}{|c||*{5}{c|}} \hline {y} & {1} & {2} & {3} & {4} & {8} \\ \hline {x} & {0} & {0.69} & {1.10} & {1.39} & {2.08} \\ \hline \end{array}


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. [Today, the natural logarithm is often  denoted by $\ln x$, read ell en of eks].  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 logarithm simply by multiplying it by \(0.4343\); or

\log_{10} x &= 0.4343 \times \ln x, \\
\ln x &= 2.3026 \times \log_{10} x.


A Useful Table of “Naperian Logarithms”
(Also called Natural Logarithms or Hyperbolic Logarithms)

\[\begin{array}{c<{\quad}|>{\ }c>{\qquad\qquad}cr<{\quad}|>{\ }c} {\text{Number}} & {\ln} && {\text{Number}} & {\ln} \\ {1} & 0.0000 && 6 & 1.7918 \\ 1.1 & 0.0953 && 7 & 1.9459 \\ 1.2 & 0.1823 && 8 & 2.0794 \\ 1.5 & 0.4055 && 9 & 2.1972 \\ 1.7 & 0.5306 && 10 & 2.3026 \\ 2.0 & 0.6931 && 20 & 2.9957 \\ 2.2 & 0.7885 && 50 & 3.9120 \\ 2.5 & 0.9163 && 100 & 4.6052 \\ 2.7 & 0.9933 && 200 & 5.2983 \\ 2.8 & 1.0296 && 500 & 6.2146 \\ 3.0 & 1.0986 && 1,000 & 6.9078 \\ 3.5 & 1.2528 && 2,000 & {7.6009} \\ 4.0 & 1.3863 && 5,000 & 8.5172 \\ 4.5 & 1.5041 && 10,000 & {9.2103} \\ 5.0 & 1.6094 && 20,000 & 9.9035 \\ \end{array}\] 

Exponential and Logarithmic Equations

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 differential of \(e^y\) with regard to \(y\) is the original function unchanged (see above), \[\frac{dx}{dy} = e^y,\] 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 \[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d(\ln x)}{dx} = x^{-1}.}\]

Note that \(x^{-1}\) is a result that we could never have got by the rule for differentiating powers. That rule (chapter 4) is to multiply by the power, and reduce the power by \(1\). Thus, differentiating \(x^3\) gave us \(3x^2\); and differentiating \(x^2\) gave \(2x^1\). But differentiating \(x^0\) does not give us \(x^{-1}\) or \(0 \times x^{-1}\), because \(x^0\) 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 \[y = \ln(x+a),\] That is, \[e^y = x+a\]

we have \(\dfrac{d(x+a)}{dy} = e^y\), since the differential of \(e^y\) remains \(e^y\).

This gives us\[ \frac{dx}{dy} = e^y = x+a; \]

hence, reverting to the original function (see chapter 13), we get \[ \frac{dy}{dx} = \frac{1}{\dfrac{dx}{dy}} = \frac{1}{x+a}. \]

Now try \[ y = \log_{10} x \]

First change to natural logarithms by multiplying by the modulus \(0.4343\). This gives us \[\begin{aligned} y &= 0.4343 \ln x; \\ \frac{dy}{dx} &= \frac{0.4343}{x} \end{aligned}\]

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

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

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 differential coefficient of the function of \(x\), so that we could have written at once, from \(\ln y = x \ln a\), \[\frac{1}{y}\, \frac{dy}{dx} = \ln a\quad\text{and}\quad \frac{dy}{dx} = a^x \ln a. %[ **”/” presumed][ **F1 – this note was placed after 1/y]\]

Let us now attempt further examples.



Example 1

Let \(-ax=z\); then \(y=e^z\). \[\frac{dy}{dz} = e^z;\quad \frac{dz}{dx} = -a;\quad\text{hence}\quad \frac{dy}{dx} = -a e^{-ax}.\]

Or thus: \[\ln y = -ax;\quad \frac{1}{y}\, \frac{dy}{dx} = -a;\quad \frac{dy}{dx} = -ay = -a e^{-ax}.\]

Example 2

Let \(\dfrac{x^2}{3}=z\); then \(y=e^z\). \[\frac{dy}{dz} = e^z;\quad \frac{dz}{dx} = \frac{2x}{3};\quad \frac{dy}{dx} = \frac{2x}{3}\, e^{\frac{x^2}{3}}.\]

Or thus: \[\ln y = \frac{x^2}{3};\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2x}{3};\quad \frac{dy}{dx} = \frac{2x}{3}\, e^{\frac{x^2}{3}}.\]

Example 3
\(y = e^{\frac{2x}{x+1}}\).

\[\begin{aligned} \ln y &= \frac{2x}{x+1},\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2(x+1)-2x}{(x+1)^2}; \\ \frac{dy}{dx} &= \frac{2}{(x+1)^2}e^{\frac{2x}{x+1}}. \end{aligned}\]

Check by writing \(\dfrac{2x}{x+1}=z\).

Example 4

\(\ln y=(x^2+a)^{\frac{1}{2}}\).

\[\frac{1}{y}\, \frac{dy}{dx} = \frac{x}{(x^2+a)^{\frac{1}{2}}}\quad\text{and}\quad \frac{dy}{dx} = \frac{x \times e^{\sqrt{x^2+a}}}{(x^2+a)^{\frac{1}{2}}}.\] For if \((x^2+a)^{\frac{1}{2}}=u\) and \(x^2+a=v\), \(u=v^{\frac{1}{2}}\), \[\frac{du}{dv} = \frac{1}{{2v}^{\frac{1}{2}}};\quad \frac{dv}{dx} = 2x;\quad \frac{du}{dx} = \frac{x}{{(x^2+a)}^{\frac{1}{2}}}.\]

Check by writing \(\sqrt{x^2+a}=z\).

Example 5

Let \((a+x^3)=z\); then \(y=\ln z\). \[\frac{dy}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 3x^2;\quad\text{hence}\quad \frac{dy}{dx} = \frac{3x^2}{a+x^3}.\]

Example 6

Let \(3x^2 + \sqrt{a+x^2}=z\); then \(y=\ln z\). \[\begin{aligned} \frac{dy}{dz} &= \frac{1}{z};\quad \frac{dz}{dx} = 6x + \frac{x}{\sqrt{x^2+a}}; \\ \frac{dy}{dx} &= \frac{6x + \dfrac{x}{\sqrt{x^2+a}}}{3x^2 + \sqrt{a+x^2}} = \frac{x(1 + 6\sqrt{x^2+a})}{(3x^2 + \sqrt{x^2+a}) \sqrt{x^2+a}}.\end{aligned}\]

Example 7
\(y=(x+3)^2 \sqrt{x-2}\).

\[\begin{aligned} \ln y &= 2 \ln(x+3)+ \tfrac{1}{2} \ln(x-2). \\ \frac{1}{y}\, \frac{dy}{dx} &= \frac{2}{(x+3)} + \frac{1}{2(x-2)}; \\ \frac{dy}{dx} &= (x+3)^2 \sqrt{x-2} \left\{\frac{2}{x+3} + \frac{1}{2(x-2)}\right\}.\end{aligned}\]

Example 8

\[\begin{aligned} \ln y &= 3 \ln(x^2+3) + \tfrac{2}{3} \ln(x^3-2); \\ \frac{1}{y}\, \frac{dy}{dx} &= 3 \frac{2x}{(x^2+3)} + \frac{2}{3} \frac{3x^2}{x^3-2} = \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2}.\end{aligned}\] For if \(y=\ln(x^2+3)\), let \(x^2+3=z\) and \(u=\ln z\). \[\frac{du}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 2x;\quad \frac{du}{dx} = \frac{2x}{x^2+3}.\] Similarly, if \(v=\ln(x^3-2)\), \(\dfrac{dv}{dx} = \dfrac{3x^2}{x^3-2}\) and \[\frac{dy}{dx} = (x^2+3)^3(x^3-2)^{\frac{2}{3}} \left\{ \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2} \right\}.\]

Example 9

\[\begin{aligned} \ln y &= \frac{1}{2} \ln(x^2+a) – \frac{1}{3} \ln(x^3-a). \\ \frac{1}{y}\, \frac{dy}{dx} &= \frac{1}{2}\, \frac{2x}{x^2+a} – \frac{1}{3}\, \frac{3x^2}{x^3-a} = \frac{x}{x^2+a} – \frac{x^2}{x^3-a} \\ \frac{dy}{dx} &= \frac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}} \left\{ \frac{x}{x^2+a} – \frac{x^2}{x^3-a} \right\}. \end{aligned}\]

Example 10
\(y=\dfrac{1}{\ln x}\)

 \[\frac{dy}{dx} = \frac{\ln x \times 0 – 1 \times \dfrac{1}{x}} {\ln^2 x} = -\frac{1}{x \ln^2x}.\]

Example 11
\(y=\sqrt[3]{\ln x} = (\ln x)^{\frac{1}{3}}\).

Let \(z=\ln x\); \(y=z^{\frac{1}{3}}\). \[\frac{dy}{dz} = \frac{1}{3} z^{-\frac{2}{3}};\quad \frac{dz}{dx} = \frac{1}{x};\quad \frac{dy}{dx} = \frac{1}{3x \sqrt[3]{\ln^2 x}}.\]

Example 12

\[\begin{aligned} \ln y &= ax(\ln 1 – \ln a^x) = -ax \ln a^x. \\ \frac{1}{y}\, \frac{dy}{dx} &= -ax \times a^x \ln a – a \ln a^x. \\ \frac{dy}{dx} &= -\left(\frac{1}{a^x}\right)^{ax} (x \times a^{x+1} \ln a + a \ln a^x). \end{aligned}\]


Try now the following exercises.

Exercises XII.

(1) Differentiate \(y=b(e^{ax} -e^{-ax})\).

(2) Find the differential coefficient with respect to \(t\) of the expression \(u=at^2+2\log_e t\).

(3) If \(y=n^t\), find \(\dfrac{d(\ln y)}{dt}\).

(4) Show that if \(y=\dfrac{1}{b}\cdot \dfrac{a^{bx}}{\ln a}\),\(\dfrac{dy}{dx}=a^{bx}\).

(5) If \(w=pv^n\), find \(\dfrac{dw}{dv}\).


(6) \(y=\ln x^n\).      (7) \(y=3e^{-\frac{x}{x-1}}\).
(8) \(y=(3x^2+1)e^{-5x}\). (9) \(y=\ln (x^a+a)\).

(10) \(y=(3x^2-1)(\sqrt{x}+1)\).

(11) \(y=\dfrac{\ln (x+3)}{x+3}\).

(12) \(y=a^x \times x^a\).

(13) It was shown by Lord Kelvin that the speed of signalling 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 \[s=ay^2 \ln \frac{1}{y};\] 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 \(y=1 \div \sqrt{e}\).

(14) Find the maximum or minimum of \[y=x^3-\ln x.\]

(15) Differentiate \(y=\ln(ax e^x)\).

(16) Differentiate \(y=(\ln ax)^3\).

Answers to Exercises
(1) \(ab(e^{ax} + e^{-ax})\). (2) \(2at + \dfrac{2}{t}\). (3) \(\ln n\).
(5) \(npv^{n-1}\). (6) \(\dfrac{n}{x}\). (7) \(\dfrac{3e^{- \frac{x}{x-1}}}{(x – 1)^2}\).

(8) \(6xe^{-5x} – 5(3x^2 + 1)e^{-5x}\).

(9) \(\dfrac{ax^{a-1}}{x^a + a}\).

(10) \(\left(\dfrac{6x}{3x^2-1} + \dfrac{1}{2\left(\sqrt x + x\right)}\right) \left(3x^2-1\right)\left(\sqrt x + 1\right)\).

(11) \(\dfrac{1 – \ln \left(x + 3\right)}{\left(x + 3\right)^2}\).

(12) \(a^x\left(ax^{a-1} + x^a \ln a\right)\).

(14) Min.: \(y = 0.7\) for \(x = 0.694\).

(15) \(\dfrac{1 + x}{x}\).

(16) \(\dfrac{3}{x} (\ln ax)^2\).

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 Fig. 40, 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, Fig. 41, 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

\[\begin{aligned} \ln y &= \ln b + x \cdot \ln p, \\ \ln y &- \ln b = x \cdot \ln p. \end{aligned}\]


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 \[y = b e^{ax}.\]

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 Fig. 42, 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 \(p=e^{-a}\), 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 \(\theta_0\) is the original excess of temperature of a hot body over that of its surroundings, \(\theta_t\) 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 \(Q_0\), 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, above, 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

\[\begin{aligned} y&=be^{-ax}\\ \ln y &= \ln b – ax \ln e = \ln b – ax,\\ \frac{1}{y}\, \frac{dy}{dx} &= -a;\\ \frac{dy}{dx} &= b e^{-ax} \times (-a) = -ay; \end{aligned}\]

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

\[ \begin{aligned} y &= bp^x; \\ \frac{dy}{dx} &= bp^x \times \ln p. \\ \ln p &= -a; \\ \frac{dy}{dx} &= y \times (-a) = -ay, \end{aligned}\]

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 \(\theta_0\) or \(Q_0\) in the preceding instances) to die away \(\dfrac{1}{e}\)th part—that is to \(0.3678\)—of its original value.

The values of \(e^x\) and \(e^{-x}\) are continually required in different branches of physics, and as they are given in very few sets of mathematical tables, some of the values are tabulated for convenience.

\[\begin{array}{| .{2,2} | .{5,4} | .{1,6} | .{1,6} |} \hline  {x} & {e^x} & {e^{-x}} & {1-e^{-x}} \\ \hline  0.00 & 1.0000 & 1.0000 & 0.0000 \\ 0.10 & 1.1052 & 0.9048 & 0.0952 \\ 0.20 & 1.2214 & 0.8187 & 0.1813 \\ 0.50 & 1.6487 & 0.6065 & 0.3935 \\ 0.75 & 2.1170 & 0.4724 & 0.5276 \\ 0.90 & 2.4596 & 0.4066 & 0.5934 \\ 1.00 & 2.7183 & 0.3679 & 0.6321 \\ 1.10 & 3.0042 & 0.3329 & 0.6671 \\ 1.20 & 3.3201 & 0.3012 & 0.6988 \\ 1.25 & 3.4903 & 0.2865 & 0.7135 \\ 1.50 & 4.4817 & 0.2231 & 0.7769 \\ 1.75 & {5.755} & 0.1738 & 0.8262 \\ 2.00 & 7.389 & 0.1353 & 0.8647 \\ 2.50 & {12.182} & 0.0821 & 0.9179 \\ 3.00 & {20.086} & 0.0498 & 0.9502 \\ 3.50 & 33.115 & 0.0302 & 0.9698 \\ 4.00 & 54.598 & 0.0183 & 0.9817 \\ 4.50 & 90.017 & 0.0111 & 0.9889 \\ 5.00 & 148.41 & 0.0067 & 0.9933 \\ 5.50 & 244.69 & 0.0041 & 0.9959 \\ 6.00 & 403.43 & 0.00248 & 0.99752 \\ 7.50 & 1808.04 & {0.00055} & 0.99947 \\ 10.00 & 22026.5 & 0.000045 & 0.999955 \\ \hline \end{array}\]

As an example of the use of this table, suppose there is a hot body cooling, and that at the beginning of the experiment ( when \(t = 0\)) it is \(72^\circ\) 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 \(\dfrac{1}{e}\) part of \(72^\circ\)), then we can calculate to what it will have fallen in any given time \(t\). For instance, let \(t\) be \(60\) minutes. Then \(\dfrac{t}{T} = 60 \div 20 = 3\), and we shall have to find the value of \(e^{-3}\), and then multiply the original \(72^\circ\) by this. The table shows that \(e^{-3}\) is \(0.0498\). So that at the end of \(60\) minutes the excess of temperature will have fallen to \(72^\circ \times 0.0498 = 3.586^\circ \).

Further Examples.

Example 1
The strength of an electric current in a conductor at a time \(t\) secs. after the application of the electromotive force producing it is given by the expression \(C = \dfrac{E}{R}\left\{1 – e^{-\frac{Rt}{L}}\right\}\).

The time constant is \(\dfrac{L}{R}\).

If \(E = 10\), \(R =1\), \(L = 0.01\); then when \(t\) is very large the term \(e^{-\frac{Rt}{L}}\) becomes \(1\), and \(C = \dfrac{E}{R} = 10\); also \[\frac{L}{R} = T = 0.01.\]

Its value at any time may be written: \[C = 10 – 10e^{-\frac{t}{0.01}},\] the time-constant being \(0.01\). This means that it takes \(0.01\) sec. for the variable term to fall by \(\dfrac{1}{e} = 0.3678\) of its initial value \(10e^{-\frac{0}{0.01}} = 10\).

To find the value of the current when \(t = 0.001~\text{sec.}\), say, \(\dfrac{t}{T} = 0.1\), \(e^{-0.1} = 0.9048\) (from table).

It follows that, after \(0.001\) sec., the variable term is \(0.9048 \times 10 = 9.048\), and the actual current is \(10 – 9.048 = 0.952\).

Similarly, at the end of \(0.1\) sec., \[\frac{t}{T} = 10;\quad e^{-10} = 0.000045;\] the variable term is \(10 \times 0.000045 = 0.00045\), the current being \(9.9995\).

Example 2
The intensity \(I\) of a beam of light which has passed through a thickness \(l\) cm. of some transparent medium is \(I = I_0e^{-Kl}\), where \(I_0\) 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 \(10\) cms. of a certain transparent medium, this means that \(82 = 100 \times e^{-K\times 10}\) or \(e^{-10K} = 0.82\), and from the table one sees that \(10K = 0.20\) very nearly; 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 \(50 = 100 \times e^{-0.02l}\), or \(0.5 = e^{-0.02l}\). It is found by putting this equation in its logarithmic form, namely, \[\log 0.5 = -0.02 \times l \times \log e,\] which gives \[l=\frac{-0.3010}{-0.02 \times 0.4343}=34.7\quad \text{ centimetres nearly.}\]

Example 3
The quantity \(Q\) of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity \(Q_0\) of the substance by the relation \(Q = Q_0 e^{-\lambda t}\), where \(\lambda\) 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 \(\lambda = 3.85 \times 10^{-3}\). Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

We have \(0.5 = e^{-0.00385t}\).

\[\begin{aligned} \log 0.5 &= -0.00385t \times \log e; \\ t &= 3\text{ minutes very nearly}. \end{aligned} \]

Exercises XIII.

(1) Draw the curve \(y = b e^{-\frac{t}{T}}\); where \(b = 12\), \(T = 8\), and \(t\) is given various values from \(0\) to \(20\).

(2) 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.

(3) Plot the curve \(y = 100(1-e^{-2t})\).

(4) The following equations give very similar curves: \[\begin{aligned} \text{(i)}\ y &= \frac{ax}{x + b}; \\ \text{(ii)}\ y &= a(1 – e^{-\frac{x}{b}}); \\ \text{(iii)}\ y &= \frac{a}{90^\circ } \arctan \left(\frac{x}{b}\right).\end{aligned}\]

Draw all three curves, taking \(a= 100\) millimetres; \(b = 30\) millimetres.

(5) Find the differential coefficient of \(y\) with respect to \(x\), if \[(\textit{a})~y = x^x;\quad (\textit{b})~y = (e^x)^x;\quad (\textit{c})~y = e^{x^x}.\]

(6) For “Thorium \(A\),” the value of \(\lambda\) 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 \(Q_0\) in the expression \[Q = Q_0 e^{-\lambda t};\] \(t\) being in seconds.

(7) A condenser of capacity \(K = 4 \times 10^{-6}\), charged to a potential \(V_0 = 20\), is discharging through a resistance of \(10,000\) ohms. Find the potential \(V\) after (a\(0.1\) second; (b\(0.01\) second; assuming that the fall of potential follows the rule \(V = V_0 e^{-\frac{t}{KR}}\).

(8) The charge \(Q\) of an electrified insulated metal sphere is reduced from \(20\) to \(16\) units in \(10\) minutes. Find the coefficient \(\mu\) of leakage, if \(Q = Q_0 \times e^{-\mu t}\); \(Q_0\) being the initial charge and \(t\) being in seconds. Hence find the time taken by half the charge to leak away.

(9) The damping on a telephone line can be ascertained from the relation \(i = i_0 e^{-\beta l}\), where \(i\) is the strength, after \(t\) seconds, of a telephonic current of initial strength \(i_0\); \(l\) is the length of the line in kilometres, and \(\beta\) is a constant. For the Franco-English submarine cable laid in 1910, \(\beta = 0.0114\). Find the damping at the end of the cable (\(40\) kilometres), and the length along which \(i\) is still \(8\)% of the original current (limiting value of very good audition).

(10) The pressure \(p\) of the atmosphere at an altitude \(h\) kilometres is given by \(p=p_0 e^{-kh}\); \(p_0\) 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.

(11) Find the minimum or maximum of \(y = x^x\).

(12) Find the minimum or maximum of \(y = x^{\frac{1}{x}}\).

(13) Find the minimum or maximum of \(y = xa^{\frac{1}{x}}\).

Answers to Exercises

(1) Let \(\dfrac{t}{T} = x\) (\(\therefore t = 8x\)), and use the Table.

(2) \(T = 34.627\); \(159.46\) minutes.

(3) Take \(2t = x\); and use the Table.

(5) (a) \(x^x \left(1 + \ln x\right)\); (b) \(2x(e^x)^x\); (c) \(e^{x^x} \times x^x \left(1 + \ln x\right)\).

(6) \(0.14\) second.

(7) (a) \(1.642\);(b) \(15.58\).

(8) \(\mu = 0.00037\), \(31^m \frac{1}{4}\).

(9) \(i\) is \(63.4\)% of \(i_0\), \(220\) kilometres.

(10) \(0.133\), \(0.145\), \(0.155\), mean \(0.144\); \(-10.2\)%, \(-0.9\)%, \(+77.2\)%.

(11) Min. for \(x = \dfrac{1}{e}\).

(12) Max. for \(x = e\).

(13) Min. for \(x = \ln a\).


[1] The original book uses the Greek letter $\epsilon$ for Euler’s number ($e$). However, today the English letter $e$ is used instead.