On true Compound Interest and the Law of Organic Growth

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.

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 $y + n\dfrac{y}{n} = 2y.$

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 Figure 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 Figure 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 $2.71828$.

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

What, then, is epsilon?

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 more about Epsilon. 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{alignat*}{2} &(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{alignat*}

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.

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 $\epsilon = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \text{etc}.\ldots$

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:

 $1.000000$ 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 $2.718281$

$\epsilon$ 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 $\epsilon^x$ when we make $n$ indefinitely great. \begin{align*} \epsilon^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!} · \frac{n^2x^2 - nx}{n^2} + \frac{1}{3!} · \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{align*}

But, when $n$ is made indefinitely great, this simplifies down to the following: $\epsilon^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \text{etc.}\dots$

This series is called the exponential series.

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{align*} y &= A + Bx + Cx^2 + Dx^3 + Ex^4 + \text{etc}.,\\ \end{align*} (in which the coefficients $A$, $B$, $C$, etc. will have to be determined), and differentiate it. \begin{align*} \dfrac{dy}{dx} &= B + 2Cx + 3Dx^2 + 4Ex^3 + \text{etc}. \end{align*}

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 $C=\dfrac{B}{2}=\dfrac{A}{1· 2}$; that $D = \dfrac{C}{3} = \dfrac{A}{1 · 2 · 3}$; that $E = \dfrac{D}{4} = \dfrac{A}{1 · 2 · 3 · 4}$, etc.

The law of change is therefore that $y = A\left(1 + \dfrac{x}{1} + \dfrac{x^2}{1 · 2} + \dfrac{x^3}{1 · 2 · 3} + \dfrac{x^4}{1 · 2 · 3 · 4} + \text{etc}.\right).$

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

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{align*} \text{when } x &= 1,\quad & y &= 2.718281 \text{ etc.}; & \text{that is, } y &= \epsilon; \\ \text{when } x &= 2,\quad & y &=(2.718281 \text{ etc.})^2; & \text{that is, } y &= \epsilon^2; \\ \text{when } x &= 3,\quad & y &=(2.718281 \text{ etc.})^3; & \text{that is, } y &= \epsilon^3; \end{align*} and therefore $\text{when } x=x,\quad y=(2.718281 \text{ etc}.)^x;\quad\text{that is, } y=\epsilon^x,$ thus finally demonstrating that $\epsilon^x = 1 + \dfrac{x}{1} + \dfrac{x^2}{1·2} + \dfrac{x^3}{1· 2· 3} + \dfrac{x^4}{1· 2· 3· 4} + \text{etc}.$

[Note.–How to read exponentials . For the benefit of those who have no tutor at hand it may be of use to state that $\epsilon^x$ is read as “epsilon to the eksth power;” or some people read it “exponential eks.” So $\epsilon^{pt}$ is read “epsilon to the pee-teeth-power” or “exponential pee tee.” Take some similar expressions:–Thus, $\epsilon^{-2}$ is read “epsilon to the minus two power” or “exponential minus two.” $\epsilon^{-ax}$ is read “epsilon to the minus ay-eksth” or “exponential minus ay-eks.”]

Of course it follows that $\epsilon^y$ remains unchanged if differentiated with respect to $y$. Also $\epsilon^{ax}$, which is equal to $(\epsilon^a)^x$, will, when differentiated with respect to $x$, be $a\epsilon^{ax}$, because $a$ is a constant.

Natural or Naperian Logarithms.

Another reason why $\epsilon$ is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If $y$ is the value of $\epsilon^x$, then $x$ is the logarithm, to the base $\epsilon$, of $y$. Or, if \begin{align*} y &= \epsilon^x, \\ \text{then}\; x &= \log_\epsilon y. \end{align*}

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

The points calculated are:

For Fig. 38:

 $x$ $0$ $0.5$ $1$ $1.5$ $2$ $y$ $1$ $1.65$ $2.71$ $4.50$ $7.39$

For Fig. 39:

 xy12348$xx$ $0$ $0.69$ $1.10$ $1.39$ $2.08$

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 $\epsilon$, 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 $\log_\epsilon a + \log_\epsilon b = \log_\epsilon ab.$ Also the rule of powers holds good; $n × \log_\epsilon a = \log_\epsilon a^n.$ 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 \begin{align*} \log_{10} x &= 0.4343 × \log_{\epsilon} x, \\ \text{ and conversely,}\; \log_{\epsilon} x &= 2.3026 × \log_{10} x. \end{align*}

A Useful Table of “Naperian Logarithms”

(Also called Natural Logarithms or Hyperbolic Logarithms)

Number $\log_{\epsilon}$ Number $\log_{\epsilon}$
$1$ $0.0000$      $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$
$3.0$ $1.0986$ $1000$$6.9078 3.5 1.2528 2000$$7.6009$
$4.0$ $1.3863$ $5000$$8.5172 4.5 1.5041 10 000$$9.2103$
$5.0$ $1.6094$ $20 000$$9.9035$

Exponential and Logarithmic Equations.

Take the equation: $y = \log_\epsilon x.$ First transform this into $\epsilon^y = x,$ whence, since the differential of $\epsilon^y$ with regard to $y$ is the original function unchanged (see here), $\frac{dx}{dy} = \epsilon^y,$ and, reverting from the inverse to the original function, $\frac{dy}{dx} = \frac{1}{\ \dfrac{dx}{dy}\ } = \frac{1}{\epsilon^y} = \frac{1}{x}.$

Now this is a very curious result. It may be written $\frac{d(\log_\epsilon x)}{dx} = x^{-1}.$

Note that $x^{-1}$ is a result that we could never have got by the rule for differentiating powers. That rule 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 $0 × x^{-1}$, because $x^0$ is itself $= 1$, and is a constant. We shall have to come back to this curious fact that differentiating $\log_\epsilon x$ gives us $\dfrac{1}{x}$ when we reach the chapter on integrating.

Now, try to differentiate \begin{align*} y &= \log_\epsilon(x+a),\\ \text{that is}\; \epsilon^y &= x+a; \end{align*} we have $\dfrac{d(x+a)}{dy} = \epsilon^y$, since the differential of $\epsilon^y$ remains $\epsilon^y$. This gives \begin{align*} \frac{dx}{dy} &= \epsilon^y = x+a; \\ \end{align*} hence, reverting to the original function, we get \begin{align*} \frac{dy}{dx} &= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{x+a}. \end{align*}

Next try \begin{align*} y &= \log_{10} x. \end{align*}

First change to natural logarithms by multiplying by the modulus $0.4343$. This gives us \begin{align*} y &= 0.4343 \log_\epsilon x; \\ \text{whence}\; \frac{dy}{dx} &= \frac{0.4343}{x}. \end{align*}

Taking the logarithm of both sides, we get \begin{align*} \log_\epsilon y &= x \log_\epsilon a, \\ \text{ or}\; x = \frac{\log_\epsilon y}{\log_\epsilon a} &= \frac{1}{\log_\epsilon a} × \log_\epsilon y. \end{align*}

Since $\dfrac{1}{\log_\epsilon a}$ is a constant, we get $\frac{dx}{dy} = \frac{1}{\log_\epsilon a} × \frac{1}{y} = \frac{1}{a^x × \log_\epsilon a};$ hence, reverting to the original function. $\frac{dy}{dx} = \frac{1}{\;\dfrac{dx}{dy}\;} = a^x × \log_\epsilon a.$

We see that, since $\frac{dx}{dy} × \frac{dy}{dx} =1\quad\text{and}\quad \frac{dx}{dy} = \frac{1}{y} × \frac{1}{\log_\epsilon a},\quad \frac{1}{y} × \frac{dy}{dx} = \log_\epsilon a.$

We shall find that whenever we have an expression such as $\log_\epsilon y =$ 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 $\log_\epsilon y = x \log_\epsilon a$, $\frac{1}{y}\, \frac{dy}{dx} = \log_\epsilon a\quad\text{and}\quad \frac{dy}{dx} = a^x \log_\epsilon a.$

Let us now attempt further examples.

Examples (1) $y=\epsilon^{-ax}$. Let $-ax=z$; then $y=\epsilon^z$. $\frac{dy}{dz} = \epsilon^z;\quad \frac{dz}{dx} = -a;\quad\text{hence}\quad \frac{dy}{dx} = -a\epsilon^{-ax}.$

Or thus: $\log_\epsilon y = -ax;\quad \frac{1}{y}\, \frac{dy}{dx} = -a;\quad \frac{dy}{dx} = -ay = -a\epsilon^{-ax}.$

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

Or thus: $\log_\epsilon y = \frac{x^2}{3};\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2x}{3};\quad \frac{dy}{dx} = \frac{2x}{3}\, \epsilon^{\frac{x^2}{3}}.$

(3) $y = \epsilon^{\frac{2x}{x+1}}$. \begin{align*} \log_\epsilon y &= \frac{2x}{x+1},\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2(x+1)-2x}{(x+1)^2}; \\ hence \frac{dy}{dx} &= \frac{2}{(x+1)^2} \epsilon^{\frac{2x}{x+1}}. \end{align*}

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

(4) $y=\epsilon^{\sqrt{x^2+a}}$. $\log_\epsilon y=(x^2+a)^{\frac{1}{2}}$. $\frac{1}{y}\, \frac{dy}{dx} = \frac{x}{(x^2+a)^{\frac{1}{2}}}\quad\text{and}\quad \frac{dy}{dx} = \frac{x × \epsilon^{\sqrt{x^2+a}}}{(x^2+a)^{\frac{1}{2}}}.$ 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 $\sqrt{x^2+a}=z$.

(5) $y=\log(a+x^3)$. Let $(a+x^3)=z$; then $y=\log_\epsilon z$. $\frac{dy}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 3x^2;\quad\text{hence}\quad \frac{dy}{dx} = \frac{3x^2}{a+x^3}.$

(6) $y=\log_\epsilon\{{3x^2+\sqrt{a+x^2}}\}$. Let $3x^2 + \sqrt{a+x^2}=z$; then $y=\log_\epsilon z$. \begin{align*} \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{align*}

(7) $y=(x+3)^2 \sqrt{x-2}$. \begin{align*} \log_\epsilon y &= 2 \log_\epsilon(x+3)+ \tfrac{1}{2} \log_\epsilon(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{align*}

(8) $y=(x^2+3)^3(x^3-2)^{\frac{2}{3}}$. \begin{align*} \log_\epsilon y &= 3 \log_\epsilon(x^2+3) + \tfrac{2}{3} \log_\epsilon(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{align*} For if $y=\log_\epsilon(x^2+3)$, let $x^2+3=z$ and $u=\log_\epsilon z$. $\frac{du}{dz} = \frac{1}{z};\quad \frac{dz}{dx} = 2x;\quad \frac{du}{dx} = \frac{2x}{x^2+3}.$ Similarly, if $v=\log_\epsilon(x^3-2)$, $\dfrac{dv}{dx} = \dfrac{3x^2}{x^3-2}$ and $\frac{dy}{dx} = (x^2+3)^3(x^3-2)^{\frac{2}{3}} \left\{ \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2} \right\}.$

(9) $y=\dfrac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}}$. \begin{align*} \log_\epsilon y &= \frac{1}{2} \log_\epsilon(x^2+a) - \frac{1}{3} \log_\epsilon(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} \\ and \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{align*}

(10) $y=\dfrac{1}{\log_\epsilon x}$ $\frac{dy}{dx} = \frac{\log_\epsilon x × 0 - 1 × \dfrac{1}{x}} {\log_\epsilon^2 x} = -\frac{1}{x \log_\epsilon^2x}.$

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

(12) $y=\left(\dfrac{1}{a^x}\right)^{ax}$. \begin{align*} \log_\epsilon y &= ax(\log_\epsilon 1 - \log_\epsilon a^x) = -ax \log_\epsilon a^x. \\ \frac{1}{y}\, \frac{dy}{dx} &= -ax × a^x \log_\epsilon a - a \log_\epsilon a^x. \\ and \frac{dy}{dx} &= -\left(\frac{1}{a^x}\right)^{ax} (x × a^{x+1} \log_\epsilon a + a \log_\epsilon a^x). \end{align*}

Try now the following exercises.

Exercises XII

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

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

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

(4) Show that if $y=\dfrac{1}{b}·\dfrac{a^{bx}}{\log_\epsilon a}$,   $\dfrac{dy}{dx}=a^{bx}$.

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

Differentiate

(6) $y=\log_\epsilon x^n$.

(7) $y=3\epsilon^{-\frac{x}{x-1}}$.

(8) $y=(3x^2+1)\epsilon^{-5x}$.

(9) $y=\log_\epsilon(x^a+a)$.

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

(11) $y=\dfrac{\log_\epsilon(x+3)}{x+3}$.

(12) $y=a^x × x^a$.

(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 $s=ay^2 \log_\epsilon \frac{1}{y};$ 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 $y=1 ÷ \sqrt{\epsilon}$.

(14) Find the maximum or minimum of $y=x^3-\log_\epsilon x.$

(15) Differentiate $y=\log_\epsilon(ax\epsilon^x)$.

(16) Differentiate $y=(\log_\epsilon ax)^3$.

(1) $ab(\epsilon^{ax} + \epsilon^{-ax})$.

(2) $2at + \dfrac{2}{t}$.

(3) $\log_\epsilon n$.

(5) $npv^{n-1}$.

(6) $\dfrac{n}{x}$.

(7) $\dfrac{3\epsilon^{- \frac{x}{x-1}}}{(x - 1)^2}$.

(8) $6x \epsilon^{-5x} - 5(3x^2 + 1)\epsilon^{-5x}$.

(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) $\dfrac{1 - \log_\epsilon \left(x + 3\right)}{\left(x + 3\right)^2}$.

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

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

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

(16) $\dfrac{3}{x} (\log_\epsilon ax)^2$.

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 Figure 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, Figure 41, with values of $\log_\epsilon y$ as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows from the equation, that \begin{align*} \log_\epsilon y &= \log_\epsilon b + x · \log_\epsilon p, \\ \text{whence }\; \log_\epsilon y &- \log_\epsilon b = x · \log_\epsilon p. \end{align*}

Now, since $\log_\epsilon p$ is a mere number, and may be written as $\log_\epsilon p=a$, it follows that $\log_\epsilon \frac{y}{b}=ax,$ and the equation takes the new form $y = b\epsilon^{ax}.$

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