On true Compound Interest and the Law of Organic Growth

$\epsilon$ 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.

(Also called Natural Logarithms or Hyperbolic Logarithms)

Exponential and Logarithmic Equations.

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*}

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

Answers

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