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 \(y_0\) 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 \(n\) 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 \(\boldsymbol{e}\)
To this mysterious number \(2.7182818\dots\), 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 \(\pi\) (called pi) stands for \(3.141592\dots\); but how many of them know that \(e\) means \(2.71828\dots\)? Yet it is an even more important number than \(\pi\)!
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 \(2.71828\ldots\) .
A little more about the number \(\boldsymbol{e}\)
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:
| \(1.000000\) | |
| 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 | \(2.718281\) |
\(e\) 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. \[\begin{align} e^x &= 1^{nx} + nx \frac{1^{nx-1} \left(\dfrac{1}{n}\right)}{1!} + nx(nx - 1) \frac{1^{nx - 2} \left(\dfrac{1}{n}\right)^2}{2!} + nx(nx - 1)(nx - 2) \frac{1^{nx-3} \left(\dfrac{1}{n}\right)^3}{3!} + \cdots\\ &= 1 + x + \frac{1}{2!} \cdot \frac{n^2x^2 - nx}{n^2} + \frac{1}{3!} \cdot \frac{n^3x^3 - 3n^2x^2 + 2nx}{n^3} + \cdots \\ &= 1 + x + \frac{x^2 -\dfrac{x}{n}}{2!} + \frac{x^3 - \dfrac{3x^2}{n} + \dfrac{2x}{n^2}}{3!} + \cdots. \end{align}\]
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:
| \(x\) | \(-2\) | \(-1\) | \(-0.5\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) |
|---|---|---|---|---|---|---|---|---|
| \(y=e^x\) | \(0.14\) | \(0.37\) | \(0.61\) | \(1\) | \(1.65\) | \(2.71\) | \(4.50\) | \(7.39\) |
| \(y\) | \(0.1\) | \(0.5\) | \(1\) | \(2\) | \(3\) | \(4\) | \(8\) |
|---|---|---|---|---|---|---|---|
| \(x=\ln y\) | \(-2.30\) | \(-0.69\) | \(0\) | \(0.69\) | \(1.10\) | \(1.39\) | \(2.08\) |
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 \(e\), 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 \(\boldsymbol{e^x}\) and \(\boldsymbol{\ln x}\)
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 (\(\log_{10}x\)). If your calculator has an \(\boxed{\text{ln}}\) button for the natural logarithm, the \(\boxed{\text{log}}\) button is likely designed to return \(\log_{10}x\).
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 \(x^3\) 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 \[\begin{align} y = \ln(x+a), \end{align}\] 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
Example 14.1. Differentiate \(y\) with respect to \(x\) if \(y=e^{-ax}\).
Solution. Let \(-ax=z\); then \(y=e^z\). \[\frac{dy}{dx} = e^z;\quad \frac{dz}{dx} = -a;\quad\text{hence}\quad \frac{dy}{dx} = -ae^{-ax}.\]
Or thus: \[\ln y = -ax;\quad \frac{1}{y}\, \frac{dy}{dx} = -a;\quad \frac{dy}{dx} = -ay = -ae^{-ax}.\]
Example 14.2. Differentiate \(y\) with respect to \(x\) if \(y=e^{\frac{x^2}{3}}\).
Solution. 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 14.3. Given \(y = e^{\frac{2x}{x+1}}\), find \(\dfrac{dy}{dx}\).
Solution. \[\begin{align} \ln y &= \frac{2x}{x+1},\quad \frac{1}{y}\, \frac{dy}{dx} = \frac{2(x+1)-2x}{(x+1)^2}; \end{align}\] hence \[\begin{align} \frac{dy}{dx} &= \frac{2}{(x+1)^2} e^{\frac{2x}{x+1}}. \end{align}\]
Check by writing \(\dfrac{2x}{x+1}=z\).
\(y=e^{\sqrt{x^2+a}}\).\(\ln y=(x^2+a)^{\frac{1}{2}}\). \[\frac{1}{y}\, \frac{dy}{dx} = \frac{x}{(x^2+a)^{\frac{1}{2}}}\quad\text{and}\quad \frac{dy}{dx} = \frac{x \times e^{\sqrt{x^2+a}}}{(x^2+a)^{\frac{1}{2}}}.\] 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 14.4. If \(y=\log(a+x^3)\), find \(\dfrac{dy}{dx}\).
Solution. 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 14.5. If \(y=\ln\left\{{3x^2+\sqrt{a+x^2}}\right\}\), find \(\dfrac{dy}{dx}\).
Solution. Let \(3x^2 + \sqrt{a+x^2}=z\); then \(y=\ln z\). \[\begin{align} \frac{dy}{dz} &= \frac{1}{z};\quad \frac{dz}{dx} = 6x + \frac{x}{\sqrt{x^2+a}}; \\ \frac{dy}{dx} &= \frac{6x + \dfrac{x}{\sqrt{x^2+a}}}{3x^2 + \sqrt{a+x^2}} = \frac{x(1 + 6\sqrt{x^2+a})}{(3x^2 + \sqrt{x^2+a}) \sqrt{x^2+a}}. \end{align}\]
Example 14.6. If \(y=(x+3)^2 \sqrt{x-2}\), find \(\dfrac{dy}{dx}\).
Solution. Taking the logarithm of both sides, we get \[\begin{align} \ln y &=\ln\left[(x+3)^2\sqrt{x-2}\right]\\ &=\ln\left[(x+3)^2\right]+\ln\left[(x-2)^{\frac{1}{2}}\right]\\ &= 2 \ln(x+3)+ \frac{1}{2} \ln(x-2). \end{align}\] Differentiating both sides, we get \[\begin{align} \frac{1}{y}\, \frac{dy}{dx} &= \frac{2}{(x+3)} + \frac{1}{2(x-2)}; \\ \frac{dy}{dx} &= (x+3)^2 \sqrt{x-2} \left\{\frac{2}{x+3} + \frac{1}{2(x-2)}\right\}. \end{align}\]
Example 14.7. If \(y=(x^2+3)^3(x^3-2)^{\frac{2}{3}}\), find \(\dfrac{dy}{dx}\).
Solution. Taking the logarithm of both sides, we get \[\begin{align} \ln y &= 3 \ln(x^2+3) + \frac{2}{3} \ln(x^3-2); \\ \frac{1}{y}\, \frac{dy}{dx} &= 3 \frac{2x}{(x^2+3)} + \frac{2}{3} \frac{3x^2}{x^3-2} = \frac{6x}{x^2+3} + \frac{2x^2}{x^3-2}. \end{align}\] For if \(u=\ln(x^2+3)\), 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 14.8. If \(y=\dfrac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}}\), find \(\dfrac{dy}{dx}\).
Solution. \[\begin{align} \ln y &= \ln\left[\frac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}}\right]\\ &=\ln \left[(x^2+a)^{\frac{1}{2}}(x^3-a)^{-\frac{1}{3}}\right]\\ &=\ln \left[(x^2+a)^{\frac{1}{2}}\right]+\ln \left[(x^3-a)^{-\frac{1}{3}}\right]\\ &= \frac{1}{2} \ln(x^2+a) - \frac{1}{3} \ln(x^3-a). \end{align}\] Differentiating \[\begin{align} \frac{1}{y}\, \frac{dy}{dx} &= \frac{1}{2}\, \frac{2x}{x^2+a} - \frac{1}{3}\, \frac{3x^2}{x^3-a} = \frac{x}{x^2+a} - \frac{x^2}{x^3-a} \end{align}\] and \[\begin{align} \frac{dy}{dx} &= \frac{\sqrt[2]{x^2+a}}{\sqrt[3]{x^3-a}} \left\{ \frac{x}{x^2+a} - \frac{x^2}{x^3-a} \right\}. \end{align}\]
Example 14.9. If \(y=\dfrac{1}{\ln x}\), find \(\dfrac{dy}{dx}\).
Solution. \[\frac{dy}{dx} = \frac{\ln x \times 0 - 1 \times \dfrac{1}{x}} {\ln^2 x} = -\frac{1}{x \ln^2x}.\tag{Quotient Rule}\]
Example 14.10. If \(y=\sqrt[3]{\ln x} = (\ln x)^{\frac{1}{3}}\), find \(\dfrac{dy}{dx}\).
Solution. 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 14.11. If \(y=\left(\dfrac{1}{a^x}\right)^{ax}\), find \(\dfrac{dy}{dx}\).
Solution. \[\begin{align} \ln y = ax(\ln 1 - \ln a^x) = -ax \ln a^x. \end{align}\] Differentiating \[\begin{align} \frac{1}{y}\, \frac{dy}{dx} = -ax \times a^x \ln a - a \ln a^x. \end{align}\] and \[\begin{align} \frac{dy}{dx} = -\left(\frac{1}{a^x}\right)^{ax} (x \times a^{x+1} \ln a + a \ln a^x). \end{align}\]
Try now the following exercises.
Exercises I
Exercise 14.1. Differentiate \(y=b(e^{ax} -e^{-ax})\).
Answer
\(ab(e ^{ax} + e ^{-ax})\).
Solution
(1)
\[y=b\left(e^{a x}-e^{-a x}\right)\]
\[\begin{align} \frac{d y}{d x}&=b\left(a e^{a x}+a e^{-a x}\right)\\ &=a b\left(e^{a x}+e^{-a x}\right) \end{align}\]
Exercise 14.2. Find the derivative of the expression \(u=at^2+2\ln t\) with respect to \(t\).
Answer
\(2at + \dfrac{2}{t}\).
Solution
\[\frac{d u}{d t}=2 a t+\frac{2}{t}\]
Exercise 14.3. If \(y=n^t\), find \(\dfrac{d(\ln y)}{dt}\).
Answer
\(\ln n\).
Solution
\[\frac{d\left(\ln n^{t}\right)}{d t}=\frac{d(t \ln n)}{d t}=\ln n\]
Exercise 14.4. Show that if \(y=\dfrac{1}{b}\cdot\dfrac{a^{bx}}{\ln a}\),\(\dfrac{dy}{dx}=a^{bx}\).
Solution
\[\frac{d y}{d x}=\frac{1}{b} \cdot \frac{b(\ln a) a^{b x}}{\ln a}=a^{b x}\]
Exercise 14.5. If \(w=pv^n\), find \(\dfrac{dw}{dv}\).
Answer
\(npv^{n-1}\).
Solution
\[\frac{d w}{d v}=n p v^{n-1}\]
Differentiate
Exercise 14.6. \(y=\ln x^n\).
Answer
\(\dfrac{n}{x}\).
Solution
\[y=\ln x^{n} \Rightarrow y=n \ln x\]
\[\frac{d y}{d x}=\frac{n}{x} \\ \]
Exercise 14.7. \(y=3e^{-\frac{x}{x-1}}\).
Answer
\(\dfrac{3e ^{- \frac{x}{x-1}}}{(x - 1)^2}\).
Solution
\[y=3 e^{-\frac{x}{x-1}}\] Let \(u=-\dfrac{x}{x-1}\). Then \(y=3e^u\), and \[\frac{dy}{du}=3 e^{u}\] \[\begin{align} \frac{d u}{d x}&=\frac{-1(x-1)-1(-x)}{(x-1)^{2}}\\ &=\frac{-x+1+x}{(x-1)^{2}}\\ &=\frac{1}{(x-1)^{2}} \end{align}\] Using the Chain Rule:
\[\frac{d y}{d x}=\frac{d y}{d x} \cdot \frac{d x}{d x}=\frac{3}{(x-1)^{2}} e^{\frac{-x}{x-1}}\]
Exercise 14.8. \(y=(3x^2+1)e^{-5x}\).
Answer
\(6x e ^{-5x} - 5(3x^2 + 1)e ^{-5x}\).
Solution
\[y=(3x^2-1)e^{-5x}\] Using the Product Rule, we get \[\begin{align} \frac{d y}{d x}&=\frac{d(3x^2-1)}{dx} e^{-5x}+(3x^2-1)\frac{d\left(e^{-5x}\right)}{dx}\\ &=6 x e^{-5 x}+\left(3 x^{2}-1\right)\left(-5 e^{-5 x}\right) \\ &=6x e ^{-5x} - 5(3x^2 + 1)e ^{-5x} \end{align}\]
Exercise 14.9. \(y=\ln\left(x^a+a\right)\).
Answer
\(\dfrac{ax^{a-1}}{x^a + a}\).
Solution
\[y=\ln \left(x^{a}+a\right)\] Let \(u=x^a+a\). Now using the Chain Rule \[\begin{align} \frac{d y}{d x}&= \frac{d(\ln u)}{du}\frac{du}{dx}\\ &=\frac{1}{u}\left(a x^{a-1}\right)\\ &=\frac{a x^{a-1}}{x^a+a} \end{align}\]
Exercise 14.10. \(y=(3x^2-1)(\sqrt{x}+1)\).
Answer
\(\left(\dfrac{6x}{3x^2-1} + \dfrac{1}{2\left(\sqrt x + x\right)}\right) \left(3x^2-1\right)\left(\sqrt x + 1\right)\).
Solution
\[y=\left(3 x^{2}-1\right) \sqrt{x+1}\] Using the Product Rule: \[\frac{d y}{d x}=6 x \sqrt{x+1}+\left(3 x^{2}-1\right) \cdot \frac{1}{2 \sqrt{x+1}}\]
Exercise 14.11. \(y=\dfrac{\ln(x+3)}{x+3}\).
Answer
\(\dfrac{1 - \ln \left(x + 3\right)}{\left(x + 3\right)^2}\).
Solution
\[y=\frac{\ln (x+3)}{x+3}\] Using the Quotient Rule: \[\begin{align} \frac{d y}{d x}&=\frac{\frac{1}{x+3}(x+3)-\ln (x+3)}{(x+3)^{2}}\\ &=\frac{1-\ln (x+3)}{(x+3)^{2}} \end{align}\]
Exercise 14.12. \(y=a^x \times x^a\).
Answer
\(a^x\left(ax^{a-1} + x^a \ln a\right)\).
Solution
\[y=a^{x} \cdot x^{a}\]
\[\begin{align} \frac{d y}{d x}&=\left(a^{x} \ln a\right) x^{a}+a x^{a-1} \cdot a^{x}\\ &=a^x\left(a x^{a-1}+x^a \ln a\right) \end{align}\]
Exercise 14.13. It was shown by Lord Kelvin that the speed of signaling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called \(y\), then the number of signals \(s\) that can be sent per minute can be expressed by the formula \[s=ay^2 \ln \frac{1}{y};\] 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 \(y=1 / \sqrt{e}\).
Solution
\[s=ay^2\ln\frac{1}{y}\qquad (y\neq 0)\] To differentiate \(\ln\frac{1}{y}=\ln\left(y^{-1}\right)\), let \(u=y^{-1}\). Then \[\begin{align} \frac{d\left(\ln\frac{1}{y}\right)}{dy}&=\frac{d(\ln u)}{du}\cdot\frac{du}{dy}\\ &=\frac{1}{u}\cdot \left(-y^{-2}\right)\\ &=-y\cdot\frac{1}{y^2}\\ &=-\frac{1}{y} \end{align}\] Now using the Product Rule, we get \[\begin{align} \frac{d s}{d y}&=2 a y\, \ln \frac{1}{y}+a y^2\left(-\frac{1}{y}\right) \\ &= -2 a y\, \ln y-a y\\ &=a y\,(-2 \ln y-1) \end{align}\]
\[\begin{align} \frac{d s}{d y}&=0 \Rightarrow-2 \ln y-1=0\\ & \Rightarrow \ln y=-\frac{1}{2} \Rightarrow y=e^{-1 / 2} \end{align}\]
\[\begin{align} \frac{d^{2} s}{d y^{2}}&=-2 a \ln y-2 a y\left(\frac{1}{y}\right)-a\\ &=-2 a \ln y-2 a-a \end{align}\]
When \(y=e^{-\frac{1}{2}}\), \[\frac{d^{2} s}{d y^{2}}=-2 a \ln e^{-\frac{1}{2}}-3 a=a \ln e-3 a=-2 a<0\] Therefore, it follows from the Second Derivative Test that \(s\) is a minimum if \(y=\dfrac{1}{\sqrt{e}}\).
Exercise 14.14. Find the maximum or minimum of \[y=x^3-\ln x.\]
Answer
Min.: \(y = 0.7\) for \(x = 0.694\).
Solution
\[y=x^{3}-\ln x\]
\[\frac{d y}{d x}=3 x^{2}-\frac{1}{x}=\frac{3 x^{3}-1}{x}\]
\[\frac{d y}{d x}=0 \Rightarrow 3 x^{3}-1=0\]
\[\Rightarrow x^{3}=\frac{1}{3} \Rightarrow x=\frac{1}{\sqrt[3]{3}}\]
To show that this specific value of \(x\) makes \(y\) a minimum, we apply the Second Derivative Test. First we differentiate \(\dfrac{dy}{dx}\) to find the second derivative \[\frac{d^2y}{dx^2}=9x^2.\] Since when \(x=\dfrac{1}{\sqrt[3]{3}}\) \[\frac{d^2y}{dx^2}=9\left(\frac{1}{\sqrt[3]{3}}\right)^2=\frac{9}{\sqrt[3]{9}}>0\] \(y=\left(\dfrac{1}{\sqrt[3]{3}}\right)^3-\ln\dfrac{1}{\sqrt[3]{3}}=\dfrac{1}{3}\left(1+\ln 3\right)\approx 0.7\) is a minimum occurring when \(x=\dfrac{1}{\sqrt[3]{3}}\).
Exercise 14.15. Differentiate \(y=\ln(axe^x)\).
Answer
\(\dfrac{1 + x}{x}\).
Solution
\[y=\ln \left(a x e^{x}\right)\]
Recall that \(\ln(AB)=\ln A+\ln B\) and \(\ln\left(A^B\right)=B\ln A\). Using these properties, we can rewrite \(y\) as \[y=\ln a+\ln x+\ln e^x=\ln a+\ln x+x\ln e\] Since \(\ln e=1\), \[y=\ln a+\ln x+x.\] Now we can differentiate term by term \[\frac{dy}{dx}=0+\frac{1}{x}+1=\frac{1+x}{x}.\]
Exercise 14.16. Differentiate \(y=(\ln ax)^3\).
Answer
\(\dfrac{3}{x} (\ln ax)^2\).
Solution
\[y=u^3\qquad\text{where}\quad u=\ln ax\] Using the Chain Rule \[\begin{align} \frac{dy}{dx}=&\frac{dy}{du}\cdot\frac{du}{dx}\\ =& 3u^2\cdot\frac{1}{ax}\\ =&3\cdot(\ln ax)^2\frac{1}{ax}\\ =&\frac{3}{ax}\left(\ln ax\right)^2 \end{align}\]
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 \(\theta_0\) 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, 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 \(\theta_0\) or \(Q_0\) in the preceding instances) to die away \(\dfrac{1}{e}\)th part—that is to \(0.3678\)—of its original value.
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 \(72~^\circ\)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 \(\dfrac{1}{e}\) 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 \(72~^\circ\text{C} \times 0.0498 = 3.586~^\circ\)C.
Further Examples
Example 14.12. The strength of an electric current in a conductor at a time \(t\) seconds after the application of the electromotive force producing it is given by the expression \(C = \dfrac{E}{R}\left\{1 - e^{-\frac{Rt}{L}}\right\}\).
The time constant is \(\dfrac{L}{R}\).
If \(E = 10\), \(R =1\), \(L = 0.01\); then when \(t\) is very large the term \(1-e^{-\frac{Rt}{L}}\) 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 \(t = 0.001~\)s, say, \(\dfrac{t}{T} = 0.1\), \(e^{-0.1} = 0.9048\).
It follows that, after \(0.001\) s, 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 \(0.1\) s, \[\frac{t}{T} = 10;\quad e^{-10} = 0.000045;\] the variable term is \(10 \times 0.000045 = 0.00045\), the current being \(9.9995\).
Example 14.13. 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 \(10\) cm of a certain transparent medium, this means that \((100-18) = 100 \times e^{-K\times10}\) or \(e^{-10K} = 0.82\). Now we take the natural logarithm of both sides \[\ln\left(e^{-10K}\right)=\ln 0.82\] or \[-10K \times \ln e =\ln 0.82\approx -0.2;\tag{$\ln e=1$}\] hence \(K\approx 0.02\).
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, \[\ln 0.5 = -0.02 \times l \times \ln e,\] which gives \[l \approx \frac{{-0.6931}}{-0.02 \times 1} \approx {34.7}~\text{centimeters}.\]
Example 14.14. 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.)
Solution. We have \(0.5 = e^{-0.00385t}\).\[\begin{align} \ln 0.5 &= -0.00385t \times \ln e; \tag{$\ln e=1$} \end{align}\] and \[\begin{align} t \approx 180.04~\text{s}\approx 3\text{ minutes very nearly}. \end{align}\]
Exercises II
Exercise 14.17. Draw the curve \(y = b e^{-\frac{t}{T}}\); where \(b = 12\), \(T = 8\), and \(t\) is given various values from \(0\) to \(20\).
Answer
Use your calculator to evaluate \(12 e^{-\frac{t}{8}}\) for different values of \(t\) within the range \(0\leq t\leq 20\). Then, connect the resulting points \((t, 12 e^{-\frac{t}{8}})\) together.
Solution
To sketch the curve by hand, we can evaluate \(y=12 e^{-\frac{t}{8}}\) at various values of \(t\) (\(0\leq t\leq 20\)). For example,
| \(t\) | \(y=12 e^{-t/8}\) |
|---|---|
| 0 | 12.000 |
| 1 | 10.590 |
| 5 | 6.423 |
| 10 | 3.438 |
| 15 | 1.840 |
| 20 | 0.985 |
Then plot each of these points \((t, y)\) on a set of axes, and connect these points with a smooth curve.
There are also multiple tools for plotting the curve \(y=12 e^{-\frac{t}{8}}\) for \(0\leq t\leq 20\). For example, you may visit WolframAlpha.com and in the search bar simply type:plot 12 e^ (-t/8) from t=0 to t=20.
Exercise 14.18. If a hot body cools so that in \(24\) minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to \(1\) per cent. of the original excess.
Answer
\(T = 34.625\); \(159.45\) minutes.
Solution
The equation of cooling is
\[\theta=\theta_{0} e^{-\frac{t}{T}}\]
After 24 minutes
\[\begin{align} \frac{\theta}{\theta_{0}} & =\frac{1}{2}=e^{-\frac{24}{T}} \\ \Rightarrow \quad \ln \frac{1}{2} & =\ln \left(e^{-\frac{24}{T}}\right) \\ -\ln 2 & =-\frac{24}{T} \\ T & =\frac{24}{\ln 2} \approx 34.6247 \end{align}\]
time-constant is approximately \(34.6247 \frac{1}{\mathrm{~min}}\).
Now we want to find \(t\) such that
\[\begin{gathered} \frac{\theta}{\theta_{0}}=0.01=e^{-\frac{t}{34.6247}} \\ \ln \left(10^{-2}\right)=\ln \left(e^{-\frac{t}{34.6247}}\right) \\ -2 \ln 10=-\frac{t}{34.6247} \\ \Rightarrow \quad t=2 \ln 10 \times 34.6247 \approx 159.453 \text { minutes } \end{gathered}\]
Exercise 14.19. Plot the curve \(y = 100(1-e^{-2t})\)
Solution
When \(x=0\), \(y=0\)
When \(x\) is a large positive number, the term \(e^{-2t}\) becomes negligible (\(\approx 0\)). Consequently, we have \[y\approx 100(1-0)=100.\]
On the other hand, when \(x\) is a large negative number, the terms \(e^{-2t}\) becomes very large positive. As a result \(y\) becomes numerically large but negative.
The curve \(y = 100(1-e^{-2t})\) is shown below.
Exercise 14.20. The following equations give very similar curves: \[\begin{align} \text{(i)}\ y &= \frac{ax}{x + b}; \\ \text{(ii)}\ y &= a(1 - e^{-\frac{x}{b}}); \\ \text{(iii)}\ y &= \frac{2a}{\pi} \arctan \left(\frac{x}{b}\right).\quad (\text{it may be written as }y = \frac{2a}{\pi} \tan^{-1} \left(\frac{x}{b}\right)) \end{align}\] Draw all three curves, taking \(a= 10\) units; \(b = 3\) units.
Solution
Let’s assume \(b>0\).
When \(x=0\), then
\(y=\frac{ax}{x + b}\) is zero.
\(y=a(1-e^{-\frac{x}{b}})\) is zero because \(e^0=1\).
\(y=\frac{2a}{\pi} \arctan \left(\frac{x}{b}\right)\) is zero because \(\arctan 0 =0\).
When \(x\) is very large positive
\(y=\dfrac{ax}{x + b}\) is \(\approx a\) because we can ignore \(b\) compared to \(x\) (for example, when \(b=3\) and \(x=1,000,000,000\), then \(x+b\approx 1,000,000,000\)).
\(y=a(1-e^{-\frac{x}{b}})\) is \(\approx a\) because \(e^{-x/b}\) is negligible for large values of \(x\).
\(y=\frac{2a}{\pi} \arctan \left(\frac{x}{b}\right)\) is \(\approx a\) because when \(x\) is large, \(\arctan(x/b)\approx \dfrac{\pi}{2}\).
For negative values of \(x\), these curves are not comparable because \(y=\dfrac{ax}{x + b}\) is very large when \(x\) is close to \(-b\).
When \(x\) is large negative,
\(e^{-x/b}\) is large positive. Therefore \(y=a(1-e^{-\frac{x}{b}})\) is large negative when \(x\) is large negative.
\(y=\dfrac{ax}{x + b}\) is \(\approx a\) because again \(b\) is negligible compared to \(x\) and thus \(x+b\approx x\).
\(y=\frac{2a}{\pi} \arctan \left(\frac{x}{b}\right)\) is \(\approx -a\) because when \(x\) is large negative, \(\arctan(x/b)\approx -\dfrac{\pi}{2}\).
These curves are shown below.
Exercise 14.21. Find the derivative of \(y\) with respect to \(x\), if \[(\text{a})~y = x^x;\quad (\text{b})~y = (e^x)^x;\quad (\text{c})~y = e^{\left(x^x\right)}.\]
Answer
a) \(x^x \left(1 + \ln x\right)\); (b) \(2x(e ^x)^x\); (c) \(e ^{x^x} \times x^x \left(1 + \ln x\right)\).
Solution
(a)
\[y=x^{x}\]
Taking the natural logarithm of both sides and recalling \(\ln \left(A^{B}\right)=B \ln A\) :
\[\begin{align} \ln y & =\ln \left(x^{x}\right) \\ & =x \ln x \end{align}\]
Now differentiating and using the Chain Rule for the left-hand side and the Product Rule for the right-hand side
\[\begin{align} \frac{1}{y} \frac{d y}{d x} & =\ln x+x \cdot \frac{1}{x} \\ \Rightarrow \quad \frac{d y}{d x} & =y(\ln x+1) \\ & =x^{x}(\ln x+1) \end{align}\]
(b)
\[y=\left(e^{x}\right)^{x}\]
Method 1)
Taking the natural logarithm of both sides and using the property \(\ln \left(A^{B}\right)=B \ln A\), we have
\[\begin{align} \ln y & =\ln \left(\left(e^{x}\right)^{x}\right) \\ & =x \ln \left(e^{x}\right) \\ & =x \cdot x=x^{2} \end{align}\]
Now let’s differentiate both sides with respect to \(x\):
\[\frac{1}{y} \cdot \frac{d y}{d x}=2 x\] Simplifying, we find \[\frac{d y}{d x}=2 x y=2 x\left(e^{x}\right)^{x}\]
Method 2) Recall that \(\left(A^{B}\right)^{C}=A^{B C}\). Therefore
\[y=\left(e^{x}\right)^{x}=e^{x \cdot x}=e^{x^{2}}\]
Let \(u=x^{2}\) and applying the Chain Rule with \(y=e^{u}\) where \(u=x^{2}\), we get
\[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =e^{u} \cdot 2 x \\ & =y \cdot 2 x \\ & =2 x\left(e^{x}\right)^{x} \end{align}\]
(c)
\[y=e^{\left(x^{x}\right)}\]
Method 1) Let \[y=e^{u} \quad \text { where } \quad u=x^{x}\] Then using the result of part (a) and the Chain Rule, we obtain \[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x}\\ &= e^{u} \cdot\left[x^{x}(\ln x+1)\right] \\ & =e^{\left(x^{x}\right)} \cdot x^{x}(\ln x+1) . \end{align}\]
Method 2) Take \(\ln\) of both sides
\[\begin{align} \ln y & =\ln \left(e^{\left(x^{x}\right)}\right)\\ & =x^{x} \ln e \\ & =x^{x} \end{align}\] Now differentiate both sides with respect to \(x\) to get \[\begin{align} \frac{1}{y} \frac{d y}{d x} & =\frac{d\left(x^{x}\right)}{d x} \\ & =x^{x}(\ln x+1) \end{align}\] Therefore \[\begin{align} \frac{d y}{d x} & =y \times x^{x}(\ln x+1) \\ & =e^{\left(x^{x}\right)} \times x^{x}(\ln x+1) . \end{align}\]
Exercise 14.22. 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.
Answer
\(0.14\) seconds.
Solution
\[Q=Q_{0} e^{-5 t}\] We need to find \(t\) such that \(\frac{Q}{Q_{0}}=\frac{1}{2}\)
\[\frac{1}{2}=e^{-5 t}\] Taking the natural logarithm of both sides \[\begin{align} \ln (0.5) & =\ln \left(e^{-5 t}\right) \\ & =-5 t \ln e \\ & =-5 t \\ \Rightarrow \quad t & =-\frac{\ln (0.5)}{5} \approx 0.1386 \text { seconds. } \end{align}\]
Exercise 14.23. 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) \(0.1\) seconds; (b) \(0.01\) seconds; assuming that the fall of potential follows the rule \(V = V_0 e^{-\frac{t}{KR}}\).
Answer
(a) \(1.642\);(b) \(15.58\).
Solution
Here \[V=20 e^{-\frac{t}{4\times 10^{-6}\times 10,000}}\] or \[V=20 e^{-25t}\]
(a) When \(t=0.1\)
\[V=20 e^{-2.5}\approx 1.6417\]
(b) When \(t=0.01\)
\[V=20 e^{-0.25}\approx 15.576\]
Exercise 14.24. 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.
Answer
\(\mu = 0.00037\), \(31\text{ min} 4 \text{ s}\).
Solution
Since \[\frac{Q}{Q_{0}}=\frac{16}{20}=e^{-\mu \times 10\times 60}\] we can find \(\mu\)
\[\begin{align} & \ln \frac{16}{20}=-\mu \times 10 \times 60 \\ & \Rightarrow \quad \mu=-\frac{1}{600} \ln 0.8 \approx 0.000372 \end{align}\]
Now we need to find \(t\) such that
\[\begin{align} \frac{1}{2} & =e^{-0.000372 t} \\ \ln (0.5) & =-0.000372 t \\ t & \approx 1863.77 \mathrm{~s} \approx 31 \mathrm{~min}\ 3.77 \mathrm{~s} \end{align}\]
Exercise 14.25. 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).
Answer
\(i\) is \(63.4\)% of \(i_0\), \(220\) kilometres.
Solution
\[\frac{i}{i_{0}}=e^{-\beta l}\]
(a)
\[\frac{i}{i_{0}}=e^{-0.0114 \times 40}\approx 0.6338=63.38\%.\]
Therefore, the strength becomes \(63.38 \%\) of the initial strength.
(b)
\[\begin{align} \frac{i}{i_{0}}=0.08 & =e^{-0.0114\, l} \\ \ln (0.08) & =-0.0114\,l \\ \ell & =-\frac{\ln (0.08)}{0.0114} \approx 221.555 \mathrm{~km} \end{align}\]
Exercise 14.26. 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.
Answer
\(0.1339\), \(0.1445\), \(0.1555\), mean \(0.1446\); \(-10.2\)%, practically nil, \(+71.9\)%.
Solution
\[\quad p=p_{0} e^{-k h}\]
Calculating \(k\) when \(h=10~\mathrm{km}\),
\[\frac{p}{p_{0}}=\frac{199.2}{760}=e^{-10 k} \Rightarrow k=-\frac{\ln \left(\frac{199.2}{760}\right)}{10} \approx 0.1339\]
Calculating \(k\) when \(h=20 \mathrm{~km}\)
\[\frac{p}{p_{0}}=\frac{42.2}{760}=e^{-20 k} \Rightarrow k=-\frac{\ln \left(\frac{42.2}{760}\right)}{20} \approx 0.1445\]
Calculating \(k\) when \(h=50 \mathrm{~km}\)
\[\frac{p}{p_0}=\frac{0.32}{760}=e^{-50 k} \Rightarrow k=-\frac{\ln \left(\frac{0.32}{760}\right)}{50} \approx 0.1555\]
Average \(k\) (the mean value of \(k\)): \[k_{\text{av}}=\frac{0.1339+0.1445+0.1555}{3} \approx 0.1446\]
Calculating \(p\) using \(k_{\text{av}}\) when \(h=10 \mathrm{~km}\):
\[\begin{align} & 760 \times e^{-10 \times 0.1446}=178.99 \\ & \text { error }=\frac{178.99-199.2}{199.2} \approx-10.1 \% \end{align}\]
Calculating \(p\) using \(k_{\text{av}}\) when \(h=20 \mathrm{~km}\):
\[\begin{align} & 760 \times e^{-20 \times 0.1446}=42.15 \\ & \text { error }=\frac{42.2-42.15}{42.2} \approx 0.12 \% \end{align}\]
Calculating \(p\) using \(k_{\text{av}}\) when \(h=50 \mathrm{~km}\):
\[\begin{array}{r} 760 \times e^{-50 \times 0.1446}=0.550 \\ \text { error }=\dfrac{0.55-0.32}{0.32} \approx 72.07 \% \end{array}\]
Exercise 14.27. Find the minimum or maximum of \(y = x^x\).
Answer
Min. for \(x = \dfrac{1}{e }\).
Solution
In Exercise S, we showed that
\[\frac{d y}{d x}=\frac{d\left(x^{x}\right)}{d x}=x^{x}(\ln x+1)\]
Setting
\[\frac{d y}{d x}=0 \Leftrightarrow x^{x}=0 \text { or } \ln x+1=0\]
Since \(x^{x} \neq 0\), at the minimum or maximum, we must have \[\ln x =-1\] Then \[\begin{align} e^{\ln x} & =e^{-1} \\ x & =e^{-1}=\frac{1}{e} . \end{align}\]
To determine if this value of \(x\) makes \(y\) a maximum or a minimum, we apply the Second Derivative Test:
\[\begin{align} \frac{d^{2} y}{d x^{2}} & =\frac{d\left(x^{x}\right)}{d x}(\ln x+1)+x^{x}\left(\frac{1}{x}\right) \\ & =x^{x}(\ln x+1)^{2}+\frac{x^{x}}{x} \end{align}\] When \(x=\dfrac{1}{e}\), then \[\begin{align} \frac{d^{2} y}{d x^{2}} & =\left(\frac{1}{e}\right)^{\frac{1}{e}} \times 0+\frac{\left(\frac{1}{e}\right)^{\frac{1}{e}}}{\frac{1}{e}} \\ & =\frac{1}{e}\left(\frac{1}{e}\right)^{\frac{1}{e}}>0 \end{align}\] Therefore, \(x=1/e\approx 0.369\) corresponds to a minimum value of \(y\).
The graph of \(y=x^{x}\) is shown below.
Exercise 14.28. Find the minimum or maximum of \(y = x^{\frac{1}{x}}\).
Answer
Max. for \(x = e\).
Solution
\[y=x^{\frac{1}{x}}\]
To differentiate, we first take the natural logarithm of both sides then use the Chain Rule:
\[\begin{align} \ln y & =\ln \left(x^{\frac{1}{x}}\right) \\ & =\frac{1}{x} \ln x \\ \frac{1}{y} \frac{d y}{d x} & =-\frac{1}{x^{2}} \ln x+\frac{1}{x} \cdot \frac{1}{x} \\ \frac{d y}{d x} & =\frac{y}{x^{2}}(1-\ln x) \end{align}\]
Since \(y=x^{\frac{1}{x}} \neq 0\),
\[\frac{d y}{d x}=0 \quad \Leftrightarrow \quad 1-\ln x=0\] or \[\frac{d y}{d x}=0 \Leftrightarrow x=e\]
To determine whether \(x=e\) makes \(y\) a minimum or a maximum, we compare the value of \(y\) at this point with the values of \(y\) at some nearby points:
When \(x=e\)
\[y=e^{\frac{1}{e}} \approx e^{\frac{1}{2.718}} \approx e^{0.368} \approx 1.444\]
When \(x=2\) \[y=2^{\frac{1}{2}}=\sqrt{2}=1.414\]
When \(x=3\)
\[y=3^{\frac{1}{3}}=\sqrt[3]{3} \approx 1.442\]
Therefore \(x=e\) makes \(y=x^{\frac{1}{x}}\) a maximum.
We can also apply the Second Derivative Test:
\[\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x} \frac{1-\ln x}{x^{2}}+y \frac{d}{d x}\left(\frac{1-\ln x}{x^{2}}\right)\]
Using the Quotient Rule
\[\frac{d}{d x}\left(\frac{1-\ln x}{x^{2}}\right)=\frac{-\frac{1}{x} \cdot x^{2}-2 x(1-\ln x)}{x^{4}}\] Therefore, \[\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x} \frac{1-\ln x}{x^{2}}-x^{\frac{1}{x}} \frac{x+2 x(1-\ln x)}{x^{2}}\]
When \(x=e\), both \(\dfrac{d y}{d x}\) and \(1-\ln x\) are zero. Therefore, when \(x=e\)
\[\frac{d^{2} y}{d x^{2}}=0 \times 0-e^{\frac{1}{e}} \frac{e+2 e(0)}{e^2}<0\]
It follows from the Second Derivative Test that \(y=e^{\frac{1}{e}}\approx 1.44\) is a maximum occurring when \(x=e\).
The graph of \(y=x^{\frac{1}{x}}\) is shown below
Exercise 14.29. Find the minimum or maximum of \(y = xa^{\frac{1}{x}}\).
Answer
Min. for \(x = \ln a\).
Solution
\[y=x a^{\frac{1}{x}}\]
Using the Product Rule
\[\frac{d y}{d x}=a^{\frac{1}{x}}+x \frac{d\left(a^{\frac{1}{x}}\right)}{d x}\]
To find \(\frac{d\left(a^{\frac{1}{x}}\right)}{d x}\), let \(u=\frac{1}{x}\) and use the Chain Rule
\[\begin{align} \frac{d a^{u}}{d x} & =\frac{d a^{u}}{d u} \cdot \frac{d u}{d x} \\ & =\left(a^{u} \ln a\right)\left(-\frac{1}{x^{2}}\right) \\ & =-\frac{\ln a}{x^{2}} a^{\frac{1}{x}} \end{align}\]
Therefore
\[\frac{d y}{d x}=a^{\frac{1}{x}}-x \frac{\ln a}{x^{2}} a^{\frac{1}{x}}=a^{\frac{1}{x}}\left(1-\frac{\ln a}{x}\right)\]
Since \(a^{\frac{1}{x}}>0\),
\[\frac{d y}{d x}=0 \Leftrightarrow 1-\frac{\ln a}{x}=0\]
or
\[\frac{d y}{d x}=0 \Leftrightarrow \boxed{x=\ln a}\]
Using the Second Derivative Test
\[\begin{gathered} \frac{d y}{d x}=y\left(1-\frac{\ln a}{x}\right) \\ \frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}\left(1-\frac{\ln a}{x}\right)+y \cdot \frac{\ln a}{x^{2}} \end{gathered}\]
When \(x=\ln a\)
\[\frac{d^{2} y}{d x^{2}}=0 \times 0+a^{\frac{1}{\ln a}} \cdot \frac{\ln a}{(\ln a)^{2}}>0\]
Therefore \(y=a^{\frac{1}{\ln a}}\) is a minimum occurring when \(x=\ln a\).
Hyperbolic Functions
Hyperbolic functions are specific combinations of exponential functions that arise frequently in various applications, leading mathematicians to give them distinctive names and thoroughly investigate their properties. Although hyperbolic functions are combined exponential functions, in some ways they resemble the trigonometric functions. As a result, the hyperbolic functions are given individual names such as hyperbolic sine, hyperbolic cosine, hyperbolic tangent, and others. Their definitions are as follows:
\[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\begin{align} \sinh x&=\frac{e^{x}-e^{-x}}{2},\\ \cosh x&=\frac{e^{x}+e^{-x}}{2},\\ \tanh x&=\frac{\sinh x}{\cosh x}\\ \text{coth } x&=\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}\\ \text{sech }x &=\frac{1}{\cosh x}\\ \text{csch }x &=\frac{1}{\sinh x} \end{align}}\]
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 \(e^{x}\pm e^{-x}\approx e^{x}\) 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 \(e^x\pm e^{-x}\approx \pm e^{-x}\), 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 \(A^2-B^2=(A-B)(A+B)\), 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 \(\cosh^2 x\), 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 \(\cosh^2 x-\sinh^2 x=1\) by \(\sinh^2 x\) gives us \[\text{coth}^2 x-\text{csch}^{2}x=1\]
Example 14.15. Prove \(\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y\).
Solution. Let’s simplify the right-hand side \[\begin{align} \sinh x\cosh y+\cosh x\sinh y= & \frac{e^{x}-e^{-x}}{2}\frac{e^{y}+e^{-y}}{2}+\frac{e^{x}+e^{-x}}{2}\frac{e^{y}-e^{-y}}{2}\\[6pt] = & \frac{1}{4}\left[e^{x+y}+\cancel{e^{x-y}}-\bcancel{e^{-x+y}}-e^{-x-y}\right]\\[6pt] & \quad+\frac{1}{4}\left[e^{x+y}-\cancel{e^{x-y}}+\bcancel{e^{-x+y}}-e^{-x-y}\right]\\[6pt] = & \frac{1}{2}\left[e^{x+y}-e^{-(x+y)}\right]\\[6pt] = & \sinh(x+y) \end{align}\]
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 \(\sinh x=\frac{1}{2}(e^x-e^{-x})\), 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 \(\dfrac{d(\cosh x)}{dx}=\sinh x\).
Example 14.16. Show \(\dfrac{d}{dx}\tanh x=\text{sech}^{2}x\).
Solution. Since \(\tanh x=\sinh x/\cosh x\), we can use the Quotient Rule: \[\begin{align} \frac{d}{dx}\tanh x= & \frac{d}{dx}\frac{\sinh x}{\cosh x}\\[6pt] = & \frac{\cosh x-\frac{d(\sinh x)}{dx}-\sinh x\frac{d(\cosh x)}{dx}}{\cos^2 x}\\[6pt] = & \frac{\cosh^2 x-\sinh^2 x}{\cosh^2 x}=\frac{1}{\cosh^{2}x} && {\small(\cosh^{2}x-\sinh^{2}x=1)}\\[6pt] = & \text{ sech}^{2}x \end{align}\]
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 \(\text{arcsinh }x\) 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 \(\text{arccosh }x\) 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 \(\text{arccosh }x\) a single-valued function, we agree that it produces only non-negative values of \(y\). Furthermore, because \(\cosh y\geq 1\), 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
We can find explicit formulas for the inverse hyperbolic functions.
Example 14.17. Show that \(\text{arcsinh } x=\sinh^{-1}x=\ln\left(x+\sqrt{x^{2}+1}\right)\).
Solution. If \(y=\text{ arcsinh }x\), then \[x=\sinh y=\frac{e^{y}-e^{-y}}{2}.\] Multiplying by \(y\) by \(2e^{y}\), we obtain \[2xe^{y}=\left(e^{y}\right)^{2}-1\] or \[\left(e^{y}\right)^{2}-2xe^{y}-1=0\] This is a quadratic equation in terms of \(e^{x}\). Therefore, by the quadratic formula, \[e^{y}=\frac{2x\pm\sqrt{4x^{2}+4}}{2}=x\pm\sqrt{x^{2}+1}\] Since \(e^{y}>0\) and \(\sqrt{x^2+1}\) is greater than \(x\), only the positive sign is acceptable. Hence, \[e^{y}=x+\sqrt{x^{2}+1}.\] Taking the natural logarithm of each side produces \[y=\ln\left(x+\sqrt{x^{2}+1}\right).\] Therefore, \[\sinh^{-1}(x)=\ln\left(x+\sqrt{x^{2}+1}\right).\]
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
Derivatives of the Inverse Hyperbolic Functions
With what we have learned so far, we can find the derivatives of the inverse hyperbolic functions.
Example 14.18. If \(y=\text{ arcsinh } x\) (also denoted by \(\sinh^{-1} x\)), find \(\dfrac{dy}{dx}\).
Solution. If \(y=\text{ arcsinh } x\), then \(x=\sinh y\), and \[\dfrac{dx}{dy}=\cosh y\] By the derivative of an inverse function (see here) \[\begin{align} \frac{dy}{dx}&=\frac{1}{\dfrac{dx}{dy}}=\frac{1}{\cosh y}=\frac{1}{\sqrt{1+\sinh^2 y}}=\frac{1}{\sqrt{1+x^2}}. \end{align}\]
We can also find the derivative of \(\text{arcsinh }x\) by differentiating \(\ln(x+\sqrt{x^2+1})\). Using the chain rule and the fact that \(\dfrac{d\left(\sqrt{x^2+1}\right)}{dx} = \dfrac{x}{\sqrt{x^2+1}}\), we get \[\begin{align} \frac{d}{dx} \ln(x+\sqrt{x^2+1}) &= \frac{1}{x+\sqrt{x^2+1}}\cdot\left(1+\frac{x}{\sqrt{x^2+1}}\right)= \frac{1}{\sqrt{x^2+1}} . \end{align}\] This confirms the result we obtained earlier.
Example 14.19. If \(y=\text{ arccosh }x\) (also denoted by \(\cosh^{-1} x\)), find \(\dfrac{dy}{dx}\).
Solution. If \(y=\text{ arccosh } x\), then \(x=\cosh y\), and \[\frac{dx}{dy}=\sinh y;\] Hence, \[\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}=\frac{1}{\sinh y}=\frac{1}{\sqrt{\cosh^2 y-1}}=\frac{1}{\sqrt{x^2-1}}.\]
Example 14.20. If \(y=\text{ arctanh } x\) (also denoted by \(\tanh^{-1} x\)), find \(\dfrac{dy}{dx}\). Show that \(\displaystyle{\frac{d}{dx}\text{arctanh }x=\frac{d}{dx}\tanh^{-1}x=\frac{1}{1-x^{2}}\quad(-1
Solution. If \(y=\text{ arctanh } x\), then \(x=\tanh y\), and \[\frac{dx}{dy}=\text{ sech}^2 y=1-\tanh^2 y.\] Thus \[\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}=\frac{1}{1-\tanh^2 y}=\frac{1}{1-x^2}.\]
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