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Derivatives of Hyperbolic Functions.

To prove that

- (a)
- Let $$\begin{array}{llll}\hfill y& =sinhu,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \Delta y& =sinh\left(u+\Delta u\right)-sinhu\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2cosh\frac{1}{2}\left(2u+\Delta u\right)sinh\frac{1}{2}\Delta u,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{\Delta y}{\Delta u}& =cosh\left(u+\frac{1}{2}\Delta u\right)\frac{sinh\frac{1}{2}\Delta u}{\frac{1}{2}\Delta u}.\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{Takethelimitofbothsides,as}\Delta u\doteq 0\text{,andput}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill lim.& \frac{\Delta y}{\Delta u}=\frac{dy}{du}=\frac{d\left(sinhu\right)}{du},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill lim.& cosh\left(u+\frac{1}{2}\Delta u\right)=coshu,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill lim.& \frac{sinh\frac{1}{2}\Delta u}{\frac{1}{2}\Delta u}=1;\phantom{\rule{2em}{0ex}}& \hfill \text{(seeArt.13)}\\ \multicolumn{3}{c}{\text{then}}\\ \\ \hfill & \frac{d\left(sinhu\right)}{du}=coshu.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$
- (b)
- Similar to (a).
- (c)
- $$\begin{array}{llll}\hfill \frac{d\left(tanhu\right)}{du}& =\frac{d}{du}\cdot \frac{sinhu}{coshu}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{cosh}^{2}u-{sinh}^{2}u}{{cosh}^{2}u}=\frac{1}{{cosh}^{2}u}={sech}^{2}u.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
- (d)
- Similar to (c).
- (e)
$$\frac{d(sechu)}{du}=\frac{d}{du}\cdot \frac{1}{coshu}=-\frac{sinhu}{{cosh}^{2}u}=-sechutanhu.$$ - (f)
- Similar to (e).

It thus appears that the functions $sinhu,coshu$ reproduce themselves in two diﬀerentiations; and, similarly, that the circular functions $sinu,cosu$ produce their opposites in two diﬀerentiations. In this connection it may be noted that the frequent appearance of the hyperbolic (and circular) functions in the solution of physical problems is chieﬂy due to the fact that they answer the question: What function has its second derivative equal to a positive (or negative) constant multiple of the function itself? (See Probs. 28–30.) An answer such as $y=coshmx$ is not, however, to be understood as asserting that $mx$ is an actual sectorial measure and $y$ its characteristic ratio; but only that the relation between the numbers $mx$ and $y$ is the same as the known relation between the measure of a hyperbolic sector and its characteristic ratio; and that the numerical value of $y$ could be found from a table of hyperbolic cosines.

- Prob. 25.
- Show that for circular functions the only modiﬁcations required are in the algebraic signs of (b), (d).
- Prob. 26.
- Show from their derivatives which of the hyperbolic and circular functions diminish as $u$ increases.
- Prob. 27.
- Find the derivative of $tanhu$ independently of the derivatives of $sinhu$, $coshu$.
- Prob. 28.
- Eliminate the constants by diﬀerentiation from the equation
$$y=Acoshmx+Bsinhmx,$$ and prove that $\frac{{d}^{2}y}{d{x}^{2}}={m}^{2}y.$

- Prob. 29.
- Eliminate the constants from the equation
$$y=Acosmx+Bsinmx,$$ - Prob. 30.
- Write down the most general solutions of the diﬀerential equations
$$\frac{{d}^{2}y}{d{x}^{2}}={m}^{2}y,\phantom{\rule{1em}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}=-{m}^{2}y,\phantom{\rule{1em}{0ex}}\frac{{d}^{4}y}{d{x}^{4}}={m}^{4}y.$$

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