Article 14
Derivatives of Hyperbolic Functions.

To prove that

(a)d(sinhu) du = coshu, (b)d(coshu) du = sinhu, (c)d(tanhu) du = sech2u, (d)d(sechu) du = sechutanhu, (e)d(cothu) du = csch2u, (f )d(cschu) du = cschucothu, (25)
Let y = sinhu, Δy = sinh u + Δu sinhu = 2cosh 1 2 2u + Δusinh 1 2Δu, Δy Δu = cosh u + 1 2Δu sinh 1 2Δu 1 2Δu .  Take the limit of both sides, as Δu0, and put lim. Δy Δu = dy du = d sinhu du , lim. cosh u + 1 2Δu = coshu, lim. sinh 1 2Δu 1 2Δu = 1;  (see Art. 13)  then  d sinhu du = coshu.
Similar to (a).
d tanhu du = d du sinhu coshu = cosh2u sinh2u cosh2u = 1 cosh2u = sech2u.
Similar to (c).
d(sechu) du = d du 1 coshu = sinhu cosh2u = sechutanhu.
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 tanh u independently of the derivatives of sinh u, cosh u.
Prob. 28.
Eliminate the constants by differentiation from the equation
y = Acosh mx + Bsinh mx,

and prove that d2y dx2 = m2y.

Prob. 29.
Eliminate the constants from the equation
y = Acos mx + Bsin mx,

and prove that d2y dx2 = m2y.

Prob. 30.
Write down the most general solutions of the differential equations
d2y dx2 = m2y,d2y dx2 = m2y,d4y dx4 = m4y.