Derivation of Hyperbolic Sine

� 2001 by Karl Hahn

This page contains optional material that you won't need to know until you cover a later section.

So you know a magic property of this hyperbolic sine function involving its relationship to its own derivative, and from that you want to know the more details about this function.

Let's start out just calling it f(u), just as we did in the main text. So f is variable that we are going to solve for, but instead of f being an unknown number that you would solve for in an algebra equation, in our context f is an unknown function. What we will solve is the equation:

              _________
   f'(u)  =  √1 + f²(u)                                           eq. 11.6a-1

and that solution will turn f from being an unknown function to a known one.

Whenever you go looking for a function based upon a relationship that it has with its derivative(s), and that relationship can be described by an equation such as the one above, we say that you are solving a differential equation. That's what 11.6a-1 is.

The first thing to do is to get rid of that nasty square root. To do that you square both sides.

   (f'(u))²  =  1 + f²(u)                                         eq. 11.6a-2

The next step is not so obvious. It is to take the derivative of both sides of the equation. Since both sides are composites, you have to use implicit differentiation on both sides to do this. You get

   2 f'(u) f"(u)  =  2 f(u) f'(u)                                 eq. 11.6a-3

You can divide 2 f'(u) out of both sides to get

   f"(u)  =  f(u)                                                 eq. 11.6a-4

All this last equation is saying is that we are looking for a function whose second derivative is equal to itself.

We already know of a function whose first derivative is equal to itself. That would be our old friend, e^u. And if its first derivative is equal to itself, its second derivative must also be equal to itself as well. Hence f(u) = e^u solves equation 11.6a-4.

But if you take e^u and multiply it by any constant, A, the new function that results still has the property that it is equal to its own first derivative and hence its own second derivative as well. So we find that if A is any constant, the function, f(u) = A e^u also solves equation 11.6a-4.

Now look at f(u) = B e^-u, where B is also any constant. We have

f'(u) = -B e^-u
and
f"(u) = B e^-u = f(u)

So this choice of f(u) also solves equation 11.6a-4. It seems we're coming up with a lot of solutions to that equation. Just one more. Suppose you let

   f(u)  =  A e^u  +  B e^-u                                        eq. 11.6a-5a

Then you have

   f'(u)  =  A e^u  -  B e^-u                                       eq. 11.6a-5a

and

   f"(u)  =  A e^u  +  B e^-u  =  f(u)                              eq. 11.6a-5c

Clearly this also solves equation 11.6a-4 for any constants, A and B. This form is called the general solution to the differential equation, f"(u) = f(u). This behavior, by the way, is typical for solutions to differential equations. Their solutions are always subject to some set of arbitrary constants. Not only that, the number of arbitrary constants that appear in the solution will always be the number of the highest derivative that appears in the differential equation. In equation 11.6a-4, the highest derivative is the second, so we expect two arbitrary constants, A and B, to appear in the general solution.

But equation 11.6a-4 was not what we originally set out to solve. It is merely an equation derived from the one we wanted to solve. The original equation was 11.6a-1, and from it we squared both sides to get equation 11.6a-2. If you substitute the general solution of f"(u) = f(u) into equation 11.6a-2 and square everything out, you get

   A² e^2u  -  2AB  +  B² e^-2u  =  1  +  A² e^2u  +  2AB  +  B² e^-2u   eq. 11.6a-6a

If you take all the cancellations offered by the above equation and simplify, you get

   -4AB  =  1                                                     eq. 11.6a-6b

Hence any combination of A and B that solves this equation can be used to make the general solution solve equation 11.6a-2. I choose to make


         1
   A  =   
         2

and



B  =  -


 1
  
 2

Why? Because the sine function, you will recall, is an odd function, meaning that sin(-x) = -sin(x). Since we are modeling the hyperbolic sine function after the sine function, we would like to make it also an odd function, and these choices for A and B do just that if we define

sinh(u) = A e^u + B e^-u =

1 2

(e^u - e^-u) eq. 11.6a-7

Can you see what other choice we could have used for A and B that would still keep hyperbolic sine an odd function? But when we define hyperbolic cosine as the derivative of hyperbolic sine, and by analogy we want hyperbolic cosine to pass through the point, (0,1), the way cosine does, that other choice for A and B is eliminated. So we are left with only the above definition for hyperbolic sine.

"But what about all those other ways we could have chosen A and B that still solve equation 11.6a-6b?" I hear you asking. Wouldn't they also work to make the substitution function for the integral

dx ~~______~~ eq. 11.6-1 √1 + x²

Indeed they would. You could have chosen

x = f(u) = A e^u -

1 4A

e^-u eq. 11.6a-8

for any A (except A = 0) and still have had a substitutable function for that integral.

"Wait a minute. When we used f(u) = sin(u) as a trig substitution function we didn't have any choices then. Why is this different?" you might ask. It's no different. We did have choices when we used f(u) = sin(u). I just didn't tell you about them then. We could have used x = sin(u + φ) with any constant φ that struck your fancy, and it would have worked on any of the examples we did using the x = sin(u) substitution. It just turns out that to save pencil lead, φ = 0 is what strikes most people's fancy. And choosing A and B the way we do for hyperbolic sine is what strikes our fancy in terms of symmetry and analogy to the sine function. That's why we do it that way.

Return to Main Text

You can email me by clicking this button: