How would i go about proving that the derivative of the hyperbolic sine of x is the hyperbolic cosine of x?

Question

Isn't it just like saying the derivative of sin x is cos x.

Answer 1

use the definition of the hyperbolic sine, it is something
with e^x and e^-x and 2 i

differentiate it and compare the result with the definition of the cosh(x)

Answer 2

Actually it is, you can use the hyperbolic identities or you can use their definitions which is

sinh(x) = (e^x - e^(-x)) / 2

Just differentiate this and you easily get

cosh(x) = (e^x + e^(-x)) / 2

The reason we don't do the same thing with cyclical sine and cosine is because their definitions involve complex numbers and you are not supposed to be doing complex calculus.

sin(x) = (e^(ix) - e^(-ix)) / 2i = -i*sinh(ix)

cos(x) = (e^(ix) + e^(-ix)) / 2 = cosh(ix)

Notice how only one of them is being divided by an imaginary number but they are both analogous to hyperbolic functions. Their identities are analogous too.

Answer 3

Use the definitions

sinh x = 1/2 [exp(x) - exp(-x)]
cosh x = 1/2 [exp(x) + exp(-x)]

Derivatives are straightforward [d exp(x) / dx = exp(x)]

d sinh x / dx = 1/2 [exp(x) - (-exp(-x))] = cosh x
d cosh x / dx = 1/2 [exp(x) + (-exp(-x))] = sinh x

Now you see why they are called hyperbolic "sine and cosine" -- they have very similar properties! Note the difference though: with regular sine and cosine, d sin x / dx = cos x, but d cos x / dx = - sin x ... note the negative sign.

Other remarkable similarities include

(cos x)^2 + (sin x)^2 = 1
(cosh x)^2 - (sinh x)^2 = 1

and, using complex numbers,

exp (ix) = cos x + i sin x
exp x = cosh x + sinh x

sin x = 1/2 [exp(ix) - exp(-ix)]
sinh x = 1/2 [exp(x) - exp(-x)]

cos x = 1/2 [exp(ix) + exp(-ix)]
cosh x = 1/2 [exp(x) + exp(-x)]