# Math2.org Math Tables: Derivatives of Hyperbolics

(Math)

## Proofs of Derivatives of Hyperbolics

Proof of sinh(x) = cosh(x) : From the derivative of ex

Given: sinh(x) = ( ex - e-x )/2; cosh(x) = (ex + e-x)/2; ( f(x)+g(x) ) = f(x) + g(x); Chain Rule; ( c*f(x) ) = c f(x).
Solve:

sinh(x)= ( ex- e-x )/2 = 1/2 (ex) -1/2 (e-x)
= 1/2 ex + 1/2 e-x = ( ex + e-x )/2 = cosh(x)       Q.E.D

Given: sinh(x) = ( ex - e-x )/2; cosh(x) = (ex + e-x)/2; ( f(x)+g(x) ) = f(x) + g(x); Chain Rule; ( c*f(x) ) = c f(x).
Solve:

cosh(x)= ( ex + e-x)/2 = 1/2 (ex) + 1/2 (e-x)
= 1/2 ex - 1/2 e-x = ( ex - e-x )/2 = sinh(x)       Q.E.D.

tanh(x)= sinh(x)/cosh(x)
= ( cosh(x) sinh(x) - sinh(x) cosh(x) ) / cosh2(x)
= ( cosh(x) cosh(x) - sinh(x) sinh(x) ) / cosh2(x) = 1 - tanh2(x)       Q.E.D.

Proof of csch(x)= -coth(x)csch(x), sech(x) = -tanh(x)sech(x), coth(x) = 1 - coth2(x) : From the derivatives of their reciprocal functions

csch(x)= 1/sinh(x)= ( sinh(x) 1 - 1 sinh(x))/sinh2(x) = -cosh(x)/sinh2(x) = -coth(x)csch(x)
sech(x)= 1/cosh(x)= ( cosh(x) 1 - 1 cosh(x))/cosh2(x) = -sinh(x)/cosh2(x) = -tanh(x)sech(x)
coth(x)= 1/tanh(x)= ( tanh(x) 1 - 1 tanh(x))/tanh2(x) = (tanh2(x) - 1)/tanh2(x) = 1 - coth2(x)