Math2.org Math Tables: Derivatives of Hyperbolics
|
(Math) |
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
Proof of Given: sinh(x) = ( ex - e-x )/2; cosh(x) = (ex + e-x)/2;
Proof of Given:
Proof of cosh(x) = sinh(x) : From the derivative of ex
( 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)= 1 - tanh2(x) : from the derivatives of sinh(x) and cosh(x)
sinh(x) = cosh(x);
cosh(x) = sinh(x); tanh(x) = sinh(x)/cosh(x); Quotient Rule.
Solve:
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.
csch(x)= -coth(x)csch(x),
sech(x) = -tanh(x)sech(x),
coth(x) = 1 - coth2(x) : From the derivatives of their reciprocal functions
Given:
sinh(x) = cosh(x);
cosh(x) = sinh(x);
tanh(x) = 1 - tanh2(x); csch(x) = 1/sinh(x); sech(x) = 1/cosh(x); coth(x) = 1/tanh(x); Quotient Rule.
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)