Higher Engineering Mathematics

Differential calculus

32

Differentiation of hyperbolic functions

32.1 Standard differential coefficients of

hyperbolic functions

From Chapter 5,

d

dx

(sinhx)=

d

dx

(

ex−e−x

2

)

=

[

ex−(−e−x)

2

]

=

(

ex+e−x

2

)

=coshx

If y=sinhax, where ‘a’ is a constant, then

dy

dx

=acoshax

d

dx

( coshx)=

d

dx

(

ex+e−x

2

)

=

[

ex+(−e−x)

2

]

=

(

ex−e−x

2

)

=sinhx

If y=coshax, where ‘a’ is a constant, then

dy

dx

=asinhax

Using the quotient rule of differentiation the deriva-

tives of tanhx, sechx, cosechxand cothxmay be

determined using the above results.

Problem 1. Determine the differential coeffi-

cient of: (a) thx(b) sechx.

(a)

d

dx

(thx)=

d

dx

(

shx

chx

)

=

(chx)(chx)−(shx)(shx)

ch^2 x

using the quotient rule

=

ch^2 x−sh^2 x

ch^2 x

=

1

ch^2 x

=sech^2 x

(b)

d

dx

(sechx)=

d

dx

(

1

chx

)

=

(chx)(0)−(1)(shx)

ch^2 x

=

−shx

ch^2 x

=−

(

1

chx

)(

shx

chx

)

=−sechxthx

Problem 2. Determine

dy

dθ

given

(a)y=cosechθ (b)y=cothθ.

(a)

d

dθ

(cosecθ)=

d

dθ

(

1

shθ

)

=

(shθ)(0)−(1)(chθ)

sh^2 θ

=

−chθ

sh^2 θ

=−

(

1

shθ

)(

chθ

shθ

)

=−cosechθcothθ

(b)

d

dθ

( cothθ)=

d

dθ

(

chθ

shθ

)

=

(shθ)(shθ)−(chθ)(chθ)

sh^2 θ

=

sh^2 θ−ch^2 θ

sh^2 θ

=

−(ch^2 θ−sh^2 θ)

sh^2 θ

=

− 1

sh^2 θ

=−cosech^2 θ

Summary of differential coefficients

yorf(x)

dy

dx

orf′(x)

sinhax acoshax

coshax asinhax

tanhax asech^2 ax

sechax −asechaxtanhax

cosechax −acosechaxcothax

cothax −acosech^2 ax