332 Higher Engineering Mathematics
(b)
d
dθ
(coth θ) =
d
dθ
ch θ
sh θ
=
(sh θ)(sh θ) − (ch θ)(ch θ)
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
y or f (x)
dy
dx
or f (x)
sinh ax
a cosh ax
cosh ax
a sinh ax
tanh ax
a sech 2 ax
sech ax
−a sech ax tanh ax
cosech ax −a cosech ax coth ax
coth ax
−a cosech 2 ax
32.2 Further worked problems on
differentiation of hyperbolic
functions
Problem 3. Differentiate the following with
respect to x:
(a) y = 4 sh 2x −
3
7
ch 3x
(b) y = 5 th
x
2
− 2 coth4x.
(a) y = 4 sh2x −
3
7
ch 3x
d y
dx
= 4(2 cosh 2x) −
3
7
(3 sinh3x)
= 8 cosh 2x −
9
7
sinh 3x
(b) y = 5 th
x
2
− 2 coth4x
d y
dx
= 5
1
2
sech
2 x
2
− 2(−4 cosech
2 4x)
=
5
2
sech
2 x
2
+ 8 cosech
2 4x
Problem 4. Differentiate the following with
respect to the variable: (a) y = 4 sin3t ch 4t
(b) y = ln (sh 3θ)− 4 ch 2 3θ.
(a) y = 4 sin3t ch 4t (i.e. a product)
dy
dx
= (4 sin3t )(4 sh 4t ) + (ch 4t )(4)(3 cos 3t )
= 16 sin 3t sh 4t + 12 ch 4t cos 3t
= 4(4 sin 3t sh 4t + 3 cos 3t ch 4t)
(b) y = ln (sh 3θ) − 4 ch
2 3θ
(i.e. a function of a function)
dy
dθ
=
1
sh 3θ
(3 ch 3θ) − (4)(2 ch 3θ)(3 sh 3θ)
= 3 coth 3θ − 24 ch 3θ sh 3θ
= 3(coth 3θ − 8 ch 3θ sh 3θ)
Problem 5. Show that the differential coefficient
of
y =
3x 2
ch 4x
is: 6x sech 4x (1 − 2x th4x).
y =
3x 2
ch 4x
(i.e. a quotient)
dy
dx
=
(ch 4x)(6x) − (3x 2 )(4 sh 4x)
(ch 4x) 2
=
6x(ch 4x − 2x sh 4x)
ch 2 4x
= 6x
ch 4x
ch
2 4x
−
2x sh 4x
ch
2 4x
= 6x
1
ch 4x
− 2x
sh 4x
ch 4x
1
ch 4x
= 6x[sech 4x − 2x th 4x sech 4x]
= 6x sech 4x (1 −2x th 4x)