332 Higher Engineering Mathematics
(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
32.2 Further worked problems on
differentiation of hyperbolic
functions
Problem 3. Differentiate the following with
respect tox:
(a)y=4sh2x−
3
7
ch3x
(b)y=5th
x
2
−2coth4x.
(a) y=4sh2x−
3
7
ch3x
dy
dx
= 4 (2cosh2x)−
3
7
(3sinh3x)
=8cosh2x−
9
7
sinh3x
(b) y=5th
x
2
−2coth4x
dy
dx
= 5
(
1
2
sech^2
x
2
)
− 2 (−4cosech^24 x)
=
5
2
sech^2
x
2
+8cosech^24 x
Problem 4. Differentiate the following with
respect to the variable: (a)y=4sin3tch4t
(b)y=ln(sh3θ)−4ch^23 θ.
(a) y=4sin3tch4t(i.e. a product)
dy
dx
=(4sin3t)(4sh4t)+(ch4t)( 4 )(3cos3t)
=16sin3tsh4t+12ch4tcos3t
= 4 (4sin3tsh4t+3cos3tch4t)
(b) y=ln(sh3θ)−4ch^23 θ
(i.e. a function of a function)
dy
dθ
=
(
1
sh3θ
)
(3ch3θ)−( 4 )(2ch3θ)(3sh3θ)
=3coth3θ−24ch3θsh3θ
= 3 (coth3θ−8ch3θsh3θ)
Problem 5. Show that the differential coefficient
of
y=
3 x^2
ch4x
is: 6xsech4x( 1 − 2 xth4x).
y=
3 x^2
ch4x
(i.e. a quotient)
dy
dx
=
(ch 4x)( 6 x)−( 3 x^2 )(4sh4x)
(ch4x)^2
=
6 x(ch 4x− 2 xsh4x)
ch^24 x
= 6 x
[
ch4x
ch^24 x
−
2 xsh4x
ch^24 x
]
= 6 x
[
1
ch4x
− 2 x
(
sh4x
ch4x
)(
1
ch4x
)]
= 6 x[sech 4x− 2 xth4xsech4x]
= 6 xsech4x(1− 2 xth4x)