Higher Engineering Mathematics, Sixth Edition

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)