Understanding Engineering Mathematics

(iii)

d^2

dx^2

(e−xcosx)=

d

dx

(−e−xcosx−e−xsinx)

=−

d

dx

(e−x(cosx+sinx))

=−[−e−x(cosx+sinx)+e−x(−sinx+cosx)]

=−e−x(−2sinx)

= 2 e−xsinx

(iv) This one requires a bit of thought if you want to avoid a mess. It

is easiest to use partial fractions in fact. Thus, to remind you of

partial fractions (62

➤

), you can check that

x+ 1

(x− 1 )(x+ 2 )

≡

2

3 (x− 1 )

+

1

3 (x+ 2 )

Then

d^2

dx^2

[

x+ 1

(x− 1 )(x+ 2 )

]

=

d^2

dx^2

[

2

3 (x− 1 )

+

1

3 (x+ 2 )

]

d

dx

[

−

2

3 (x− 1 )^2

−

1

3 (x+ 2 )^2

]

=

4

3 (x− 1 )^3

+

2

3 (x+ 2 )^3

B. This sort of example brings together a lot of what we have already
done. First, note that

d^2 y

dx^2

    =

d^2 y/dt^2

d^2 x/dt^2

We must be more subtle. Start with

d^2 y

dx^2

=

d

dx

(

dy

dx

)

Now with parametric differentiation

dy

dx

will be a function oft–in

this case, withx=t^2 +1,y=t−1wehave

dy

dx

=

dy/dt

dx/dt

=

1

2 t

If we were to differentiate now with respect toxwe would have to

express

1

2 t

in terms ofx– possible but messy. Instead, we use the