(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