# 18.01 SPRING 2008 PROBLEM SET 2 SOLUTIONS Graded ```18.01 SPRING 2008
PROBLEM SET 2
SOLUTIONS
See attached photocopies.
1. (a) Let y = xp/q , so by definition y q = xp . Differentiate both sides of this with respect to x, obtaining
qy q−1
dy
= pxp−1 .
dx
Note that we’re allowed to use the power rule for this because both p and q are assumed to be
dy
, this becomes
positive integers. Solving for dx
p
dy
p xp−1
p
p
p xp−1
p xp−1
=
= xp−1−(p− q ) = xp/q−1 .
=
=
p
q−1
p/q
q−1
(q−1)
dx
qy
q (x )
q xq
q
q
(b) Assume n is a positive rational number and y = x−n . Then yxn = 1, and differentiating both
sides with respect to x yields
dy n
x + ynxn−1 = 0.
dx
Here we’ve used the power rule to differentiate xn since n is positive. Solving for
dy
dx ,
we find
yxn−1
x−n xn−1
dy
= −n n = −n
= −nx−n−1 .
dx
x
xn
The upshot is that from now on you can feel free to use the power rule without worrying whether n is
positive, negative, rational or whatever!
d
2. The 4th derivative of sin x is itself, thus the same is true for dx
n sin x whenever n is any multiple of
4. For sin ax we have to use the chain rule, which introduces a new factor of a in front every time
d2
dn
2
n
we differentiate, i.e. dx
2 sin ax = −a sin ax and so forth; in particular, dxn sin ax = a sin ax if n is
divisible by 4. Since 172 is divisible by 4, we have
n
d2 d172
d2 172
d174
sin
ax
=
sin
ax
=
a sin ax = −a174 sin ax.
dx174
dx2 dx172
dx2
1
```