1. Dierentiate
(tan x)1/x .
Let f (x) = (tan x)1/x . Then ln f = x1 ln(tan x) and (ln f )0 = − x12 ln(tan x)+
1
(1 + tan2 x) so
x tan x
Solution:
1
1
1 + tan2 x
1
x
(tan x) x −1 .
f (x) = f (x) (ln f (x)) = − 2 (tan x) ln(tan x) +
x
x
0
0
2. A ball is thrown up. At time t it is at height 5t − 10t2 . When is it at rest?
2
Solution: If the position is y(t) = 5t − 10t then the velocity at time t is v(t) =
5 − 20t and this is zero when t = 14 .
dy
dt
=
3. A freshley brewed cup of coee has temperature 95◦ C is a20◦ C room. After 30 minutes
its temperature is 70◦ C. When it is cooling at the rate of 1◦ C/min?
Solution: We suppose that the temperature follows Newton's law of cooling, that is
that for some constant k, T 0 = k(T − 20) where T is the temperature of the cup in
degrees celsius. Switching to the variable y = T − 20 we have y 0 = T 0 by the sum rule
so that y 0 = ky . It follows that y decays exponentially: y = Cekt . If t = 0 is the time
of the brewing then y(0) = T (0) − 20 = 75 so C = 75. At t = 30 minutes we have
y(30) = 75e30k and also y(30) = T (30) − 20 = 50. It follows that
e30k =
so
k=
2
50
=
75
3
1
2
ln .
30 3
0
Finally, we need to nd t such that T 0 = −1. Since T 0 = y 0 and y 0 = Cekt = Ckekt
we need to nd t such that
Ckekt = −1
or
kt = ln(−
1
) = − ln(−Ck)
Ck
(note that k is negative!). It follows that the time is
ln
ln( 75
ln 32 )
ln(−Ck)
ln(−Ck)
30
t=−
= 30
=
30
= 30
3
3
k
ln 2
ln( 2 )
1
5
2
+ ln ln
ln
3
2
3
2
.