MATH 100 V1A
October 17th – Practice problems
Hints and Solutions
1. In an ideal environment, a cell culture grows at a rate proportional to the number of
cells present. Suppose a culture has 500 cells initially and 900 cells after 24 hours.
How many cells will there be after an additional 10 hours?
Solution: Let P (t) denote the number of cells in the culture at time t. We are told that
the cell culture grows at a rate proportional to the number of cells present, i.e.
dP
= kP,
dt
for some constant k (to be determined). From class, we know that the solutions to
this differential equation are of the form P (t) = Aekt , for some other constant A (also
to be determined). We are also told that the cell culture initially has 500 cells, so
500 = P (0) = Ae0 = A. Therefore our equation simplifies to P (t) = 500ekt . Since
there are 900 cells after 24 hours, we must have that 900 = P (24) = e24k . Solving this
log( 9 )
for k gives us that k = 245 . So the equation that describes the number of cells in
the culture as a function of time is given by
log
P (t) = 500e
9
t
5 .
Now we can use this to solve
the problem: at t = 34 hours, the number of cells in the
log
culture will be P (34) = e
9
·34
5
≈ 1150.
2. Make a conjecture for the nth derivative of f (x) = xeax . How might you prove this
conjecture?
Hint: The first three derivatives are given by:
f 0 (x) = eax + xaeax = eax + af (x)
f 00 (x) = aeax + af 0 (x) = aeax + a(eax + xaeax ) = 2aeax + a2 f (x)
f 000 (x) = 2a2 eax + a2 f 0 (x) = 2a2 eax + a2 (eax + xaeax ) = 3a2 eax + a3 f (x).
Can you see a pattern?
Note: It is not necessary to write the derivatives in terms of f (x) – I just find that it
makes finding the second and third derivatives easier.
3. Differentiate the following functions:
(a) f (x) = e−x sin(x).
(b) f (x) =
ex
.
ex +e3x
(c) f (x) = log(2x) cos(x).
(d) f (x) = log(log(log(x))).
Hint: The answers are:
(a) f 0 (x) = e−x (cos(x) − sin(x)).
2x
2e
(b) (Simplify the equation before differentiating) f 0 (x) = − (1+e
2x )2 .
(c) f 0 (x) =
cos(x)
x
(d) f 0 (x) =
1
.
x log(x) log(log(x))
− log(2x) sin(x).
4. Using properties of the function ex , prove the following identities:
(a) log(ab) = log(a) + log(b) for a, b > 0.
(b) log(ax ) = x log(a) for a > 0.
Solution to (a): We want to use the property that ex+y = ex ey , and that ex and log(x)
are inverse functions, i.e. elog(x) = x for all x in the domain of the logarithm function
and log(ex ) = x for all x. Note that this tells us that a = elog(a) and b = elog(b) for
a, b > 0. So,
log(ab) = log elog(a) elog(b) = log elog(a)+log(b) (using the first property)
= log(a) + log(b). (using the fact that they’re inverse functions)
Proving identity (b) is similar.
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