Core Mathematics for Modellers
Problem Sheet for Lecture 4 (solutions)
Differentiation of Standard Functions and Composite Functions
1. (a)
(b)
(c)
(d)
(e)
(f)
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
= 2 − 3x2
= 2ax + b − dx−2
= −4 + 8x3
= x−2
= 16x3 − 24x
= x + x−2 /2
2. (a) f 0 (x) = 4e−2x + 2 cos(2x) + 3 sin(3x),
√ 3 x
0
x
(b) f (x) = 4 e −
+ sin x,
2
1
35
1
(c) f 0 (x) = − e−x + 8 + 3x ln 3 − ,
2
x
x
3. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
d2 y
dx2
d2 y
dx2
d2 y
dx2
d2 y
dx2
d2 y
dx2
d2 y
dx2
= −6x
= 2a + 2dx−3
= 24x2
= −2x−3
= 48x2 − 24
= 1 − x−3
f 0 (π/2) = 4e−π − 5.
f 0 (0) = 4.
f 0 (1) = −
1
+ 34 + ln 27.
2e
18(3x − 7)5 ,
3(3 − x)−4 ,
3(5 − x)−2 ,
3
12x2 e4x ,
−6x + 21
,
(x2 − 7x + 2)2
1 −4
1
2
−3 2x + 2
x −
,
3x
3x
1
,
x(ln x)2
1
1 −2/3 1 −3/2
x
− x
.
2
x1/3 + x−1/2 3
4. (a) Differentiating (2) and equating with the right-hand side of (1) we get that r = 1.5/day.
(b) To find T we’re looking for the solution of N (t + T ) = 2N (t). Thus
N0 er(t+T ) = 2N0 ert
erT = 2
ln 2
T =
.
r
Here’s the plot of T (r):
(c) The function now takes the form
N (t) = N0 e(b−d)t = N0 ed(R0 −1)t ,
hence:
• The population will grow if R0 > 1.
• The population will decline if R0 < 1.
• The population will remain constant if R0 = 1.
5. (a) Here’s the plot of CD4 decline over time:
raKe−r(t−t0 ))
dCD4(t)
=
dt
(1 + ae−r(t−t0 ) ))2
dCD4(t)
dCD4(t)
(c)
|t=4 ≈ −62 and
|t=8 ≈ −10.
dt
dt
Hence, the rate of CD4 decline is greater at t = 4.
(b)
6. The definition of a derivative is
dy(x)
y(x + ∆x) − y(x)
= lim
.
∆x→0
dx
∆x
Hence,
d(4x3 )
4(x + ∆x)3 − 4x3
= lim
∆x→0
dx
∆x
4x3 + 12x2 ∆x + 12x∆x2 + 4∆x3 − 4x3
= lim
∆x→0
∆x
= 4 lim 3x2 + 3x∆x + 4∆x2 = 12x2
∆x→0
7. y(x) = sin(x), y 0 (x) = cos(x), y 00 (x) = − sin(x), y 000 (x) == − cos(x), y 0000 (x) = sin(x).
Hence, the 4th, 8th, 12th, ..., 200th derivative of sin x is sin x.
8. (a) (1 − x)−1 = 1 + x + x2 + x3 + ... valid for −1 < x < 1.
(b)
(c)
d
2
3
dx (1 + x + x + x + ...) = 1 + 2x
d
−1 = (1 − x)−2 . Also:
dx (1 − x)
1
=
(1 − x)2
1
1−x
2
+ 3x2 + ... valid for −1 < x < 1.
= (1 + x + x2 + x3 + ...)(1 + x + x2 + x3 + ...) =
1 + x + x2 + x3 + ...
x + x2 + x3 + ...
x2 + x3 + ...
x3 + ...
+...
= 1 + 2x + 3x2 + 4x3 + ...
(d) They’re the same! This suggests that it should be possible to differentiate infinite series termby-term in some cases.