OKAN ÜNİVERSİTESİ
MÜHENDİSLİK FAKÜLTESİ
MÜHENDİSLİK TEMEL BİLİMLERİ BÖLÜMÜ
2016–17 Spring
MATH115 Basic Mathematics – Homework 3 Solutions
11. (a) We calculate that
N. Course
ds
8
d
2
=
−2t−1 + 4t−2 = 2t−2 − 8t−3 = 2 − 3 .
dt
dt
t
t
(b) Note first that
w = (z + 1)(z − 1)(z 2 + 1) = (z 2 − 1)(z 2 + 1) = z 4 − 1.
Therefore w0 =
d
4
dz (z
− 1) = 4z 3 and w00 =
d
3
dz 4z
= 12z 2 .
(c) By the product rule
d
dy
=
(2x + 3)ex = (2x + 3)0 ex + (2x + 3)(ex )0 = 2ex + (2x + 3)ex = (2x + 5)ex .
dx
dx
d sin x + cos x
ds
d d
(sin x + cos x) sec x =
=
=
(tan x + 1) = sec2 x.
12. (a)
dx
dx
dx
cos x
dx
dr
d
θ sin θ + cos θ = sin θ + θ cos θ − sin θ = θ cos θ.
(b) By the product rule
=
dθ
dθ
13. By the quotient rule,
db
u0 v − uv 0
(x2 − 1)0 (x2 + x − 2) − (x2 − 1)(x2 + x − 2)0
=
=
dx
v2
(x2 + x − 2)2
2
2
2x(x + x − 2) − (x − 1)(2x + 1)
=
(x2 + x − 2)2
3
2
(2x + 2x − 4x) − (2x3 + x2 − 2x − 1)
=
(x2 + x − 2)2
2
x − 2x + 1
(x − 1)2
1
= 2
=
=
.
(x + x − 2)2
(x − 1)2 (x + 2)2
(x + 2)2
14. (a) By the Chain Rule,
ds
d
=
dt
dt
(b) First note that
we find that
d
dt
−10
−11
−11
t
t
d t
t
−1
= −10
−1
− 1 = −5
−1
.
2
2
dt 2
2
5 sin
t
3
=
5
3
cos
dy
d
d
=
(cos u)
dt
du
dt
t
3
by the Chain Rule. Using the Chain Rule a second time,
t
5
t
t
5 sin
= − sin 5 sin
cos
.
3
3
3
3
15. (a) First we calculate that
d p
d √ d
1
2 − 2x
1−x
g 0 (x) =
2x − x2 =
u (2x − x2 ) = √ (2 − 2x) = √
=√
.
2
dx
du
dx
2 2
2 2x − x
2x − x2
We can see that g 0 exists on (0, 23 ) and we can see that g 0 (x) = 0 if and only if x = 1.
Therefore x = 1 is the only critical point of g.
√
√
(b) We calculate g(0) = 0 − 0 = 0, g(1) = 2 − 1 = 1 and g
3
2
=
q
3−
9
4
=
q
3
4
≈ 0.866.
Therefore the absolute maximum value of g is 1 which occurs at x = 1; and the absolute minimum
value of g is 0 which occurs at x = 0.
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