1
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.1 Differentiate the following functions from first principles (use the rules of differentiation to check
your answer):
(a)
(d)
y = x2 + x
1
y= 2
x
(b)
(e)
y = x3 + 1
x+1
y=
x
(c)
(f)
1
2x − 1
√
y = 9−x
y=
Outline Solution
(a)
y = x2 + x
y + ∆y = (x + ∆x)2 + x + ∆x
∆y = (x + ∆x)2 + (x + ∆x) − (x2 + x)
= x2 + 2x∆x + (∆x)2 + ∆x − x2
= 2x∆x + (∆x)2 + ∆x
∆y
∆x
∆y
lim
∆x→0 ∆x
= 2x + ∆x + 1
=
dy
= x2 + 1
dx
(b) From first principles:
f (x) = x3 + 1
f (x + h) = (x + h)3 + 1
f (x + h) − f (x) = (x + h)3 − x3
= x3 + 3x2 h + 3xh2 + h3 − x3
2
2
3
= 3x h + 3xh + h
f (x + h) − f (x)
lim
= lim 3x2 + 3xh + h2
h→0
h→0
h
= 3x2
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
(c) From first principles:
f (x) =
f (x + h) =
f (x + h) − f (x) =
=
=
=
lim
h→0
f (x + h) − f (x)
h
=
=
1
2x − 1
1
2(x + h) − 1
1
1
−
2(x + h) − 1 2x − 1
(2x − 1) − [2(x + h) − 1]
[2(x + h) − 1] (2x − 1)
2x − 1 − 2x − 2h + 1
[2(x + h) − 1] (2x − 1)
−2h
[2(x + h) − 1] (2x − 1)
−2
lim
h→0 [2(x + h) − 1] (2x − 1)
−2
(2x − 1)2
(d)
y =
y + ∆y =
∆y =
=
=
=
∆y
∆x
∆y
lim
∆x→0 ∆x
=
=
1
x2
1
(x + ∆x)2
1
1
− 2
2
(x + ∆x)
x
2
x − (x + ∆x)2
x2 (x + ∆x)2
2
x − x2 + 2x∆x + (∆x)2
x2 (x + ∆x)2
−2x∆x
−2∆x
=
2
2
x (x + ∆x)
x(x + ∆x)2
−2
x(x + ∆x)2
dy
−2
= 3
dx
x
2
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
(e) From first principles:
f (x) =
f (x + h) =
f (x + h) − f (x) =
=
=
lim
h→0
f (x + h) − f (x)
h
=
=
x+1
x
x+h+1
x+h
x+h+1 x+1
−
x+h
x
x(x + h + 1) − (x + 1)(x + h)
(x + h) x
2
x + xh + x − x2 − x − xh − h
(x + h) x
−1
lim
h→0 (x + h) x
−1
x2
(f) From first principles:
f (x) =
f (x + h) =
f (x + h) − f (x) =
=
=
=
f (x + h) − f (x)
lim
h→0
h
=
=
√
9−x
p
9 − (x + h)
√
√
9−x−h− 9−x
h√
i √9 − x − h + √9 − x √
√
9−x−h− 9−x · √
9−x−h+ 9−x
(9 − x − h) − (9 − x)
√
√
9−x−h+ 9−x
−h
√
√
9−x−h+ 9−x
−1
√
lim √
h→0
9−x−h+ 9−x
−1
√
2 9−x
3
4
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.2 Differentiate the following functions with respect to x.
(a)
y = x4 + 10x3 + 4
(b)
(c)
y = 19x4 + 45x2 − 23x + 1
1
x + ln x −
x
(d)
√ 3
1
y = x0.1 + √ + 4 x
x
y = 35x5 + 4x2 − 29
(f)
y = sin x + cos x + tan x
(e)
Outline Solution
(a)
y = x4 + 10x3 + 4
dy
= 4x3 + 30x2
dx
(b)
√ 3
1
y = x0.1 + √ + 4 x
x
1
dy
dx
3
= x0.1 + x− 2 + x 4
1 3 3 1
= 0.1x−0.9 − x− 2 + x− 4
2
4
(c)
f (x) = 19x4 + 45x2 − 23x + 1 =⇒ f 0 (x) = 76x3 + 90x − 23
(d)
f (x) = 35x5 + 4x2 − 29 =⇒ f 0 (x) = 175x4 + 8x
(e)
y = x + ln x −
1
x
⇒
dy
1
1
1+ + 2
dx
x x
(f)
y = sin x + cos x + tan x
⇒
dy
= cos x − sin x + sec2 x
dx
5
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.3 Differentiate the following functions:
(a)
y = x2 − 5x (x + 4)
(b)
(d)
y = ex [cos(x) + kx]
(e)
1
2
4
y=x
2 + 3x+ x
1
3
y=
− 4
z + 5z 3
2
z
z
(c)
y = x3 + 2x ex
(f)
y = sin(x) ln(5x)
Outline Solution
(a) Using the product rule for y = x2 − 5x (x + 4):
du
= 2x − 5
v =x+4
dx
dv
du
= u
+v
= x2 − 5x · 1 + (x + 4) · (2x − 5)
dx
dx
= x2 − 5x + 2x2 + 3x − 20
x2 − 5x
u =
dy
dx
⇒
⇒
dv
=1
dx
= 3x2 − 2x − 20
1
(b) Using the product rule for y = x 4 2 + 3x + x2 :
du
1 3
= x− 4
v = 2 + 3x + x2 ⇒
dx
4
1 3
1
dy
dv
du
= u
+v
= 2 + 3x + x2 · x− 4 + x 4 · (3 + 2x)
dx
dx
dx
4
1
1 −3
2
=
x 4 2 + 3x + x + x 4 (3 + 2x)
4
(c) Using the product rule for y = x3 + 2x ex :
1
u = x4
⇒
du
= 3x2 + 1
v = ex
dx
dv
du
= u
+v
= x3 + 2x · ex + ex · 3x2 + 2
dx
dx
= ex x3 + 3x2 + 2x + 2
u = x3 + 2x
dy
dx
⇒
⇒
dv
= 3 + 2x
dx
dv
= ex
dx
(d) Using the product rule for y = ex [cos(x) + kx]:
dv
du
= ex
v = cos(x) + kx ⇒
= − sin(x) + k
dx
dx
dy
dv
du
= u
+v
= ex [− sin(x) + k] + [cos(x) + kx] · ex
dx
dx
dx
= ex [cos(x) − sin(x) + kx + k]
1
3
z + 5z 3 :
(e) Using the product rule for y =
− 4
2
z
z
u = ex
⇒
1
3
du
2
12
−
= z −2 − 3z −4 ⇒
= −2z −3 + 12z −5 = − 3 + 5
z2 z4
dz
z
z
dv
v = z + 5z 3 ⇒
= 1 + 15z 2
dz
dy
dv
du
1
3
2
12
2
3
= u
+v
=
−
· 1 + 15z + z + 5z · − 3 + 5
dz
dz
dz
z2 z4
z
z
14
9
= 5+ 2 + 4
z
z
u =
6
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
(f) Using the product rule for y = sin(x) ln(5x):
du
= cos(x)
v = ln(5x)
dx
dv
du
1
= u
+v
= sin(x) · + ln(5x) · cos(x)
dx
dx
x
1
=
sin(x) + cos(x) ln(5x)
x
u = sin(x)
dy
dx
⇒
⇒
dv
5
1
=
=
dx
5x
x
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MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.4 Differentiate the following functions with respect to x.
(a)
(d)
1+x
2+x
ez − e−z
y= z
e + e−z
y=
(b)
(e)
x2 + 1
x
1
1
x 3 + x− 3
y=
tan 2x
y=
(c)
(f)
ln x
cos x
1 + sin x
y=
1 + cos x
y=
Outline Solution
(a) Using the quotient rule for y =
u = 1+x
dy
dx
⇒
dv
− u dx
v2
1
(2 + x)2
v
=
=
du
dx
(b) Using the quotient rule for y =
u = x2 + 1
dy
dx
du
dx
(c) Using the quotient rule for y =
=
=
=
⇒
dv
=1
dx
⇒
dv
=1
dx
ln x
:
cos x
v = cos x
1
x
⇒
dv
= − sin x
dx
− ln x(− sin x)
cos2 x
ez − e−z
:
ez + e−z
du
dv
= ez + e−z
v = ez + e−z ⇒
= ez − e−z
dx
dx
dv
v du
(ez + e−z ) · (ez + e−z ) − (ez − e−z ) · (ez − e−z )
dx − u dx
=
v2
(ez + e−z )2
e2z + 2 + e−2z − e2z − 2 + e−2z
u = ez − e−z
=
du
= 2x
v=x
dx
x · 2x − x2 + 1 · 1
=
x2
1
du
=
dx
x
du
dv
cos x
v dx − u dx
=
v2
cos x + x ln x sin x
x cos2 x
(d) Using the quotient rule for y =
dy
dx
⇒
⇒
u = ln x
=
du
=1
v =2+x
dx
(2 + x) · 1 − (1 + x) · 1
=
(2 + x)2
x2 + 1
:
x
dv
− u dx
v2
2
x −1
x2
v
=
=
dy
dx
1+x
:
2+x
⇒
(ez + e−z )2
(ez
4
+ e−z )2
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MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
1
(e) Using the quotient rule for y =
1
1
u = x 3 + x− 3
dy
dx
v
=
du
dx
⇒
dv
− u dx
=
v2
tan 2x x
− 23
1 2 1 4
du
= x− 3 − x− 3
v = tan(2x) ⇒
dx h 3
3 i 2
4
1
1
tan 2x 13 x− 3 − 31 x− 3 − x 3 + x− 3 2 sec2 2x
− 34
−x
=
1
x 3 + x− 3
:
tan 2x
dv
= 2 sec2 (2x)
dx
2
tan
2x
1
− 6 x + x− 3 sec2 2x
1
3
3 tan2 2x
(f) Using the quotient rule for y =
du
= cos x
v = 1 + cos x ⇒
dx
dv
v du
(1 + cos x)(cos x) − (1 + sin x)(− sin x)
dx − u dx
=
2
v
(1 + cos x)2
cos x + cos2 x + sin x + sin2 x
(1 + cos x)2
cos x + sin x + 1
(1 + cos x)2
u = 1 + sin x
dy
dx
=
=
=
1 + sin x
:
1 + cos x
⇒
dv
= − sin x
dx
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MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.5 Differentiate the following functions with respect to x.
(a)
(d)
7
5
y = 4x + x − 1
r
1−x
y=
1+x
9
1 −2
5
y= x −
x
√ y = ln tan x
(b)
(e)
(c)
y = ex
(f)
y=
2
cos(x)
ex
ln [sin(x)]
Outline Solution
9
(a) Using the Chain Rule for y = 4x7 + x5 − 1 :
du
= 28x6 + 5x4
dx
dy du
=
·
= 28x3 + 5x4 · 9u8
du dx
9
= 9 x4 28x2 + 5 4x7 + x5 − 1
u = 4x7 + x5 − 1
dy
dx
⇒
(b) Using the Chain Rule for y =
1
du
1
⇒
= 5x4 + 2
x
x
dx
dy du
1
=
·
= 5x4 + 2 · −2u−3
du dx
x
1
1 −3
4
5
= −2 5x + 2
x −
x
x
2
(c) Using the Chain Rule for y = ex
cos(x)
⇒
dy
= 9u8
du
y = u−2
⇒
dy
= −2u−3
du
:
du
= 2x cos(x) − x2 sin(x)
dx
dy
dy du
=
·
= eu · 2x cos(x) − x2 sin(x)
dx
du dx
2
= ex cos(x) 2x cos(x) − x2 sin(x)
r
1−x
:
(d) Using the Chain Rule and Quotient Rule
1+x
u = x2 cos(x)
⇒
1 −2
:
x5 −
x
u = x5 −
dy
dx
y = u9
y = eu
⇒
1−x
dw
(1 + x)(−1) − (1 − x)(+1)
−2
⇒
=
=
1+x
dx
(1 + x)2
(1 + x)2
√
dy
1
y =
w ⇒
= √
dw
2 w
dy
dy dw
1
−2
=
·
= √ ·
dx
dw dx
2 w (1 + x)2
r
1 1+x
−2
=
·
2 1 − x (1 + x)2
−1
= p
(1 − x)(1 + x)3
w =
dy
= eu
du
10
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
√ (e) Using the Chain Rule for y = ln tan x :
√
du
1
v = tan u
= √
dx
2 x
1
dy
=
y = ln v ⇒
dv v 1
dy
dy dv du
1
2
√
· sec u ·
=
·
·
=
dx
dv du dx
v
2 x
√
sec2 ( x)
√
√
=
2 x tan ( x)
u =
⇒
x
(f) We will use the Chain Rule and Quotient Rule for y =
⇒
dv
= sec2 u
du
ex
. Firstly, the Chain Rule
ln [sin(x)]
for v = ln [sin(x)]:
dw
= cos(x)
du
dv dw
1
=
·
=
· cos(x)
dw dx
w
cos x
=
sin x
= cot(x)
w = sin(x)
dv
dx
⇒
v = ln(w)
⇒
dv
1
=
dw
w
Secondly, using the Quotient Rule:
du
= ex
v = ln [sin(x)]
dx
dv
v du
ln {sin(x)} ex − ex cot(x)
dx − u dx
=
v2
[ln {sin(x)}]2
ex [ln {sin(x)} − cot(x)]
[ln {sin(x)}]2
u = ex
dy
dx
=
=
⇒
⇒
dv
= cot(x)
dx
11
MS121: IT Mathematics — Semester 2 — Tutorial Sheet 3
3.6 Find the tangent to the curve y = f (x) at the given values of x:
(a)
f (x) = x2 + 3x, x = 2
(b)
(c)
f (x) = x2 − x, x = −4
5
f (x) = x4 − 2 , x = 2
x
(d)
(e)
(f)
f (x) = x3 + 5, x = −1
1
f (x) = x3 +
, x = −1
3x
1
f (x) = x + , x = 4
x
Outline Solution
(a) f (x) = x2 + 3x, x = 2
f 0 (x) = 2x + 3, f 0 (2) = 7, f (2) = 10.
The point slope equation of the line through (2, 10) with slope 7 is
y = 10 + 7(x − 2) = 7x − 4.
(b) f (x) = x3 + 5, x = −1,
f 0 (x) = 3x2 , f 0 (−1) = 3, f (−1) = 4.
The point slope equation of the line through (−1, 4) with slope 3 is
y = 4 + 3(x + 1) = 3x + 7
(c) f (x) = x2 − x, x = 4.
f 0 (x) = 2x − 1, f 0 (4) = 7, f (4) = 12.
The point slope equation of the line through (4, 12) with slope 7 is
y = 12 + 7(x − 4) = 7x − 16
(d) f (x) = x3 +
1
, x = −1
3x
8
4
1
, f 0 (−1) = , f (1) = − .
3x2
3
3
The point slope equation of the line through (−1, −4/3) with slope 8/3 is
4 8
8
4
y = − + (x + 1) = x +
3 3
3
3
5
(e) f (x) = x4 − 2 , x = 2,
x
10
133
59
f 0 (x) = 4x3 + 3 , f 0 (2) =
, f (2) = .
x
4
4
The point slope equation of the line through (2, 59/4) with slope 133/4 is
59 133
133
207
y=
+
(x − 2) =
x−
4
4
4
4
1
(f) f (x) = x + , x = 4.
x
1
15
17
f 0 (x) = 1 − 2 , f 0 (4) = , f (4) = .
x
16
4
The point slope equation of the line through (4, 17/4) with slope 15/16 is
17 15
15
1
y=
+ (x − 4) = x +
4
16
16
2
f 0 (x) = 3x2 −
JC
MS121: IT Mathematics
Semester 2 Tutorial Sheet 3
15/03/15