MA1301 Introductory Mathematics
Tutorial 3 Solution
TAN BAN PIN
National University of Singapore
TAN BAN PIN MA1301 Introductory Mathematics 1 / 23
Question 1
Differentiate the following functions with respect to x. Simplify your
answers.
√ 2
(b) y = 2xx2+3x+15 ;
(a) y = tan5 ( x); +x+5
√
(c) y = cos−1 (ln x); (d) y = x sin(ex );
(e) y = csc(ln x) − esec 2x ; (f) y = cot5 √1x ;
1+sin x
(g) y = 1−sin x ; (h) y = e2x −e−2x
.
e2x +e−2x
TAN BAN PIN MA1301 Introductory Mathematics 2 / 23
Question 1(a) Suggested Solution
dy √ d √
= 5 tan4 ( x) tan( x)
dx dx
4 √ √ d√
= 5 tan ( x) sec2 ( x) x
dx
√ √ 1
= 5 tan4 ( x) sec2 ( x) √
2 x
TAN BAN PIN MA1301 Introductory Mathematics 3 / 23
Question 1(b) Suggested Solution
d 2 d
dy + 3x + 15)) · (x2 + x + 5) − (2x2 + 3x + 15) ·
dx ((2x dx (x
2 + x + 5)
=
dx (x2 + x + 5)2
(4x + 3)(x2 + x + 5) − (2x2 + 3x + 15)(2x + 1)
=
(x2 + x + 5)2
−x2 − 10x
= 2
(x + x + 5)2
TAN BAN PIN MA1301 Introductory Mathematics 4 / 23
Question 1(c) Suggested Solution
dy 1 d
= −p ln x
dx 1 − (ln x)2 dx
1 1
= −p
1 − ln2 x x
1
=− p
x 1 − ln2 x
TAN BAN PIN MA1301 Introductory Mathematics 5 / 23
Question 1(d) Suggested Solution
dy d√ √ d
= x · sin(ex ) + x · sin(ex )
dx dx dx
1 √ d
= √ sin(ex ) + x · cos(ex ) (ex )
2 x dx
x
sin(e ) √
= √ + xex cos(ex )
2 x
TAN BAN PIN MA1301 Introductory Mathematics 6 / 23
Question 1(e) Suggested Solution
dy d d
= − csc(ln x) cot(ln x) ln x − esec 2x sec 2x
dx dx dx
1 d
= − csc(ln x) cot(ln x) − esec 2x sec 2x tan 2x 2x
x dx
1
= − csc(ln x) cot(ln x) − 2esec 2x sec 2x tan 2x
x
TAN BAN PIN MA1301 Introductory Mathematics 7 / 23
Question 1(f) Suggested Solution
dy 1 d 1
= 5 cot4 √ cot √
dx x dx x
4 1 2 1 d 1
= 5 cot √ (−1) csc √ √
x x dx x
4 1 2 1 1 −3/2
= 5 cot √ (−1) csc √ − x
x x 2
5 1 1 1
= cot4 √ csc2 √ √
2 x x x x
5 4 1 2 1
= √ cot √ csc √
2x x x x
TAN BAN PIN MA1301 Introductory Mathematics 8 / 23
Question 1(g) Suggested Solution
d d
dy + sin x) · (1 − sin x) − (1 + sin x) ·
dx (1 dx (1 − sin x)
=
dx (1 − sin x)2
cos x(1 − sin x) − (1 + sin x)(− cos x)
=
(1 − sin x)2
2 cos x
=
(1 − sin x)2
TAN BAN PIN MA1301 Introductory Mathematics 9 / 23
Question 1(h) Suggested Solution
dy d
dx (e− e−2x ) · (e2x + e−2x ) − (e2x − e−2x ) · dx
2x d
(e2x + e−2x )
=
dx (e2x + e−2x )2
(2e + 2e )(e + e ) − (e2x − e−2x )(2e2x − 2e−2x )
2x −2x 2x −2x
=
(e2x + e−2x )2
8
= 2x
(e + e−2x )2
TAN BAN PIN MA1301 Introductory Mathematics 10 / 23
Question 2
dy
Using implicit differentiation, find dx in terms of x and y.
(a) x2 + 4xy + y 2 = 20; (b) sin(x2 y) + ex−2y = 7;
(c) ln x + ln y + xy = 3.
TAN BAN PIN MA1301 Introductory Mathematics 11 / 23
Question 2(a) Suggested Solution
Differentiate x2 + 4xy + y 2 = 20 with respect to x:
dy dy
0 = 2x + 4 1 · y + x · + 2y
dx dx
dy
= (2x + 4y) + (4x + 2y) .
dx
Then
dy 2x + 4y x + 2y
=− =− .
dx 4x + 2y 2x + y
TAN BAN PIN MA1301 Introductory Mathematics 12 / 23
Question 2(b) Suggested Solution
Differentiate sin(x2 y) + ex−2y = 7 with respect to x:
d 2 d
0 = cos(x2 y) (x y) + ex−2y (x − 2y)
dx
dx
2 2 dy x−2y dy
= cos(x y) 2x · y + x +e 1−2
dx dx
dy
= (2xy cos(x2 y) + ex−2y ) + (x2 cos(x2 y) − 2ex−2y ) .
dx
Then
dy 2xy cos(x2 y) + ex−2y
=− 2 .
dx x cos(x2 y) − 2ex−2y
TAN BAN PIN MA1301 Introductory Mathematics 13 / 23
Question 2(c) Suggested Solution
Differentiate ln x + ln y + xy = 3 with respect to x:
1 1 dy dy 1 1 dy
0= + +1·y+x· = +y + +x .
x y dx dx x y dx
Then
1 1+xy 1
dy +y y
= − x1 =− x
1+xy =− x
1 =− .
dx y +x y y
x
TAN BAN PIN MA1301 Introductory Mathematics 14 / 23
Question 3
Using logarithmic differentiation, differentiate the following with respect to
x.
(a) y = xln x ; (b) y = (sin x)tan x ;
√
e−2x sin 4x
(c) y = (x2 +2)3
.
TAN BAN PIN MA1301 Introductory Mathematics 15 / 23
Question 3(a) Suggested Solution
Take logarithmic function to y = xln x :
ln y = ln(xln x ) = ln x · ln x = ln2 x.
Differentiate with respect to x:
1 dy d 1
= 2 ln x (ln x) = 2 ln x · .
y dx dx x
Then
dy 2 ln x 2 ln x
=y· = xln x = 2xln x−1 ln x.
dx x x
TAN BAN PIN MA1301 Introductory Mathematics 16 / 23
Question 3(b) Suggested Solution
Take logarithmic function to y = (sin x)tan x :
ln y = ln[(sin x)tan x ] = tan x · ln(sin x).
Differentiate with respect to x:
1 dy 1 d
= sec2 x · ln(sin x) + tan x · (sin x)
y dx sin x dx
1
= sec2 x · ln(sin x) + tan x · cos x = sec2 x · ln(sin x) + 1.
sin x
Then
dy
= y[sec2 x ln(sin x) + 1] = (sin x)tan x [sec2 x ln(sin x) + 1].
dx
TAN BAN PIN MA1301 Introductory Mathematics 17 / 23
Question 3(c) Suggested Solution
Take logarithmic function:
√
e−2x sin 4x 1
ln y = ln = ln(e−2x sin 4x) − 3 ln(x2 + 2)
(x2 + 2)3 2
1
= [ln(e−2x ) + ln(sin 4x)] − 3 ln(x2 + 2)
2
1
= −x + ln(sin 4x) − 3 ln(x2 + 2).
2
Differentiate with respect to x:
1 dy 1 1 d 3 d 2
= −1 + (sin 4x) − 2 (x + 2)
y dx 2 sin 4x dx x + 2 dx
1 1 3
= −1 + 4 cos 4x − 2 · 2x
2 sin 4x x +2
6x
= −1 + 2 cot 4x − 2 .
x +2
TAN BAN PIN MA1301 Introductory Mathematics 18 / 23
Question 3(c) Suggested Solution
Then
dy 6x
= y −1 + 2 cot 4x − 2
dx x +2
√
−2x
e sin 4x 6x
= · −1 + 2 cot 4x − 2 .
(x2 + 2)3 x +2
TAN BAN PIN MA1301 Introductory Mathematics 19 / 23
Question 4
Find the first and second derivatives of y with respect to x. Simplify your
answers.
(a) x = et + 1, y = tet , t is the parameter.
(b) x = sin θ, y = 1 − 2 cos θ, θ is the parameter.
(c) x = t − sin t, y = 1 − cos t, t is the parameter.
TAN BAN PIN MA1301 Introductory Mathematics 20 / 23
Question 4(a) Suggested Solution
dx dy
dt = et and dt = 1 · et + t · et = (1 + t)et . Then
dy dy/dt (1 + t)et
= = = 1 + t,
dx dx/dt et
and
d2 y
1 d dy 1 d
= · = · (1 + t) = e−t · 1 = e−t .
dx2 dx/dt dt dx et dt
TAN BAN PIN MA1301 Introductory Mathematics 21 / 23
Question 4(b) Suggested Solution
dx dy
dθ = cos θ and dθ = −2(− sin θ) = 2 sin θ. Then
dy dy/dθ 2 sin θ
= = = 2 tan θ,
dx dx/dθ cos θ
and
d2 y
1 d dy 1 d
2
= · = · (2 tan θ) = sec θ · 2 sec2 θ = 2 sec3 θ.
dx dx/dθ dθ dx cos θ dθ
TAN BAN PIN MA1301 Introductory Mathematics 22 / 23
Question 4(c) Suggested Solution
dx dy
dt = 1 − cos t and dt = sin t. Then
dy dy/dt sin t
= = ,
dx dx/dt 1 − cos t
and
d2 y
1 d dy 1 d sin t
2
= · = ·
dx dx/dt dt dx 1 − cos t dt 1 − cos t
d d
1 sin t · (1 − cos t) − sin t · dx
dx (1 − cos t)
= · 2
1 − cos t (1 − cos t)
1 cos t · (1 − cos t) − sin t · sin t
= ·
1 − cos t (1 − cos t)2
1 cos t − cos2 t − sin2 t
= ·
1 − cos t (1 − cos t)2
1 cos t − 1 1
= · 2
=− .
1 − cos t (1 − cos t) (1 − cos t)2
TAN BAN PIN MA1301 Introductory Mathematics 23 / 23
