10350 Tutorial Week 06 - Set 01
Name
Section
1. A differentiable function g(x) is such that
g 0 (2) = 3,
g(2) = −2,
1a. If C(x) = [g(x)]4
g(3) = 10
?
C 0 (2) =
find
and
g 0 (3) = −4
−96
C 0 (x) = 4 · [g(x)]3 · g 0 (x)
C 0 (2) = 4 · [g(2)]3 · g 0 (2)
= 4 · (−2)3 · 3
= −96
1b. If P (x) = x · eg(x)
?
P 0 (3) =
find
−11e10
P 0 (x) = eg(x) + x · eg(x) · g 0 (x)
P 0 (3) = eg(3) + 3 · eg(3) · g 0 (3)
= e10 + 3 · e10 · (−4)
= −11e10
1c. If Q(x) =
3
(g(x) + 1)4
Q0 (x) = (
find
?
Q0 (2) =
36
3
)0
(g(x) + 1)4
= (3 · (g(x) + 1)−4 )0
= 3 · (−4) · (g(x) + 1)−5 · g 0 (x)
= 3 · (−4) ·
g 0 (x)
(g(x)+1)5
Q0 (2) = 3 · (−4) ·
g 0 (2)
(g(2)+1)5
= −12 ·
3
(−2+1)5
= 36
1
10350 Tutorial Week 06 - Set 02
Name
Section
sin x
2. Assuming the limit lim
= 1, find the values of the following limits showing your steps VERY
x→0 x
CLEARLY.
sin(5x)
=
x→0
7x
2a. lim
lim
x→0
sin(5x)
sin(5x) 1
sin(5x) 5 1
= lim
·
= lim
· ·
x→0
x→0
7x
x
7
5x
1 7
= lim
x→0
5
sin(5x) 5
·
=
5x
7
7
sin(2x)
=
x→0 sin(3x)
2b. lim
sin(2x)
sin(2x)
1
sin(2x)
3x
1
= lim
·
= lim
· 2x ·
·
x→0 sin(3x)
x→0
x→0
1
sin(3x)
2x
sin(3x) 3x
lim
sin(2x)
3x
2x
2
2
·
·
= 1·1· =
x→0
2x
sin(3x) 3x
3
3
= lim
tan(6x)
=
x→0 tan x
2c. lim
tan(6x)
sin(6x) cos x
sin(6x)
1
cos x
= lim
·
= lim
·
·
x→0 tan x
x→0 cos(6x) sin x
x→0
1
sin x cos(6x)
lim
= lim
x→0
sin(6x)
x
1 cos x
sin(6x) 6x
x
cos x
· 6x ·
· ·
= lim
·
·
·
x→0
6x
sin x x cos(6x)
6x
x sin x cos(6x)
= 1·6·1·1 = 6
x2
2d. lim
=
x→0 sin 5x
x2
x
5x
1
= lim x ·
= lim x ·
·
x→0
x→0
sin 5x
sin 5x
sin 5x 5
= 0·1·
1
= 0
5
2
10350 Tutorial Week 06 - Set 03
Name
Section
3. Consider the function

sin(x − 1)


+2

(x − 1)
f (x) =



−1
x 6= 1
x=1
3a. Using limits describe the kind of discontinuity at x = 1.
sin(x − 1)
+2
x→1 (x − 1)
lim f (x) = lim
x→1
z =x−1
=
sin z
+2
z→0 z
lim
=3
So, we have
lim f (x) = 3,
x→1
f (1) = −1
Thus, removable discontinuity.
3b. Is it possible to redefine f (1) so that f (x) is continuous for all x?
Yes, redefine f (1) = 3.
3
10350 Tutorial Week 06 - Set 04
Name
Section
Find the derivatives of the following functions:
4. f (x) = (2x2 + π)4
f 0 (x) = 4(2x2 + π)3 · 4x
= 16x(2x2 + π)3
5. g(x) = ex
2 +2x
g 0 (x) = ex
2 +2x
· (2x + 2)
= (2x + 2)ex
2 +2x
6. h(x) = x cos(2x)
h0 (x) = cos(2x) + x · (− sin(2x)) · 2
= cos(2x) − 2x sin(2x)
7. y =
e2x − 1
.
e2x + 1
y0 =
Simplify the expression you get.
2 · e2x · (e2x + 1) − (e2x − 1) · 2 · e2x
(e2x + 1)2
=
2e4x + 2e2x − 2e4x + 2e2x
(e2x + 1)2
=
4e2x
(e2x + 1)2
4