MATH 104-184 Quiz #3
October 23
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Differentiate y = cos(ex ).
Answer: y 0 = − sin ex · ex
(b) The cost function is given by C(q) = 2q 2 + 7q + 330. Estimate the cost of producing the
10th unit.
Answer: $43
Solution: C 0 (q) = 4q + 7 ⇒ C 0 (9) = 36 + 7 = 43
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find the slope of the tangent line to 3x + sin(xy) + y 3 = 9 at the point (0, 2)
Answer: −
2 + ln 3
12
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
We differentiate implicitly to get
3x ln 3 + cos(xy) · [y + xy 0 ] + 3y 2 y 0 = 0
Subbing in x = 0 and y = 2 yields
ln 3 + 1 · [2 + 0] + 3(4)y 0 = 0
3
from which we see that y 0 = − 2+ln
12
√
(b) Find the derivative of the function g(x) = sin3 ( 3 x).
Answer:
g 0 (x) = 3 sin2 (x1/3 ) · cos(x1/3 ) · 31 x−2/3
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
3
g(x) = sin(x1/3 )
d sin(x1/3 )
dx
d 1/3 (x )
g 0 (x) = 3 sin2 (x1/3 ) · cos(x1/3 ) ·
dx
1
g 0 (x) = 3 sin2 (x1/3 ) · cos(x1/3 ) · x−2/3
3
g 0 (x) = 3 sin2 (x1/3 ) ·
Long answer question — you must show your work
3. 4 marks Determine the point(s) where the tangent line to
h(x) = x3/x , x > 0
is horizontal.
Answer: (e, e3/e )
Solution: Marking scheme: 1pt for taking the log of both sides; 1pt differentiating
properly;1pt for setting h’=0; 1pt for getting x = e; don’t worry if they didn’t find the
point (but do make a note of it).
Use logarithmic differentiation:
h(x) = x3/x
ln h = ln x3/x
3
ln h = · ln x
x
1 0 −3
3 1
· h = 2 · ln x + ·
h
x x x
−3
ln
x
3
h0 = h ·
+
x2
x2
1 − ln x
h0 = 3h ·
x2
From here, we note that h 6= 0, since x > 0. This means that h0 (x) = 0 if and only if
ln x = 1. Therefore, x = e is the only solution. Hence the point is (e, e3/e ).
MATH 104-184 Quiz #3
October 23
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Differentiate y = sin(ex ).
Answer: y 0 = cos ex · ex
(b) The cost function is given by C(q) = 4q 2 + 8q + 330. Estimate the cost of producing the
7th unit.
Answer: $56
Solution: C 0 (q) = 8q + 8 ⇒ C 0 (6) = 48 + 8 = 56
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find the slope of the tangent line to 2x + sin(xy) + y 3 = 9 at the point (0, 2)
Answer: −
2 + ln 3
12
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
We differentiate implicitly to get
2x ln 2 + cos(xy) · [y + xy 0 ] + 3y 2 y 0 = 0
Subbing in x = 0 and y = 2 yields
ln 2 + 1 · [2 + 0] + 3(4)y 0 = 0
2
from which we see that y 0 = − 2+ln
12
√
(b) Find the derivative of the function g(x) = cos3 ( 3 x).
Answer:
g’(x)=3 cos2 (x1/3 )·(− sin(x1/3 ))· 13 x−2/3
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
3
g(x) = cos(x1/3 )
g 0 (x) = 3 cos2 (x1/3 ) ·
d cos(x1/3 )
dx
d 1/3 (x )
dx
1
g 0 (x) = 3 cos2 (x1/3 ) · (− sin(x1/3 )) · x−2/3
3
g 0 (x) = 3 cos2 (x1/3 ) · (− sin(x1/3 )) ·
Long answer question — you must show your work
3. 4 marks Determine the point(s) where the tangent line to
h(x) = x2/x , x > 0
is horizontal.
Answer: (e, e3/e )
Solution: Marking scheme: 1pt for taking the log of both sides; 1pt differentiating
properly;1pt for setting h’=0; 1pt for getting x = e; don’t worry if they didn’t find the
point (but do make a note of it).
Use logarithmic differentiation:
h(x) = x2/x
ln h = ln x2/x
2
ln h = · ln x
x
1 0 −2
2 1
· h = 2 · ln x + ·
h
x x x
−2
ln
x
2
h0 = h ·
+
x2
x2
1 − ln x
h0 = 2h ·
x2
From here, we note that h 6= 0, since x > 0. This means that h0 (x) = 0 if and only if
ln x = 1. Therefore, x = e is the only solution. Hence the point is (e, e2/e ).
MATH 104-184 Quiz #3
October 23
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Differentiate y = esin x .
Answer: y 0 = esin x · cos x
(b) The cost function is given by C(q) = 5q 3 + 80q + 330. Estimate the cost of producing the
11th unit.
Answer: $1580
Solution: C 0 (q) = 15q 2 + 80 ⇒ C 0 (10) = 1500 + 80 = 1580
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find the slope of the tangent line to 4x + sin(xy) + y 3 = 9 at the point (0, 2)
Answer: −
2 + ln 4
12
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
We differentiate implicitly to get
4x ln 4 + cos(xy) · [y + xy 0 ] + 3y 2 y 0 = 0
Subbing in x = 0 and y = 2 yields
ln 4 + 1 · [2 + 0] + 3(4)y 0 = 0
4
from which we see that y 0 = − 2+ln
12
√
(b) Find the derivative of the function g(x) = cos4 ( 4 x).
Answer:
4 cos3 (x1/4 ) · (− sin(x1/4 )) · 14 x−3/4
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
4
g(x) = cos(x1/4 )
g 0 (x) = 4 cos3 (x1/4 ) ·
d cos(x1/4 )
dx
d 1/4 (x )
dx
1
g 0 (x) = 4 cos3 (x1/4 ) · (− sin(x1/4 )) · x−3/4
4
g 0 (x) = 4 cos3 (x1/4 ) · (− sin(x1/4 )) ·
Long answer question — you must show your work
3. 4 marks Determine the point(s) where the tangent line to
h(x) = x4/x , x > 0
is horizontal.
Answer: (e, e4/e )
Solution: Marking scheme: 1pt for taking the log of both sides; 1pt differentiating
properly;1pt for setting h’=0; 1pt for getting x = e; don’t worry if they didn’t find the
point (but do make a note of it).
Use logarithmic differentiation:
h(x) = x4/x
ln h = ln x4/x
4
ln h = · ln x
x
1 0 −4
4 1
· h = 2 · ln x + ·
h
x x x
−4
ln
x
4
h0 = h ·
+
x2
x2
1 − ln x
h0 = 4h ·
x2
From here, we note that h 6= 0, since x > 0. This means that h0 (x) = 0 if and only if
ln x = 1. Therefore, x = e is the only solution. Hence the point is (e, e4/e ).
MATH 104-184 Quiz #3
October 23
Grade:
First Name:
Last Name:
Student-No:
Section:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Differentiate y = ecos x .
Answer: y 0 = ecos x · (− sin x)
(b) he cost function is given by C(q) = 5q 2 + 90q + 330. Estimate the cost of producing the
14th unit.
Answer: $220
Solution: C 0 (q) = 10q + 90 ⇒ C 0 (13) = 130 + 90 = 220
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find the slope of the tangent line to 5x + sin(xy) + y 3 = 9 at the point (0, 2)
Answer: −
2 + ln 5
12
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
We differentiate implicitly to get
5x ln 5 + cos(xy) · [y + xy 0 ] + 3y 2 y 0 = 0
Subbing in x = 0 and y = 2 yields
ln 5 + 1 · [2 + 0] + 3(4)y 0 = 0
5
from which we see that y 0 = − 2+ln
12
√
(b) Find the derivative of the function g(x) = sin4 ( 4 x).
Answer:
g 0 (x) = 4 sin3 (x1/4 ) · cos(x1/4 ) · 41 x−3/4
Solution: Marking scheme: 1pt for a decent step towards the first line of differentiation. 2pt for a correct final answer (or near perfect solution)
4
g(x) = sin(x1/4 )
d sin(x1/4 )
dx
d 1/4 (x )
g 0 (x) = 4 sin3 (x1/4 ) · cos(x1/4 ) ·
dx
1
g 0 (x) = 4 sin3 (x1/4 ) · cos(x1/4 ) · x−3/4
4
g 0 (x) = 4 sin3 (x1/4 ) ·
Long answer question — you must show your work
3. 4 marks Determine the point(s) where the tangent line to
h(x) = x5/x , x > 0
is horizontal.
Answer: (e, e5/e )
Solution: Marking scheme: 1pt for taking the log of both sides; 1pt differentiating
properly;1pt for setting h’=0; 1pt for getting x = e; don’t worry if they didn’t find the
point (but do make a note of it).
Use logarithmic differentiation:
h(x) = x5/x
ln h = ln x5/x
5
ln h = · ln x
x
1 0 −5
5 1
· h = 2 · ln x + ·
h
x x x
−5
ln
x
5
h0 = h ·
+
x2
x2
1 − ln x
h0 = 5h ·
x2
From here, we note that h 6= 0, since x > 0. This means that h0 (x) = 0 if and only if
ln x = 1. Therefore, x = e is the only solution. Hence the point is (e, e5/e ).