100/180 Quiz 3.1
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.1
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Evaluate the following limit by interpreting it as a derivative:
tan(x) − 1
lim
x→π/4
x − π/4
You may use the derivative rules we have learned so far in class
— but do not use L’Hopital’s rule.
Answer: 2
Solution: This limit represents the derivative computed at x = π/4 of the
function f (x) = tan(x). Since the derivative of f (x) is sec2 (x), then its value at
x = π/4 is exactly 2.
(b) Let y = arctan (log(x)). Compute
(Recall: log x = loge x = ln x.)
dy
.
dx
Answer:
Solution:
d
dy
=
arctan (log x)
dx
dx
1
1
=
·
2
1 + (log x) ) x
1
x(1 + log(x)2 )
100/180 Quiz 3.1
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (x2 + 1)sin(x) .
Solution: We use logarithmic differentiation; so
log f (x) = log(x2 + 1) · sin x
Then differentiate to obtain
f 0 (x)
d 2x sin x
=
log(x2 + 1) · sin x = cos x · log(x2 + 1) + 2
f (x)
dx
x +1
In conclusion:
2x sin x
2
f (x) = f (x) · cos x · log(x + 1) + 2
or equivalently
x +1
2x sin x
2
sin(x)
2
= (x + 1)
· cos x · log(x + 1) + 2
x +1
0
(b) Let f (x) be a function differentiable at x = 3 and let g(x) = x · f (x). The line
tangent to the curve y = f (x) at x = 3 has slope 2 while the line tangent to the
curve y = g(x) at x = 3 has slope 5. What is f (3)?
Solution:
• We know f 0 (3) = 2 and g 0 (3) = 5 and g 0 (x) = xf 0 (x) + f (x)
• This means 5 = g 0 (3) = 3f 0 (3) + f (3) = 6 + f (3)
• Hence f (3) = −1.
100/180 Quiz 3.1
End of week 7
Page 4 of 4
Long answer question — you must show your work
3. 4 marks If x2 y + 4 = 4y 2 ex , then find
your answer.
dy
dx
at all points where x = 0. You must justify
Solution:
• First find the y-coordinates where x = 0:
0 + 4 = 4y 2 e0
1 = y2
Hence y = ±1.
• Now we use implicit differentiation to get y 0 in terms of x, y:
x2 y 0 + 2xy = 8yy 0 ex + 4y 2 ex
• Now set x = 0 to get
0 = 8yy 0 + 4y 2
y 0 = −y/2
• So at (x, y) = (0, 1) we have y 0 = −1/2,
• and at (x, y) = (0, −1) we have y 0 = 1/2.
100/180 Quiz 3.2
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.2
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Evaluate the following limit by interpreting it as a derivative:
3/2
x − 43/2
lim
x→4
x−4
You may use the derivative rules we have learned so far in class
— but do not use L’Hopital’s rule.
Answer: 3
Solution: This limit represents the derivative computed at x = 4 of the function
f (x) = x3/2 . Since the derivative of f (x) is (3/2) · x1/2 , then its value at x = 4
is exactly (3/2) · 41/2 = 3.
(b) Let y = arcsin (ex ). Compute
dy
.
dx
Answer: √
Solution:
d
dy
=
arcsin (ex )
dx
dx
1
=√
· ex
1 − e2x
ex
1 − e2x
100/180 Quiz 3.2
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (x4 + 1)cos(x) .
Solution: We use logarithmic differentiation; so
log f (x) = log(x4 + 1) · cos x
Then differentiate to obtain
f 0 (x)
d 4x3 cos x
=
log(x4 + 1) · cos x = − sin x · log(x4 + 1) + 4
f (x)
dx
x +1
In conclusion:
4x3 cos x
or equivalently
f (x) = f (x) · − sin x · log(x + 1) + 4
x +1
4x3 cos x
4
cos(x)
4
= (x + 1)
· − sin x · log(x + 1) + 4
x +1
0
4
(b) Let f (x) be a function differentiable at x = 1 and let g(x) = x · f (x). The line
tangent to the curve y = f (x) at x = 1 has slope 3 while the line tangent to the
curve y = g(x) at x = 1 has slope 4. What is f (1)?
Solution:
• We know f 0 (1) = 3 and g 0 (1) = 4 and g 0 (x) = xf 0 (x) + f (x)
• This means 4 = g 0 (1) = f 0 (1) + f (1) = 3 + f (1)
• Hence f (1) = 1.
100/180 Quiz 3.2
End of week 7
Page 4 of 4
Long answer question — you must show your work
3. 4 marks If xy 2 + 8 = 2y 2 ex , then find
your answer.
dy
dx
at all points where x = 0. You must justify
Solution:
• First find the y-coordinates where x = 0:
0 + 8 = 2y 2 e0
4 = y2
Hence y = ±2.
• Now we use implicit differentiation to get y 0 in terms of x, y:
x(2yy 0 ) + y 2 = 4yy 0 ex + 2y 2 ex
• Now set x = 0 to get
y 2 = 4yy 0 + 2y 2
y 0 = −y/4
• So at (x, y) = (0, 2) we have y 0 = −1/2,
• and at (x, y) = (0, −2) we have y 0 = 1/2.
100/180 Quiz 3.3
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.3
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Evaluate the following limit by interpreting it as a derivative:
2x
e − e20
lim
x→10
x − 10
You may use the derivative rules we have learned so far in class
— but do not use L’Hopital’s rule.
Answer: 2e20
Solution: This limit represents the derivative computed at x = 10 of the function f (x) = e2x . Since the derivative of f (x) is 2e2x , then its value at x = 10 is
exactly 2e20 .
√
(b) Let y = arctan ( x). Compute
dy
.
dx
Answer:
Solution:
√ dy
d
=
arctan x
dx
dx
1
1
=
√ 2· √
1 + ( x) 2 x
1
= √
2 x (1 + x)
1
√
2 x(1 + x)
100/180 Quiz 3.3
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (cos x)x
4 +1
.
Solution: We use logarithmic differentiation; so
log f (x) = (x4 + 1) log cos x
Then differentiate to obtain
f 0 (x)
d 4
− sin x
=
(x + 1) log cos x = 4x3 log cos(x) + (x4 + 1)
f (x)
dx
cos x
4
(x + 1) sin x
= 4x3 log cos(x) −
cos x
In conclusion:
(x4 + 1) sin x
f (x) = f (x) · 4x log cos(x) −
or equivalently
cos x
(x4 + 1) sin x
x4 +1
3
= (cos x)
· 4x log cos(x) −
cos x
0
3
(b) Let f (x) be a function differentiable at x = 2 and let g(x) = x · f (x). The line
tangent to the curve y = f (x) at x = 2 passes through the point (2, 3). The line
tangent to the curve y = g(x) at x = 2 has slope 5. What is f 0 (2)?
Solution:
• We know f (2) = 3, g 0 (2) = 5, and g 0 (x) = xf 0 (x) + f (x).
• This means 5 = g 0 (2) = 2f 0 (2) + f (2) = 2f 0 (2) + 3
• Hence f 0 (2) = 1.
100/180 Quiz 3.3
End of week 7
Long answer question — you must show your work
dy
3. 4 marks If 8 log x + 2xy = y 2 , then find dx
at all points where x = 1.
(Recall: log x = loge x = ln x.) You must justify your answer.
Solution:
• First find the y-coordinates where x = 1:
8 log(1) + 2(1)y = y 2
0 + 2y = y 2
2y − y 2 = 0
y(2 − y) = 0
Hence y = 0 or y = 2.
• Now we use implicit differentiation to get y 0 in terms of x, y:
8
+ 2x · y 0 + 2y = 2y · y 0
x
• Now set x = 1 and y = 0 to get
8
+ 2y 0 + 0 = 0
1
2y 0 = −8
y 0 = −4
So at (x, y) = (1, 0), y 0 = −4
• Now set x = 1 and y = 2 to get
8
+ 2y 0 + 2(2) = 2(2)y 0
1
12 + 2y 0 = 4y 0
12 = 2y 0
y0 = 6
So at (x, y) = (1, 2), y 0 = 6
Page 4 of 4