Yanimov Almog
Math-1550-11-Fall2009
WeBWorK assignment number Sections 3.1 3.2 is due : 08/13/2017 at 03:21pm CDT.
The
(* replace with url for the course home page *)
for the course contains the syllabus, grading policy and other information.
This file is /conf/snippets/setHeader.pg you can use it as a model for creating files which introduce each problem set.
The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making
some kind of mistake. Usually you can attempt a problem as many times as you want before the due date. However, if you are
having trouble figuring out your error, you should consult the book, or ask a fellow student, one of the TA’s or your professor for
help. Don’t spend a lot of time guessing – it’s not very efficient or effective.
Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers,
you can if you wish enter elementary expressions such as 2 ∧ 3 instead of 8, sin(3 ∗ pi/2)instead of -1, e ∧ (ln(2)) instead of 2,
(2 + tan(3)) ∗ (4 − sin(5)) ∧ 6 − 7/8 instead of 27620.3413, etc. Here’s the list of the functions which WeBWorK understands.
You can use the Feedback button on each problem page to send e-mail to the professors.
1. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.7.pg
From Rogawski ET section 3.1, exercise 7.
Determine f 0 (a) for a = 1, 2, 4, 7.
f 0 (1) =
f 0 (2) =
f 0 (4) =
f 0 (7) =
Solution: Remember that the value of the derivative of f at
x = a can be interpreted as the slope of the line tangent to the
graph of y = f (x) at x = a. From the figure, we see that the graph
of y = f (x) is a horizontal line (that is, a line with zero slope)
on the interval 0 ≤ x ≤ 3. Accordingly, f 0 (1) = f 0 (2) = 0. On
the interval 3 ≤ x ≤ 5, the graph of y = f (x) is a line of slope
0.5; thus, f 0 (4) = 0.5. Finally, the line tangent to the graph of
y = f (x) at x = 7 is horizontal, so f 0 (7) = 0.
Calculate the slope of the secant line through the points on the
graph where x = 1 and x = 3.
slope =
Solution:
The slope of the secant line is
f (3)− f (1)
3−1
=
0−(−2)
2
=1
2. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.11.pg
3. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.13.pg
From Rogawski ET section 3.1, exercise 11.
Let f (x) be the function whose graph is shown below.
From Rogawski ET section 3.1, exercise 13.
Let f (x) be the function whose graph is shown below.
1
7. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.57.pg
From Rogawski ET section 3.1, exercise 57.
The limit below represents a derivative f 0 (a). Find f (x) and
a.
44+h − 256
lim
h→0
h
f (x) =
a=
Solution: The difference quotient
Which is larger?
44+h − 256
h
• A. f 0 (5.5)
• B. f 0 (6.5)
Solution: The line tangent to the graph of y = f (x) at x = 6.5
has a larger slope than the line tangent to the graph of y = f (x)
at x = 5.5. Therefore, f 0 (6.5) is larger than f 0 (5.5).
has the form
f (a + h) − f (a)
h
where f (x) = 4x and a = 4.
8. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function/3.2.11.pg
4. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.15.pg
From Rogawski ET section 3.2, exercise 11.
Use the Power Rule to compute the derivative:
d 2/3
t
|t=6 =
dt
From Rogawski ET section 3.1, exercise 15.
Use the definition of the derivative to find the derivative of:
f (x) = 11x − 15.
f 0 (x) =
Solution:
f (x)
lim f (x+h)−
= lim
h
h→0
h→0
11x+11h−15−11x+15
h
11h
h→0 h
= lim
Solution:
d 2/3
= 23 t −1/3 , so
dt t
d 2/3
|t=6
dt t
= 23 (6)−1/3 =
2
√
.
336
9. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function/3.2.13.pg
=
lim 11 = 11
From Rogawski ET section 3.2, exercise 13.
Use the power rule to compute the derivative.
d 0.6
x
dx
f 0 (x) =
Solution:
d 0.6
x = 0.6(x0.6−1 ) = 0.6x−0.4 .
dx
h→0
5. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.19.pg
From Rogawski ET section 3.1, exercise 19.
let f (x) = 1x . Compute the difference quotient for f (x) at
x = 2 with h = 0.5
difference quotient =
Solution:
1 −1
f (x)
2
The difference quotient is f (x+h)−
= 2+0.5
= −0.2.
h
0.5
10. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function/3.2.23.pg
6. (1 pt) Rogawski-ET/3 Differentiation/3.1 Definition of the Derivative/3.1.53.pg
From Rogawski ET section 3.2, exercise 23.
Find the derivative of the function f (x) = x3 + x2 − 14.
f 0 (x) =
From Rogawski ET section 3.1, exercise 53.
The limit below represents a derivative f 0 (a). Find f (x) and
a.
(5 + h)3 − 125
lim
h→0
h
f (x) =
a=
Solution: The difference quotient
Solution:
d
3
2
2
dx (x + x − 14) = 3x + 2x.
11. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function/3.2.27.pg
From Rogawski ET section 3.2, exercise 27.
Find the derivative of the function f (x) = 8x−3 + x2 + 14.
0
f (x) =
(5 + h)3 − 125
h
has the form
f (a + h) − f (a)
h
where f (x) = x3 and a = 5.
Solution:
d
−3 + x2 + 14) = −3 · 8x−4 + 2x =
dx (8x
2
−24
x4
+ 2x.
12. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function/3.2.55.pg
From Rogawski ET section 3.2, exercise 55.
Find all values of x where the tangent lines to y = x6 and
y = x7 are parallel.
x=
Solution: Let f (x) = x6 and let g(x) = x7 . The two graphs
have parallel tangent lines at all x where f 0 (x) = g0 (x).
f 0 (x) = g0 (x)
6x5 = 7x6
6x5 − 7x6 = 0
Its derivative does not exist at x = −1. At that value of x there
is a sharp point.
5
x (6 − 7x) = 0
15.
(1 pt)
DerivAsFunct/2-9-23.pg
hence, x = 0 or x = 76 .
Library/UVA-Stew5e/setUVA-Stew5e-C02S09-
Use the definition of the derivative (don’t be tempted to take
shortcuts!) to find the derivative of the function
13. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a Function√
f (x) = 2 + 4x.
/3.2.57.pg
From Rogawski ET section 3.2, exercise 57.
Then state the domain of the function and the domain of the
Determine coefficients a and b such that p(x) = x2 + ax + b
derivative.
satisfies p(1) = −7 and p0 (1) = 1.
Note: When entering interval notation in WeBWorK, use I for
a=
∞, -I for −∞, and U for the union symbol. If the set is empty,
b=
enter (0,0) .
Solution: Let p(x) = x2 + ax + b satisfy p(1) = −7 and
p0 (1) = 1. Since p0 (x) = 2x + a, this implies −7 = p(1) =
f 0 (x) =
1 + a + b and 1 = p0 (1) = 2 + a; i.e., a = −1 and b = −7.
Domain of f (x) =
14. (1 pt) Rogawski-ET/3 Differentiation/3.2 The Derivative as a FunctionDomain of f 0 (x) =
/3.2.77.pg
From Rogawski ET section 3.2, exercise 77.
16. (1 pt) Library/ASU-topics/setDerivativeFunction/3-3-05.pg
Find the points c (if any) such that f 0 (c) does not exist.
Suppose that
f (x + h) − f (x) = 8hx2 − 3hx + 2h2 x − 2h2 − 4h3 .
f (x) = |x + 1|
Find f 0 (x).
c=
Solution: Here is the graph of f (x) = |x + 1|.
f 0 (x) =
c
Generated by the WeBWorK system WeBWorK
Team, Department of Mathematics, University of Rochester
3
Yanimov Almog
Math-1550-11-Fall2009
WeBWorK assignment number Sections 3.3 3.4 is due : 08/14/2017 at 09:57am CDT.
The
(* replace with url for the course home page *)
for the course contains the syllabus, grading policy and other information.
This file is /conf/snippets/setHeader.pg you can use it as a model for creating files which introduce each problem set.
The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making
some kind of mistake. Usually you can attempt a problem as many times as you want before the due date. However, if you are
having trouble figuring out your error, you should consult the book, or ask a fellow student, one of the TA’s or your professor for
help. Don’t spend a lot of time guessing – it’s not very efficient or effective.
Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers,
you can if you wish enter elementary expressions such as 2 ∧ 3 instead of 8, sin(3 ∗ pi/2)instead of -1, e ∧ (ln(2)) instead of 2,
(2 + tan(3)) ∗ (4 − sin(5)) ∧ 6 − 7/8 instead of 27620.3413, etc. Here’s the list of the functions which WeBWorK understands.
You can use the Feedback button on each problem page to send e-mail to the professors.
3. (1 pt) Rogawski-ET/3 Differentiation/3.3 Product and Quotient Rules1. (1 pt) Rogawski-ET/3 Differentiation/3.3 Product and Quotient Rules/3.3.53.pg
/3.3.5.pg
From Rogawski ET section 3.3, exercise 53.
Calculate F 0 (0), where
From Rogawski ET section 3.3, exercise 5.
Use the Product Rule to compute the derivative:
d
2
dt ((t + 1)(t + 9))|t=12 =
F(x) =
Solution:
Let y = (t 2 + 1)(t + 9). Then
dy
d 2
d
2
2
dt = ( dt (t +1))(t +9)+(t +1) dt (t +9) = 2t(t +9)+(t +
2
1) = 3t + 18t + 1.
Therefore,
9x6 + 4x5 − 10x4 + 3x
.
10x2 − (8x9 + 5x4 ) + 5
Hint:
Do not calculate F 0 (x).
Instead, write
F(x) = f (x)/g(x) and express F 0 (0) directly in terms of
f (0), f 0 (0), g(0), g0 (0).
F 0 (0) =
Solution: Taking the hint, let
dy
|t=12 = 3(12)2 + 18 · 12 + 1 = 649.
dt
f (x) = 9x6 + 4x5 − 10x4 + 3x
and let
g(x) = 10x2 − 8x9 + 5x4 + 5.
2. (1 pt) Rogawski-ET/3 Differentiation/3.3 Product and Quotient Rules/3.3.19.pg
Then F(x) =
From Rogawski ET section 3.3, exercise 19.
Compute the derivative:
d
dx
1
x+1
=
Now,
f 0 (x) = 9 · 6x5 + 4 · 5x4 − 10 · 4x3 + 3
and
1
x + 1 x=1
g0 (x) = 10 · 2x − 8 · 9x8 + 5 · 4x3
Moreover, f (0) = 0, f 0 (0) = 3, g(0) = 5, and g0 (0) = 0.
Using the quotient rule:
The derivative is:
Solution:
Using the Quotient Rule:
d
dx
f (x)
g(x) .
F 0 (0) =
(x + 1)(0) − 1 · (1)
1
=−
2
(x + 1)
(x + 1)2
4.
(1
g(0) f 0 (0) − f (0)g0 (0) 15
=
= 0.6.
(g(0))2
25
pt)
Library/UVA-Stew5e/setUVA-Stew5e-C03S02-
ProdQuotRules/3-2-31b.pg
Consider the functions f (x) and g(x), for which f (0) = 7,
g(0) = 9, f 0 (0) = −8, and g0 (0) = −7.
f (x)
.
Find h0 (0) for the function h(x) = g(x)
0
h (0) =
Therefore,
d
dx
1
1
1
=−
=− .
x + 1 x=1
(1 + 1)2
4
1
5.
(1 pt)
ProdQuotRules/3-2-23.pg
t=
(D) Use interval notation to indicate when the particle is moving
in the positive direction. (If the particle is never moving in the
positive direction, enter (0,0).)
Answer =
Library/UVA-Stew5e/setUVA-Stew5e-C03S02-
Find an equation for the line tangent to the graph of
5x
f (x) =
x+6
at the point (2, 1.25).
y=
9.
(1 pt)
RatesofChange/3-3-16.pg
Library/UVA-Stew5e/setUVA-Stew5e-C03S03-
6. (1 pt) Rogawski-ET/3 Differentiation/3.4 Rates of Change/3.4.5.pg
From Rogawski ET section 3.4, exercise 5.
Calculate the rate of change dV
dr where V is the volume of a
cylinder whose height is equal to 3 times its radius. (The volume
of a cylinder of height h and radius r is πr2 h).
The rate of change is
Solution: To find the rate of change dV
dr , first apply the condition that h = 3r to simplify the volume formula. We get
V = π × r2 × (3r) = 3πr3 . The derivative is given by the power
2
2
rule - dV
dr = (3)(3)πr = 9πr .
If a tank holds 4200 gallons of water, which drains from the
bottom of the tank in 43 minutes, then Torricelli’s Law gives the
volume V of water remaining in the tank after t minutes as
t 2
,
0 ≤ t ≤ 43.
V = 4200 1 −
43
7. (1 pt) Rogawski-ET/3 Differentiation/3.4 Rates of Change/3.4.13.pg
From Rogawski ET section 3.4, exercise 13.
A stone is tossed vertically upward with an initial velocity of
30 ft/s from the top of a 23 ft building.
(a) What is the height of the stone after 0.42 s?
(b) Find the velocity of the stone after 1 s.
(c) When does the stone hit the ground?
(a)
(b)
(c)
Solution: We employ Galileo’s formula,
10. (1 pt) Library/ma122DB/set4/s3 3 27.pg
Suppose that the cost, in dollars, for a company to produce x
pairs of a new line of jeans is
Find the rate at which the water is draining from the tank after:
(A) 4 minutes → Rate of change =
(B) 9 minutes → Rate of change =
(C) 16 minutes → Rate of change =
C(x) = 1700 + 5x + 0.01x2 + 0.0002x3 .
(a) Find the marginal cost function.
Answer:
(b) Find the marginal cost at x = 100.
Answer:
s(t) = s0 + v0t − 1/2gt 2 = 23 + 30t − 16t 2
(c) Find the cost at x = 100.
Answer:
where the time t is in seconds (s) and the height s is in feet (ft).
(a) The height of the stone after 0.42 seconds is s(0.42) =
32.7776.
(b) The velocity at time t is s0 (t) = 30 − 32t. When t = 1, this
is -2 ft/s.
(c) When the stone hits the ground, its height is zero. Solve
23 + 30t − 16t 2 = 0 to obtain t = −30±45.793
or t = 2.45947. (We
−32
discard the other solution, since it represents a time before the
stone was thrown.)
8.
(1 pt)
RatesofChange/3-3-01.pg
11.
(1 pt) Library/UVA-Stew5e/setUVA-Stew5e-C03S02ProdQuotRules/3-2-06a.pg
Suppose that
f (x) =
ex
.
x2 + 14
Find f 0 (1).
Library/UVA-Stew5e/setUVA-Stew5e-C03S03-
f 0 (1) =
Suppose that a particle moves according to the law of motion
s = t 2 − 9t + 22,
t ≥ 0.
12.
(1 pt)
ProdQuotRules/3-2-25.pg
(A) Find the velocity at time t.
v(t) =
(B) What is the velocity after 3 seconds?
Velocity after 3 seconds =
(C) Find all values of t for which the particle is at rest. (If there
are no such values, enter 0. If there are more than one value, list
them separated by commas.)
Library/UVA-Stew5e/setUVA-Stew5e-C03S02-
Find an equation for the line tangent to the graph of
f (x) = 2xex
at the point (a, f (a)) for a = 1.
y=
2
13.
(1 pt)
ProdQuotRules/3-2-35.pg
Find f 0 (5).
Library/UVA-Stew5e/setUVA-Stew5e-C03S02-
15.
(1 pt) Rogawski-ET/3 Differentiation/3.4 Rates of Change-
/3.4.31.pg
From Rogawski ET section 3.4, exercise 31.
What is the velocity of an object dropped from a height of
110 m when it hits the ground?
Solution: We employ Galileo’s formula,
s(t) = s0 − 1/2gt 2 = 110 − 4.9t 2
where the time, t, is in seconds (s) and the height, s, is in meters (m). When the ball hits the ground its height is 0. Solve
s(t) = 110 − 4.9t 2 = 0 to obtain t = 4.73804 s. (We discard the
negative time, which took place before the ball was dropped.)
The velocity at time t is given by s0 (t) = −2 · 4.9 · t, and so the
velocity at impact is
Note: Click on graph for larger version in new browser
window.
The graphs of the function f (given in blue) and g (given
in red) are plotted above. Suppose that u(x) = f (x)g(x) and
v(x) = f (x)/g(x). Find each of the following:
u0 (1) =
v0 (1) =
v(4.73804) = −9.8(4.73804) = −46.4328m/s
This signifies that the ball is falling at 46.4328 m/s.
16. (1 pt) Library/ASU-topics/setProductQuotientRule/3-5-39.pg
Let
x
.
f (x) = 2
x + 16
0
Find all the values of x for which f (x) = 0. What is the product
of all these values?
(For example, if f 0 (x) was equal to zero at the points 1, 2, and
3, then the answer would be 1*2*3 = 6.)
Product =
14. (1 pt) Library/Rochester/setDerivatives2Formulas/s2 2 12.pg
If
f (x) =
3 − x2
8 + x2
find f 0 (x).
c
Generated by the WeBWorK system WeBWorK
Team, Department of Mathematics, University of Rochester
3
Yanimov Almog
Math-1550-11-Fall2009
WeBWorK assignment number Sections 3.5 3.6 is due : 08/14/2017 at 11:41am CDT.
The
(* replace with url for the course home page *)
for the course contains the syllabus, grading policy and other information.
This file is /conf/snippets/setHeader.pg you can use it as a model for creating files which introduce each problem set.
The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making
some kind of mistake. Usually you can attempt a problem as many times as you want before the due date. However, if you are
having trouble figuring out your error, you should consult the book, or ask a fellow student, one of the TA’s or your professor for
help. Don’t spend a lot of time guessing – it’s not very efficient or effective.
Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers,
you can if you wish enter elementary expressions such as 2 ∧ 3 instead of 8, sin(3 ∗ pi/2)instead of -1, e ∧ (ln(2)) instead of 2,
(2 + tan(3)) ∗ (4 − sin(5)) ∧ 6 − 7/8 instead of 27620.3413, etc. Here’s the list of the functions which WeBWorK understands.
You can use the Feedback button on each problem page to send e-mail to the professors.
1.
(1 pt) Rogawski-ET/3 Differentiation/3.5 Higher Derivatives-
4.
(1 pt) Rogawski-ET/3 Differentiation/3.5 Higher Derivatives-
/3.5.3.pg
/3.5.29.pg
From Rogawski ET section 3.5, exercise 3.
Calculate the second and third derivatives.
From Rogawski ET section 3.5, exercise 29.
Calculate y(k) (0) for 0 ≤ k ≤ 5, where y = 6x4 + ax3 + bx2 +
cx + d (with a,b,c,d the constants)
y(0) (0) =
y(1) (0) =
y(2) (0) =
y(3) (0) =
y(4) (0) =
y(5) (0) =
y = 6x4 − 9x2 + 5x
y” =
y”’ =
Solution: Let y = 6x4 − 9x2 + 5x. Then y0 = 24x3 − 18x + 5,
00
y = 72x2 − 18, and y000 = 144x.
2.
(1 pt) Rogawski-ET/3 Differentiation/3.5 Higher Derivatives-
/3.5.17.pg
From Rogawski ET section 3.5, exercise 17.
d2y
−3 + 6x2
Calculate the derivative indicated. dx
2 |x=3 , y = 7x
Solution: Let y = 7x−3 + 6x2 . Then
and
d2y
dx2
dy
dx
Solution:
Applying the power, constant multiple, and sum rules at each
stage, we get (note that y0 = y by convention)
= (−21)x−4 + 12x
= 84x−5 + 12. Hence
k
0
1
2
3
4
5
d2y
|x=3 = 84(3)−5 + 12 = 12.3457
dx2
3.
(1 pt) Rogawski-ET/3 Differentiation/3.5 Higher Derivatives-
/3.5.19.pg
From Rogawski
ET section 3.5, exercise 19.
√
If h(x) = x
then h000 (4) =
y(k)
6x4 + ax3 + bx2 + cx + d
24x3 + 3ax2 + 2bx + c
72x2 + 6ax + 2b
144x + 6a
144
0
from which we get:
y(0) (0) = d,
y(1) (0) = c,
y(2) (0) = 2b,
y(3) (0) = 6a,
y(4) (0) = 144,
y(5) (0) = 0.
Solution:
√
Let h(x) = x = x1/2 . Then,
h0 (x) = 21 x−1/2 ,
h00 (x) = − 14 x−3/2 ,
h000 (x) = 38 x−5/2 ,
Thus,
3
h000 (4) = · 4−5/2 = 0.01171875
8
.
5. (1 pt) Library/Rochester/setDerivatives13Higher/hder1.pg
Find the 48 th derivative of the function f (x) = cos(x).
1
The answer is function
10. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.9.pg
6. (1 pt) Library/UVA-Stew5e/setUVA-Stew5e-C03S07-HigherDerivs/3-7-47.pg
From Rogawski ET section 3.6, exercise 9.
Use the Product Rule to find the derivative of f .
f (x) = x6 tan(x)
f 0 (x) =
Solution: Let f (x) = x6 tan(x). Then f 0 (x) = x6 sec2 (x) +
5
6x tan(x) .
Suppose that the equation of motion for a particle (where s is in
meters and t in seconds) is
s = t 4 − 5t 3 + 5
.
(a) Find the time (other than at t = 0 when the acceleration is 0.
Acceleration is 0 at time t =
(b) Find the position and velocity at this time.
Position =
Velocity =
11. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.15.pg
From Rogawski ET section 3.6, exercise 15.
Use the Product Ruleto find the derivative of f .
f (x) = 4 + 3x − 3x2 sin(x)
f 0 (x) =
Solution:
Let f (x) = 4 + 3x − 3x2 sin(x). Then
f 0 (x) = 4 + 3x − 3x2 cos(x) + (3 − 6x) sin(x) .
7. (1 pt) Library/maCalcDB/setDerivatives1/nsc2s10p3.pg
12. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.17.pg
From Rogawski ET section 3.6, exercise 17.
If
sec(x)
f (x) =
x8
then f 0 (x) =
Solution:
Applying the Quotient rule:
Identify the graphs A (blue), B( red) and C (green) as the
graphs of a function and its derivatives:
is the graph of the function
is the graph of the function’s first derivative
is the graph of the function’s second derivative
sec x 0
x8
=
0
(sec x)0 x8 − x8 sec x x8 sec x tan x − 8x7 sec x sec x(x tan x
=
=
(x8 )2
x16
x9
13. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.21.pg
From Rogawski ET section 3.6, exercise 21.
Use the Quotient Rule to find the derivative of f .
8. (1 pt) Library/maCalcDB/setDerivatives4Trig/s2 7 32.pg
Find the 29th derivative of sin(x) by finding the first few derivatives and observing the pattern that occurs.
(sin(x))(29) =
f (x) =
x
tan(x) + 3
f 0 (x) =
Solution:
Let f (x) be as above. Then the Quotient Rule says
9. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.5.pg
From Rogawski ET section 3.6, exercise 5.
Use the Product Rule to find the derivative of f .
f (x) = csc(x) tan(x)
f 0 (x) =
Solution:
Let f (x) = csc(x) tan(x). Then
f 0 (x) =
d
d
(tan(x) + 3) · dx
[x] − x · dx
[tan(x) + 3]
.
2
(tan(x) + 3)
d
Since dx
[x] = 1 and
see that
f 0 (x) = sec2 (x) csc(x) − csc(x) cot(x) tan(x)
d
dx [tan(x) + 3]
f 0 (x) =
2
=
d
dx [tan(x)]
tan(x) + 3 − x sec2 (x)
(tan(x) + 3)2
= sec2 (x), we
14. (1 pt) Rogawski-ET/3 Differentiation/3.6 Trigonometric Functions/3.6.39.pg
15. (1 pt) Library/270/setDerivatives4Trig/s2 4 27b.pg
Let
−7x
f (x) =
sin x + cos x
From Rogawski ET section 3.6, exercise 39.
The ratio
d
dx 7 · cot(x)
csc2 (x)
is a constant number.
.
Its value is
f 0 (π) =
16. (1 pt) Library/UVA-Stew5e/setUVA-Stew5e-C03S04-DerivsTrig/34-25.pg
Solution:
Find the equation of the tangent line to the curve
d
d cos x
sin x(− sin x) − cos x(cos x)
7 · cot(x) = 7
= 7·
dx
dx sin x
sin2 x
y = 4x cos x
−(sin2 x + cos2 x)
−7
=7
=
= −7 csc2 x
2
sin x
sin2 x
at the point (π, −4π).
The equation of this tangent line can be written in the form
y = mx + b where
m=
and b =
Then,
d
dx 7 · cot(x)
csc(x)2
=
−7 csc2 x
= −7
csc2 x
c
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