E3_1 (6415722)
Due:
Mon Oct 6 2014 10:00 PM AKDT
Question
1.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47
Question Details
SCalc7 2.3.008. [1875245]
-
SCalc7 2.3.009. [1874861]
-
SCalc7 2.3.014. [1875139]
-
SCalc7 2.3.017.MI. [1874979]
-
Differentiate the function.
h(x) = (x − 5)(3x + 9)
h'(x) =
Solution or Explanation
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2.
Question Details
Differentiate the function.
g(t) = 4t−3/8
g'(t) =
Solution or Explanation
g(t) = 4t−3/8
3.
g'(t) = 4 −
Question Details
3 −11/8
3
t
= − t−11/8
8
2
Differentiate the function.
y=
x (x − 10)
y' =
Solution or Explanation
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4.
Question Details
Differentiate the function.
y=
4x2 + 2x + 2
x
y' =
Solution or Explanation
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5.
Question Details
SCalc7 2.3.019. [1874373]
-
SCalc7 2.3.027.MI. [1874747]
-
SCalc7 2.3.035. [1875267]
-
SCalc7 2.3.043. [1874746]
-
SCalc7 2.3.053. [1874935]
-
Differentiate the function.
H(x) = (x + x−1)3
H'(x) =
Solution or Explanation
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6.
Question Details
Differentiate.
F(y) =
1
y2
−
3
y4
(y + 9y3)
Solution or Explanation
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7.
Question Details
Differentiate.
y=
t2 + 5
t4 − 2t2 + 3
y' =
Solution or Explanation
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8.
Question Details
Differentiate.
f(v) =
v
v+ c
v
f '(v) =
Solution or Explanation
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9.
Question Details
(a) The curve y = 1/(1 + x2) is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the
point
y=
1,
1
.
2
(b) Illustrate part (a) by graphing the curve and tangent line on the same screen.
Solution or Explanation
(a)
y = f(x) =
1,
(b)
1
2
1
1 + x2
is f '(1) = −
f '(x) =
2
22
=−
(1 + x2)(0) − 1(2x)
(1 + x2)2
=
−2x
. So the slope of the tangent line at the point
(1 + x2)2
1
1
1
1
and its equation is y −
= − (x − (1)) or y = − x + 1.
2
2
2
2
10.
Question Details
SCalc7 2.3.070. [1680629]
-
SCalc7 2.3.080. [1874454]
-
If h(2) = 2 and h'(2) = −4, find
d
dx
h(x)
x
x = 2.
Solution or Explanation
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11.
Question Details
Find equations of the tangent lines to the curve
y=
x−1
x+1
that are parallel to the line x − 2y = 2.
y =
(smaller y-intercept)
y =
(larger y-intercept)
Solution or Explanation
y=
x−1
x+1
y' =
(x + 1)(1) − (x − 1)(1)
(x + 1)2
=
2
(x + 1)2
. If the tangent intersects the curve when x = a, then its slope is
2/(a + 1)2. But if the tangent is parallel to x − 2y = 2, that is, y =
2
(a + 1)2
=
1
2
(a + 1)2 = 4
a+1=±2
1
1
x − 1, then its slope is
. Thus,
2
2
a = 1 or −3. When a = 1, y = 0 and the equation of the tangent is
1
1
1
1
(x − 1) or y = x − . When a = −3, y = 2 and the equation of the tangent is y − 2 = (x + 3) or
2
2
2
2
1
7
y= x+ .
2
2
y−0=
12.
Question Details
SCalc7 2.3.081. [1874376]
Find an equation of the normal line to the parabola y = x2 − 8x + 7 that is parallel to the line x − 4y = 4.
y=
Solution or Explanation
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-
13.
Question Details
SCalc7 2.3.089. [1875022]
-
Find a cubic function y = ax3 + bx2 + cx + d whose graph has horizontal tangents at the points (−2, 8) and (2, 2).
y=
Solution or Explanation
y = f(x) = ax3 + bx2 + cx + d
f(−2) = 8
f '(x) = 3ax2 + 2bx + c. The point (−2, 8) is on f, so
−8a + 4b − 2c + d = 8 (1). The point (2, 2) is on f, so f(2) = 2
8a + 4b + 2c + d = 2 (2). Since
there are horizontal tangents at (−2, 8) and (2, 2), f '(±2) = 0.
f '(−2) = 0
8b = 0
12a − 4b + c = 0 (3) and f '(2) = 0
becomes 8a + 4(0) + 2(−12a) + 5 = 2
is y =
14.
12a + 4b + c = 0 (4). Subtracting equation (3) from (4) gives
b = 0. Adding (1) and (2) gives 8b + 2d = 10, so d = 5 since b = 0. From (3) we have c = −12a, so (2)
3 = 16a
3 3
9
x − x + 5.
16
4
Question Details
a=
3
3
. Now c = −12
16
16
=−
9
and the desired cubic function
4
SCalc7 2.3.095. [1874800]
Consider the following function.
f(x) = |x2 − 4|
(a) Find a formula for f '.
if |x| >
f '(x) =
if |x| <
For what values of x is the function not differentiable? (Enter your answers as a comma-separated list.)
x=
(b) Sketch the graph of f.
-
Sketch the graph of f '
Solution or Explanation
(a) Note that x2 − 4 < 0 for x2 < 4 ⇔ |x| < 2 ⇔ −2 < x < 2. So
x2 − 4
f(x) =
−x2 + 4
x2 − 4
if x ≤ −2
if −2 < x < 2
f '(x) =
if x ≥ 2
To show that f '(2) does not exist we investigate
2x
−2x
2x
lim
if x < −2
if −2 < x < 2
if x > 2
=
2x
−2x
if |x| > 2
if |x| < 2
f(2 + h) − f(2)
by computing the left- and right-hand derivatives.
h
h→0
+ h)2 +
f(2 + h) − f(2)
[−(2
4] − 0
lim
and
= lim
= h → 0− (−4 − h) = −4
h
h
h → 0−
2
2
f(2 + h) − f(2)
lim [(2 + h) − 4] − 0 = lim 4h + h = lim (4 + h) = 4.
f '+(2) = lim
=
h
h
h
h → 0+
h → 0+
h → 0+
h → 0+
f(2 + h) − f(2)
Since the left and right limits are different, lim
does not exist, that is, f '(2) does not exist. Similarly,
h
h→0
f '−(2) =
lim
h → 0−
f '(−2) does not exist. Therefore, f is not differentiable at 2 or at −2.
(b)
15.
Question Details
7
Find the value of c such that the line y = x + 7 is tangent to the curve y = c
4
c=
SCalc7 2.3.099. [1874853]
-
SCalc7 2.4.002. [1874427]
-
SCalc7 2.4.005. [1875296]
-
x.
Solution or Explanation
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16.
Question Details
Differentiate.
f(x) = 5
x sin x
f '(x) =
Solution or Explanation
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17.
Question Details
Differentiate.
y = sec θ tan θ
y' =
Solution or Explanation
y = sec θ tan θ
y' = sec θ(sec2θ) + tan θ(sec θ tan θ) = sec θ(sec2 θ + tan2 θ). Using the identity
1 + tan2 θ = sec2 θ, we can write alternative forms of the answer as sec θ(1 + 2 tan2 θ) or sec θ(2 sec2 θ − 1).
18.
Question Details
SCalc7 2.4.011. [1874885]
-
SCalc7 2.4.025. [1875290]
-
Differentiate.
f(θ) =
sec θ
2 + sec θ
f '(θ) =
Solution or Explanation
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19.
Question Details
(a) Find an equation of the tangent line to the curve y = 16x sin x at the point (π/2, 8π).
y=
(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
Solution or Explanation
(a)
y = 16x sin x
y' = 16(x cos x + sin x · 1). At π , 8π , y' = 16 π cos π + sin π
2
2
2
2
equation of the tangent line is y − 8π = 16 x − π , or y = 16x.
2
(b)
= 16(0 + 1) = 16, and the
20.
Question Details
SCalc7 2.4.029. [1874566]
-
SCalc7 2.4.034. [3043889]
-
If H(θ) = θ sin θ, find H'(θ) and H''(θ).
H'(θ) =
H''(θ) =
Solution or Explanation
H(θ) = θ sin θ
H'(θ) = θ(cos θ) + (sin θ) · 1 = θ cos θ + sin θ
H''(θ) = θ(−sin θ) + (cos θ) · 1 + cos θ = −θ sin θ + 2 cos θ
21.
Question Details
Find the values of x on the curve y =
answers as a comma-separated list.)
x=
Solution or Explanation
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cos x
at which the tangent is horizontal. (Let n be an integer. Enter your
2 + sin x
22.
Question Details
SCalc7 2.4.035.MI. [1874337]
-
A mass on a spring vibrates horizontally on a smooth level surface (see the figure). Its equation of motion is
x(t) = 8 cos t, where t is in seconds and x is in centimeters.
(a) Find the velocity and acceleration at time t.
v(t) =
a(t) =
(b) Find the position, velocity, and acceleration of the mass at time t = 5π/6.
x 5π
6
=
v 5π
6
=
a 5π
6
=
In what direction is it moving at that time?
Since v 5π
6
?
0, the particle is moving to the ---Select---
.
Solution or Explanation
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23.
Question Details
SCalc7 2.4.040. [1681299]
Find the limit.
lim
x→0
sin 8x
sin 9x
Solution or Explanation
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-
24.
Question Details
SCalc7 2.4.041. [1681073]
-
SCalc7 2.4.045. [1680875]
-
SCalc7 2.4.047. [1875243]
-
SCalc7 2.5.001. [1874329]
-
Find the limit.
tan 8t
sin 4t
lim
t→0
Solution or Explanation
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25.
Question Details
Find the limit.
sin 7θ
θ + tan 4θ
lim
θ→0
Solution or Explanation
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26.
Question Details
Find the limit.
lim
x → π/4
8 − 8 tan x
sin x − cos x
Solution or Explanation
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27.
Question Details
Write the composite function in the form f(g(x)). [Identify the inner function u = g(x) and the outer function y = f(u).]
y=
3
1 + 7x
(g(x), f(u)) =
Find the derivative dy/dx.
dy
=
dx
Solution or Explanation
Let u = g(x) = 1 + 7x and y = f(u) =
3
u . Then
dy
dy du
=
=
dx
du dx
1 −2/3
u
(7) =
3
3
7
3
(1 + 7x)2
.
28.
Question Details
SCalc7 2.5.012. [1927539]
-
SCalc7 2.5.016. [1927629]
-
SCalc7 2.5.017. [1874966]
-
SCalc7 2.5.021. [1875192]
-
Find the derivative of the function.
f(t) =
4
1 + tan t
f '(t) =
Solution or Explanation
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29.
Question Details
Find the derivative of the function.
y = 6 cot nθ
y' =
Solution or Explanation
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30.
Question Details
Find the derivative of the function.
f(x) = (2x − 5)4(x2 + x + 1)5
f '(x) =
Solution or Explanation
f(x) = (2x − 5)4(x2 + x + 1)5
f '(x) = (2x − 5)4 · 5(x2 + x + 1)4(2x + 1) + (x2 + x + 1)5 · 4(2x − 5)3 · 2
= (2x − 5)3(x2 + x + 1)4[(2x − 5) · 5(2x + 1) + (x2 + x + 1) · 8]
= (2x − 5)3(x2 + x + 1)4(20x2 − 40x − 25 + 8x2 + 8x + 8)
= (2x − 5)3(x2 + x + 1)4(28x2 − 32x − 17)
31.
Question Details
Find the derivative of the function.
y=
x2 + 6
4
x2 − 6
y' =
Solution or Explanation
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32.
Question Details
SCalc7 2.5.025. [1927541]
-
SCalc7 2.5.027. [1875237]
-
SCalc7 2.5.031. [1874572]
-
SCalc7 2.5.034. [1927703]
-
Find the derivative of the function.
z−9
z+9
F(z) =
F '(z) =
Solution or Explanation
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33.
Question Details
Find the derivative of the function.
y=
r
r2 + 5
y' =
Solution or Explanation
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34.
Question Details
Find the derivative of the function.
y = sin(tan 5x)
y' =
Solution or Explanation
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35.
Question Details
Find the derivative of the function.
y = x sin
9
x
y'(x) =
Solution or Explanation
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36.
Question Details
SCalc7 2.5.036. [1927577]
-
SCalc7 2.5.040. [1874753]
-
SCalc7 2.6.001. [1874671]
-
Find the derivative of the function.
f(t) =
t
t2 + 3
f '(t) =
Solution or Explanation
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37.
Question Details
Find the derivative of the function.
y = cos(cos(cos x))
y' =
Solution or Explanation
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38.
Question Details
Consider the following equation.
5x2 − y2 = 4
(a) Find y' by implicit differentiation.
y' =
(b) Solve the equation explicitly for y and differentiate to get y' in terms of x.
y' = ±
Solution or Explanation
(a)
d
d
(5x2 − y2) =
(4)
dx
dx
(b)
5x2 − y2 = 4
From part (a), y' =
10x − 2yy' = 0
y2 = 5x2 − 4
5x
=
±
y
5x
5x2 − 4
y=±
2yy' = 10x
5x2 − 4 , so y' = ±
y' =
5x
y
1
(5x2 − 4)−1/2(10x) = ±
2
, which agrees with part (b).
5x
5x2 − 4
.
39.
Question Details
SCalc7 2.6.005.MI. [1874651]
-
SCalc7 2.6.007. [1874864]
-
SCalc7 2.6.009. [1874484]
-
SCalc7 2.6.013. [1874661]
-
Find dy/dx by implicit differentiation.
x4 + y 5 = 7
y' =
Solution or Explanation
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40.
Question Details
Find dy/dx by implicit differentiation.
6x2 + 7xy − y2 = 9
y' =
Solution or Explanation
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41.
Question Details
Find dy/dx by implicit differentiation.
x7(x + y) = y2(8x − y)
y' =
Solution or Explanation
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42.
Question Details
Find dy/dx by implicit differentiation.
2 cos x sin y = 1
y' =
Solution or Explanation
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43.
Question Details
SCalc7 2.6.024. [1874468]
-
Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.
y sec x = 7x tan y
x' =
Solution or Explanation
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44.
Question Details
SCalc7 2.6.027. [1874628]
-
SCalc7 2.6.029. [1874903]
-
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2 + xy + y2 = 3, (1, 1)
(ellipse)
y=
Solution or Explanation
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45.
Question Details
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2 + y2 = (3x2 + 4y2 − x)2
(0, 0.25)
(cardioid)
y=
Solution or Explanation
Click to View Solution
46.
Question Details
SCalc7 2.6.030. [1874991]
-
SCalc7 2.6.032. [1874717]
-
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2/3 + y2/3 = 4
(−3 3 , 1)
(astroid)
y=
Solution or Explanation
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47.
Question Details
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y2(y2 − 4) = x2(x2 − 5)
(0, −2)
(devil's curve)
y=
Solution or Explanation
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Assignment Details
Name (AID): E3_1 (6415722)
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Author: Frith, Russell ( [email protected] )
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Last Saved: Oct 2, 2014 07:55 AM AKDT
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