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E3_1 (6415722)

Due: Mon Oct 6 2014 10:00 PM AKDT

1. -Question Details SCalc7 2.3.008. [1875245]

Differentiate the function.

Solution or Explanation

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h(x) = (x − 5)(3x + 9)

h'(x) =

2. -Question Details SCalc7 2.3.009. [1874861]

Differentiate the function.

g'(t) =

Solution or Explanation

g(t) = 4t−3/8

g(t) = 4t−3/8 g'(t) = 4 − t−11/8 = − t−11/838

32

3. -Question Details SCalc7 2.3.014. [1875139]

Differentiate the function.

Solution or Explanation

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y = (x − 10)x

y' =

4. -Question Details SCalc7 2.3.017.MI. [1874979]

Differentiate the function.

y' =

Solution or Explanation

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y = 4x2 + 2x + 2

x

5. -Question Details SCalc7 2.3.019. [1874373]

Differentiate the function.

Solution or Explanation

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H(x) = (x + x−1)3

H'(x) =

6. -Question Details SCalc7 2.3.027.MI. [1874747]

Differentiate.

Solution or Explanation

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F(y) = − (y + 9y3)1y2

3y4

7. -Question Details SCalc7 2.3.035. [1875267]

Differentiate.

y' =

Solution or Explanation

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y = t2 + 5t4 − 2t2 + 3

8. -Question Details SCalc7 2.3.043. [1874746]

Differentiate.

Solution or Explanation

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f(v) = v

v + cv

f '(v) =

9. -Question Details SCalc7 2.3.053. [1874935]

(a) The curve is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the

point

y =

y = 1/(1 + x2)

1, .12

(b) Illustrate part (a) by graphing the curve and tangent line on the same screen.

Solution or Explanation

(a) So the slope of the tangent line at the point

is and its equation is

(b)

y = f(x) = f '(x) = = .11 + x2

(1 + x2)(0) − 1(2x)(1 + x2)2

−2x(1 + x2)2

1, 12

f '(1) = − = − 222

12

y − = − (x − (1)) or y = − x + 1.12

12

12

10. -Question Details SCalc7 2.3.070. [1680629]

If find

Solution or Explanation

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h(2) = 2 and h'(2) = −4,

ddx

h(x)x

x = 2.

11. -Question Details SCalc7 2.3.080. [1874454]

Find equations of the tangent lines to the curve

that are parallel to the line

Solution or Explanation

If the tangent intersects the curve when then its slope is

But if the tangent is parallel to that is, then its slope is Thus,

When and the equation of the tangent is

or When and the equation of the tangent is or

y = x − 1x + 1

x − 2y = 2.

y = (smaller y-intercept)

y = (larger y-intercept)

y = y' = = .x − 1x + 1

(x + 1)(1) − (x − 1)(1)(x + 1)2

2(x + 1)2

x = a,

2/(a + 1)2. x − 2y = 2, y = x − 1,12

.12

= (a + 1)2 = 4 a + 1 = ± 2 a = 1 or −3.2(a + 1)2

12

a = 1, y = 0

y − 0 = (x − 1)12

y = x − .12

12

a = −3, y = 2 y − 2 = (x + 3)12

y = x + .12

72

12. -Question Details SCalc7 2.3.081. [1874376]

Find an equation of the normal line to the parabola that is parallel to the line

y =

Solution or Explanation

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y = x2 − 8x + 7 x − 4y = 4.

13. -Question Details SCalc7 2.3.089. [1875022]

Find a cubic function whose graph has horizontal tangents at the points

y =

Solution or Explanation

The point is on f, so The point is on f, so Since

there are horizontal tangents at Subtracting equation gives

Adding (1) and (2) gives so d = 5 since b = 0. From (3) we have so (2)

becomes Now and the desired cubic function

is

y = ax3 + bx2 + cx + d (−2, 8) and (2, 2).

y = f(x) = ax3 + bx2 + cx + d f '(x) = 3ax2 + 2bx + c. (−2, 8)f(−2) = 8 −8a + 4b − 2c + d = 8 (1). (2, 2) f(2) = 2 8a + 4b + 2c + d = 2 (2).

(−2, 8) and (2, 2), f '(±2) = 0.f '(−2) = 0 12a − 4b + c = 0 (3) and f '(2) = 0 12a + 4b + c = 0 (4). (3) from (4)8b = 0 b = 0. 8b + 2d = 10, c = −12a,

8a + 4(0) + 2(−12a) + 5 = 2 3 = 16a a = .316

c = −12 = − 316

94

y = x3 − x + 5.316

94

14. -Question Details SCalc7 2.3.095. [1874800]

Consider the following function.

(a)

if |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.

f(x) = |x2 − 4|

Find a formula for f '.

f '(x) =

Sketch the graph of f '

Solution or Explanation

(a) Note that for

To show that does not exist we investigate by computing the left- and right-hand derivatives.

Since the left and right limits are different, does not exist, that is, does not exist. Similarly,

does not exist. Therefore, f is not differentiable at 2 or at −2.

(b)

x2 − 4 < 0 x2 < 4 ⇔ |x| < 2 ⇔ −2 < x < 2. So

f(x) = f '(x) = = x2 − 4 if x ≤ −2

−x2 + 4 if −2 < x < 2

x2 − 4 if x ≥ 2

2x if x < −2−2x if −2 < x < 2

2x if x > 2

2x if |x| > 2−2x if |x| < 2

f '(2) lim h → 0

f(2 + h) − f(2)h

f '−(2) = = = (−4 − h) = −4 and

f '+(2) = = = = (4 + h) = 4.

lim h → 0−

f(2 + h) − f(2)h lim

h → 0−

[−(2 + h)2 + 4] − 0h

lim h → 0−

lim h → 0+

f(2 + h) − f(2)h lim

h → 0+

[(2 + h)2 − 4] − 0h

lim h → 0+

4h + h2

hlim

h → 0+

lim h → 0

f(2 + h) − f(2)h

f '(2)

f '(−2)

15. -Question Details SCalc7 2.3.099. [1874853]

Find the value of c such that the line is tangent to the curve

c =

Solution or Explanation

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y = x + 774

y = c .x

16. -Question Details SCalc7 2.4.002. [1874427]

Differentiate.

f '(x) =

Solution or Explanation

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f(x) = 5 sin xx

17. -Question Details SCalc7 2.4.005. [1875296]

Differentiate.

y' =

Solution or Explanation

Using the identity we can write alternative forms of the answer as

y = sec θ tan θ

y = sec θ tan θ y' = sec θ(sec2θ) + tan θ(sec θ tan θ) = sec θ(sec2 θ + tan2 θ).1 + tan2 θ = sec2 θ, sec θ(1 + 2 tan2 θ) or sec θ(2 sec2 θ − 1).

18. -Question Details SCalc7 2.4.011. [1874885]

Differentiate.

Solution or Explanation

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f(θ) = sec θ2 + sec θ

f '(θ) =

19. -Question Details SCalc7 2.4.025. [1875290]

(a) Find an equation of the tangent line to the curve at the point (π/2, 8π).

y =

y = 16x sin x

(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

Solution or Explanation

(a) and the

equation of the tangent line is

(b)

y = 16x sin x y' = 16(x cos x + sin x · 1). At , 8π ,π2

y' = 16 cos + sin = 16(0 + 1) = 16,π2

π2

π2

y − 8π = 16 x − , or y = 16x.π2

20. -Question Details SCalc7 2.4.029. [1874566]

Solution or Explanation

If H(θ) = θ sin θ, find H'(θ) and H''(θ).

H'(θ) =

H''(θ) =

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 SCalc7 2.4.034. [3043889]

Find the values of x on the curve at which the tangent is horizontal. (Let n be an integer. Enter your

answers as a comma-separated list.)

Solution or Explanation

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y = cos x2 + sin x

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 where t is in seconds and x is in centimeters.

(a) Find the velocity and acceleration at time t.

(b) Find the position, velocity, and acceleration of the mass at time t = 5π/6.

=

=

=

In what direction is it moving at that time?

Since ? 0, the particle is moving to the ---Select--- .

Solution or Explanation

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x(t) = 8 cos t,

v(t) =

a(t) =

x 5π6

v 5π6

a 5π6

v 5π6

23. -Question Details SCalc7 2.4.040. [1681299]

Find the limit.

Solution or Explanation

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lim x→0

sin 8xsin 9x

24. -Question Details SCalc7 2.4.041. [1681073]

Find the limit.

Solution or Explanation

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lim t→0

tan 8tsin 4t

25. -Question Details SCalc7 2.4.045. [1680875]

Find the limit.

Solution or Explanation

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lim θ→0

sin 7θθ + tan 4θ

26. -Question Details SCalc7 2.4.047. [1875243]

Find the limit.

Solution or Explanation

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lim x→π/4

8 − 8 tan xsin x − cos x

27. -Question Details SCalc7 2.5.001. [1874329]

Write the composite function in the form [Identify the inner function and the outer function

Find the derivative dy/dx.

Solution or Explanation

f(g(x)). u = g(x) y = f(u).]

y = 1 + 7x3

(g(x), f(u)) =

= dydx

Let u = g(x) = 1 + 7x and y = f(u) = . Then = = u−2/3 (7) = .u3 dydx

dydu

dudx

13

73 (1 + 7x)2

3

28. -Question Details SCalc7 2.5.012. [1927539]

Find the derivative of the function.

Solution or Explanation

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f(t) = 1 + tan t4

f '(t) =

29. -Question Details SCalc7 2.5.016. [1927629]

Find the derivative of the function.

Solution or Explanation

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y = 6 cot nθ

y' =

30. -Question Details SCalc7 2.5.017. [1874966]

Find the derivative of the function.

Solution or Explanation

f(x) = (2x − 5)4(x2 + x + 1)5

f '(x) =

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 SCalc7 2.5.021. [1875192]

Find the derivative of the function.

y' =

Solution or Explanation

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y = x2 + 6x2 − 6

4

32. -Question Details SCalc7 2.5.025. [1927541]

Find the derivative of the function.

Solution or Explanation

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F(z) = z − 9z + 9

F '(z) =

33. -Question Details SCalc7 2.5.027. [1875237]

Find the derivative of the function.

y' =

Solution or Explanation

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y = r

r2 + 5

34. -Question Details SCalc7 2.5.031. [1874572]

Find the derivative of the function.

y' =

Solution or Explanation

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y = sin(tan 5x)

35. -Question Details SCalc7 2.5.034. [1927703]

Find the derivative of the function.

Solution or Explanation

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y = x sin 9x

y'(x) =

36. -Question Details SCalc7 2.5.036. [1927577]

Find the derivative of the function.

Solution or Explanation

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f(t) = t

t2 + 3

f '(t) =

37. -Question Details SCalc7 2.5.040. [1874753]

Find the derivative of the function.

y = cos(cos(cos x))

y' =

Solution or Explanation

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38. -Question Details SCalc7 2.6.001. [1874671]

Consider the following equation.

(a) Find by implicit differentiation.

(b) Solve the equation explicitly for y and differentiate to get in terms of x.

Solution or Explanation

(a)

(b)

From part (a), which agrees with part (b).

5x2 − y2 = 4

y'

y' =

y'

y' = ±

(5x2 − y2) = (4) 10x − 2yy' = 0 2yy' = 10x y' = ddx

ddx

5xy

5x2 − y2 = 4 y2 = 5x2 − 4 y = ± , so y' = ± (5x2 − 4)−1/2(10x) = ± .5x2 − 4 12

5x

5x2 − 4

y' = = ,5xy

5x± 5x2 − 4

39. -Question Details SCalc7 2.6.005.MI. [1874651]

Find dy/dx by implicit differentiation.

Solution or Explanation

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x4 + y5 = 7

y' =

40. -Question Details SCalc7 2.6.007. [1874864]

Find dy/dx by implicit differentiation.

Solution or Explanation

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6x2 + 7xy − y2 = 9

y' =

41. -Question Details SCalc7 2.6.009. [1874484]

Find dy/dx by implicit differentiation.

Solution or Explanation

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x7(x + y) = y2(8x − y)

y' =

42. -Question Details SCalc7 2.6.013. [1874661]

Find dy/dx by implicit differentiation.

Solution or Explanation

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2 cos x sin y = 1

y' =

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.

Solution or Explanation

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y sec x = 7x tan y

x' =

44. -Question Details SCalc7 2.6.027. [1874628]

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

(ellipse)

y =

Solution or Explanation

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x2 + xy + y2 = 3, (1, 1)

45. -Question Details SCalc7 2.6.029. [1874903]

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

y =

Solution or Explanation

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x2 + y2 = (3x2 + 4y2 − x)2(0, 0.25)(cardioid)

Name (AID): E3_1 (6415722)Submissions Allowed: 2Category: ExamCode:Locked: YesAuthor: Frith, Russell ( afrgf@uaa.alaska.edu )Last Saved: Oct 2, 2014 07:55 AM AKDTPermission: ProtectedRandomization: PersonWhich graded: Last

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46. -Question Details SCalc7 2.6.030. [1874991]

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

y =

Solution or Explanation

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x2/3 + y2/3 = 4(−3 , 1)(astroid)

3

47. -Question Details SCalc7 2.6.032. [1874717]

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Solution or Explanation

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y2(y2 − 4) = x2(x2 − 5)(0, −2)(devil's curve)

y =

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