MA 302 Grantham University Calculus Differentiations Questions

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Grantham University

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1. –/5 points LARCALC11 2.4.013. My Notes Ask Your Teacher My Notes Ask Your Teacher My Notes Ask Your Teacher My Notes Ask Your Teacher Find the derivative of the function. h(s) = −2 5s2 + 6 h'(s) = −10s√5s2+6 2. –/5 points LARCALC11 2.4.027. Find the derivative of the function. y'(x) = 3. –/5 points LARCALC11 2.4.037. Find the derivative of the trigonometric function. g(x) = 4 tan(5x) g'(x) = 4. –/5 points LARCALC11 2.5.010. Find dy/dx by implicit differentiation. 4 x**2y + 9 y**2x = -1 dy/dx = –/5 points 5. LARCALC11 2.5.025. Find dy/dx by implicit differentiation. Then find the slope of the graph at the given point. xy = 35, dy dx = At (-7, -5): y' = (-7, -5) My Notes Ask Your Teacher LARCALC11 2.5.055. –/5 points 6. Find an equation of the tangent line to the graph at the given point. x+ y = 4, (9, 1) y= Use a graphing utility to graph the equation and the tangent line in the same viewing window. My Notes Ask Your Teacher 7. LARCALC11 2.6.007. –/5 points My Notes A point is moving along the graph of the given function at the rate dx/dt. Find dy/dt for the given values of x. y = 4x2 + 3; (a) dx dt = 3 centimeters per second x = −1 cm/sec (b) x =0 cm/sec (c) x =1 cm/sec Ask Your Teacher 8. –/5 points LARCALC11 2.6.011. My Notes Ask Your Teacher The radius r of a circle is increasing at a rate of 5 centimeters per minute. Find the rate of change of the area when r = 34 centimeters. cm2/min 9. –/5 points LARCALC11 2.6.017.MI. My Notes Ask Your Teacher At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 20 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 2 feet 1 high? (Hint: The formula for the volume of a cone is V = πr2h.) 3 h' = ft/min 10. LARCALC11 2.4.076. –/5 points My Notes Ask Your Teacher Consider the following. Function y = sin 3x Point π 4 , 2 2 (a) Find an equation of the tangent line to the graph of the function at the given point. (Let x be the independent variable and y be the dependent variable.) (b) Use a graphing utility to graph the function and its tangent line at the point. Submit Assignment Home Save Assignment Progress My Assignments Copyright Request Extension 2020 Cengage Learning, Inc. All Rights Reserved
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QUESTION 1.
𝒅
(−𝟐√𝟓𝒔𝟐 + 𝟔)
𝒅𝒔
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
= −2 (√5𝑠 2 + 6)
𝑑𝑠
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=

𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = √𝑢, 𝑢 = 5𝑠 2 + 6
𝑑
𝑑
= −2
(√𝑢) (5𝑠 2 + 6)
𝑑𝑢
𝑑𝑠
1
= −2 ⋅
⋅ 10𝑠
2 √𝑢
Substitute back 𝑢 = 5𝑠 2 + 6
1
= −2 ⋅
⋅ 10𝑠
2√5𝑠 2 + 6
=−

1 ⋅ 2 ⋅ 10𝑠
2√5𝑠 2 + 6

=−

𝟏𝟎𝒔
√𝟓𝒔𝟐 + 𝟔

𝑑
1
(√𝑢) =
𝑑𝑢
2 √𝑢
𝑑
(5𝑠 2 + 6) = 10𝑠
𝑑𝑠

QUESTION 2.
𝒅
𝟐𝒙
(
)
𝒅𝒙 √𝒙𝟐 + 𝟗
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
𝑥
=2 (
)
𝑑𝑥 √𝑥 2 + 9
𝑓
𝑓 ′ ⋅ 𝑔 − 𝑔′
Apply the Quotient Rule: ( )′ =
𝑔
𝑔2
𝑑
𝑑
(𝑥)√𝑥 2 + 9 −
(√𝑥 2 + 9)𝑥
𝑑𝑥
𝑑𝑥
= 2⋅
(√𝑥 2 + 9)2
=

1

⋅ 2𝑥
2√𝑢
Substitute back 𝑢 = 𝑥 2 + 9
1
=
⋅ 2𝑥
2√𝑥 2 + 9
=

𝑥
√𝑥 2 + 9

1 ⋅ √𝑥 2 + 9 −
= 2⋅

=

𝑥
+9

√𝑥 2

(√𝑥 2 + 9)2
𝟏𝟖
(𝒙𝟐 + 𝟗)√𝒙𝟐 + 𝟗

𝑑
(𝑥) = 1
𝑑𝑥

𝑑
(√𝑥 2 + 9)
𝑑𝑥
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=

𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = √𝑢, 𝑢 = 𝑥 2 + 9
𝑑
𝑑
=
(√𝑢) (𝑥 2 + 9)
𝑑𝑢
𝑑𝑥
𝑑
1
(√𝑢) =
𝑑𝑢
2√𝑢
𝑑 2
(𝑥 + 9)
𝑑𝑥
Apply the Sum/Difference Rule: (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔
𝑑 2
𝑑
=
(𝑥 ) +
(9)
𝑑𝑥
𝑑𝑥
𝑑 2
(𝑥 ) = 2𝑥
𝑑𝑥
𝑑
(9) = 0
𝑑𝑥
𝑑 2
(𝑥 + 9) = 2𝑥
𝑑𝑥

QUESTION 3

𝒅
(𝟒 𝐭𝐚𝐧(𝟓𝒙))
𝒅𝒙
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
= 4 (tan(5𝑥))
𝑑𝑥
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=

𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = tan (𝑢), 𝑢 = 5𝑥
𝑑
𝑑
=4
(tan (𝑢)) (5𝑥)
𝑑𝑢
𝑑𝑥
= 4sec 2 (𝑢) ⋅ 5
Substitute back 𝑢 = 5
= 4sec 2 (5𝑥) ⋅ 5
Simplify
= 𝟐𝟎𝐬𝐞𝐜 𝟐 (𝟓𝒙)

𝑑
(tan (𝑢)) = sec 2 (𝑢)
𝑑𝑢
𝑑
(5𝑥) = 5
𝑑𝑥

QUESTION 4

𝑑
(4𝑥 2 𝑦 + 9𝑦 2 𝑥)
𝑑𝑥
𝒅
𝒅
(𝟒𝒙𝟐 𝒚 + 𝟗𝒚𝟐 𝒙) =
(−𝟏)
Apply the Sum/Difference Rule: (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔′
𝒅𝒙
𝒅𝒙
𝑑
𝑑
=
(4𝑥 2 𝑦) +
(9𝑦 2 𝑥)
𝑑
𝑑𝑥
𝑑𝑥
(−1) = 0
𝑑
𝑑
𝑑𝑥
(4𝑥 2 𝑦) = 4(2𝑥𝑦 +
(𝑦)𝑥 2 )
𝑑
𝑑
𝑑
𝑑𝑥
𝑑𝑥
(−1) = 0 = 4(2𝑥𝑦 +
(𝑦)𝑥 2 ) + 9(2𝑥𝑦 (𝑦) + 𝑦 2 )
𝑑
𝑑
𝑑𝑥
𝑑...


Anonymous
Excellent resource! Really helped me get the gist of things.

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