a. Answer The value of d d x u . v is 13 . Explanation Given information: The value of u 0 = 0 , u ′ 0 = − 3 , v 0 = − 1 , v ′ 0 = 2 Concept used: The product rule is d d x u . v = u . d d x v + v d d x u Calculation: Differentiate with respect to x . d d x u . v = u 0 v ′ 0 + v 0 u ′ 0 = 5 × 2 + − 1 × − 3 = 10 + 3 = 13 Conclusion: The value of d d x u . v is 13 . b. To determine To calculate: The value of d d x u v , when x is differentiable at 0 Answer The value of d d x u v is − 7 Explanation Given information: The value of u 0 = 0 , u ′ 0 = − 3 , v 0 = − 1 , v ′ 0 = 2 . Concept used: The quotient formula is d d x u v = v . d d x u − u . d d x v v 2 . Calculation: Differentiate with respect to x . d d x u v = v 0 . u ′ 0 − u 0 . v ′ 0 v 2 0 = − 1 . − 3 − 5. 2 − 1 2 = 3 − 10 1 = − 7 Conclusion: The value of d d x u v is 7 25 c. To determine To calculate: The value of d d x v u , when x is differentiable at 0 Answer The value of d d x v u is − 7 Explanation Given information: The value of u 0 = 0 , u ′ 0 = − 3 , v 0 = − 1 , v ′ 0 = 2 . Concept used: The quotient formula is d d x v u = u . d d x v − v . d d x u u 2 . Calculation: Differentiate with respect to x . d d x v u = u 0 . v ′ 0 − v 0 . u ′ 0 u 2 0 = 5.2 − − 1 . − 3 5 2 = 10 − 3 25 = 7 25 Conclusion: The value of d d x v u is 7 25 d. To determine To calculate: The value of d d x 7 v − 2 u , when x is differentiable at 0 Answer The value of d d x 7 v − 2 u is 20 . Explanation Given information: The value of u 0 = 0 , u ′ 0 = − 3 , v 0 = − 1 , v ′ 0 = 2 . Calculation: Differentiate with respect to x . d d x 7 v − 2 u = 7 v ′ 0 − 2 u ′ 0 = 7 × 2 − 2 − 3 = 14 + 6 = 20 Conclusion: The value of d d x 7 v − 2 u is 20

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020

6th Edition

ISBN: 9781418300203

Author: Prentice Hall

Publisher: Prentice Hall

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Question

Chapter 2.3, Problem 23E

To determine

To calculate: The value of ddxu.v , when x is differentiable at 0

Expert Solution

Answer to Problem 23E

The value of ddxu.v is 13 .

Explanation of Solution

Given information:

The value of u0=0 , u′0=−3 , v0=−1 , v′0=2

Concept used:

The product rule is ddxu.v=u.ddxv+vddxu

Calculation:

Differentiate with respect to x .

ddxu.v=u0v′0+v0u′0=5×2+−1×−3=10+3=13

Conclusion: The value of ddxu.v is 13 .

To determine

To calculate: The value of ddxuv , when x is differentiable at 0

Expert Solution

Answer to Problem 23E

The value of ddxuv is −7

Explanation of Solution

Given information:

The value of u0=0 , u′0=−3 , v0=−1 , v′0=2 .

Concept used:

The quotient formula is ddxuv=v.ddxu−u.ddxvv2 .

Calculation:

Differentiate with respect to x .

ddxuv=v0.u′0−u0.v′0v20=−1.−3−5.2−12=3−101=−7

Conclusion: The value of ddxuv is 725

To determine

To calculate: The value of ddxvu , when x is differentiable at 0

Expert Solution

Answer to Problem 23E

The value of ddxvu is −7

Explanation of Solution

Given information:

The value of u0=0 , u′0=−3 , v0=−1 , v′0=2 .

Concept used:

The quotient formula is ddxvu=u.ddxv−v.ddxuu2 .

Calculation:

Differentiate with respect to x .

ddxvu=u0.v′0−v0.u′0u20=5.2−−1.−352=10−325=725

Conclusion: The value of ddxvu is 725

To determine

To calculate: The value of ddx7v−2u , when x is differentiable at 0

Expert Solution

Answer to Problem 23E

The value of ddx7v−2u is 20 .

Explanation of Solution

Given information:

The value of u0=0 , u′0=−3 , v0=−1 , v′0=2 .

Calculation:

Differentiate with respect to x .

ddx7v−2u=7v′0−2u′0=7×2−2−3=14+6=20

Conclusion: The value of ddx7v−2u is 20

Chapter 2.3, Problem 22E

Chapter 2.3, Problem 24E

Chapter 2 Solutions

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

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