BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.3, Problem 27E
To determine

To find: The value of H(θ) and H(θ).

Expert Solution

Answer to Problem 27E

The value of H(θ)=θcosθ+sinθ and the value of H(θ)=θsinθ+2cosθ.

Explanation of Solution

Given:

The function is H(θ)=θsinθ.

Formula used:

Product Rule:

If f(θ) and g(θ) are differentiable functions, then the product rule is,

ddθ[f(θ)g(θ)]=f(θ)ddθ[g(θ)]+g(θ)ddθ[f(θ)] (1)

Power Rule:

If n is positive integer, then the power rule is,

ddx(xn)=nxn1 (2)

Sum Rule:

If f(θ) and g(θ) are differentiable function, then the sum rule is,

ddθ[f(θ)+g(θ)]=ddθ[f(θ)]+ddθ[g(θ)] (3)

Calculation:

Obtain the first derivative of H(θ) that is H(θ) as follows.

Apply Product Rule as shown in equation (1).

Substitute θ for f(θ) and sinθ for g(θ) in equation (1) as,

ddθ[θsinθ]=θddθ[sinθ]+sinθddθ[θ]

Apply Power Rule as shown in equation (2) as,

ddθ[θsinθ]=θ(cosθ)+sinθ(θ11)      =θcosθ+sinθ

Thus, the first derivative of H(θ) is H(θ)=θcosθ+sinθ.

Obtain the second derivative of H(θ) that is H(θ).

Apply Sum Rule as shown in equation (3).

ddθ[θcosθ+sinθ]=ddθ[θcosθ]+ddθ[sinθ]

Apply Product Rule as shown in equation (1).

ddθ[θcosθ+sinθ]=[θddθ[cosθ]+cosθddθ[θ]]+ddθ[sinθ]=[θ(sinθ)+cosθ(1θ11)]+[cosθ]=[θsinθ+cosθ]+[cosθ]=θsinθ+2cosθ

Thus, the second derivative of H(θ) is H(θ)=θsinθ+2cosθ.

Therefore, the value of H(θ)=θcosθ+sinθ and the value of H(θ)=θsinθ+2cosθ.

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