   Chapter 3.4, Problem 47E

Chapter
Section
Textbook Problem

Find y′ and y″.y = cos(sin 3θ)

To determine

To find: The first and second derivatives of y. That is,y and y.

Explanation

Given:

The function is y=cos(sin3θ).

Result used: Chain Rule:

If h is differentiable at θ and g is differentiable at h(x), then the composite function F=gh defined by F(θ)=g(h(θ)) is differentiable at θ and F is given by the product

F(θ)=g(h(θ))h(θ) (1)

Product Rule:

If f(θ). and g(θ) are both differentiable function, then

ddθ[f(θ)g(θ)]=f(θ)ddθ[g(θ)]+g(θ)ddθ[f(θ)] (2)

Calculation:

Obtain the first derivative of y.

y=ddθ(y)=ddθ(cos(sin3θ))

Let h(θ)=sin3θ and g(u)=cosu  where u=h(θ).

Apply the chain rule as shown in equation (1),

y=g(h(θ))h(θ) (3)

The derivative g(h(θ)) is computed as follows,

g(h(θ))=g(u)=ddu(g(u))=ddu(cosu)=sinu

Substitute u=sin3θ in the above equation,

g(h(θ))=sin(sin3θ)

Thus, the derivative g(h(θ)) is g(h(θ))=sin(sin3θ).

The derivative of h(θ) is computed as follows,

h(θ)=ddθ(sin3θ)=ddθ(sin3θ)(ddθ(3θ))=cos3θ(3)=3cos3θ

Thus, the derivative of h(θ) is h(θ)=3cos3θ.

Substitute sin(sin3θ) for g(h(θ)) and 3cos3θ for h(θ) in equation (3),

g(h(θ))h(θ)=sin(sin3θ)(3cos3θ)=3cos3θsin(sin3θ)

Therefore, the derivative of y=cos(sin3θ) is y=3cos3θsin(sin3θ)_.

Obtain the second derivative of y.

y=ddθ(y)=ddθ(3cos3θsin(sin3θ)_)

\

Apply the product rule as shown in equation (2),

y=3cos3θddθ(sin(sin3θ))+sin(sin3θ)ddθ(3cos3θ)=3cos3θddθ(sin(sin3θ))3sin(sin3θ)(sin3θ3)

=3cos3θddθ(sin(sin3θ))+9sin(sin3θ)(sin3θ) (4)

Obtain the derivative ddθ(sin(sin3θ))

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