Chapter 3, Problem 50RE

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Calculate y'.50. y = sin 2 ( cos sin π x )

To determine

To calculate: The derivative of y.

Explanation

Given:

The function is y=sin2(cossinπx).

Result used: Chain Rule

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

F(x)=g(h(x))h(x) (1)

Calculation:

Obtain the derivative of y.

y=ddx(y)=ddx(sin2(cossinπx))

Let h(x)=cossinπx and g(u)=sin2u  where u=h(x).

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

y=g(h(x))h(x) (2)

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

g(h(x))=g(u)=ddu(g(u))=ddu(sin2u)=ddu(sinu)2

Simplify further and obtain the derivative,

g(h(x))=2(sinu)21ddx(sinu)=2(sinu)(cosu)=2sinucosu

Substitute u=cossinπx in the above equation,

g(h(x))=2sin(cossinπx)cos(cossinπx)

Therefore, the derivative g(h(x)) is g(h(x))=2sin(cossinπx)cos(cossinπx) .

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

h(x)=ddx(cossinπx)

Let s(x)=sinπx and k(u)=cosu  where u=s(x) .

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

h(x)=k(s(x))s(x) (3)

The derivative k(s(x)) is computed as follows,

k(s(x))=k(u)=ddu(k(u))=ddu(cosu)=sinu

Substitute u=sinπx in the above equation,

k(s(x))=sin(sinπx)

Thus, the derivative k(s(x)) is k(s(x))=sin(sinπx)

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