   Chapter 3.6, Problem 4E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Differentiate the function.f(x) = ln(sin2x)

To determine

To find: The derivative of the function f(x)=ln(sin2x).

Explanation

Given:

The function f(x)=ln(sin2x).

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 f(x).

f(x)=ddx(f(x))=ddx(ln(sin2x))

Let h(x)=sin2x and g(u)=lnu  where u=h(x)

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

f(x)=g(h(x))h(x) (2)

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

g(h(x))=g(u)=ddu(g(u))=ddu(lnu)=1u

Substitute u=sin2x in the above equation,

g(h(x))=1sin2x

Thus, the derivative g(h(x)) is g(h(x))=1sin2x

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 