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Q23E

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Essential Calculus: Early Transcendentals
Found in: Page 120
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the derivative of the function\(y = sin\left( {xcosx} \right)\)

The derivative of the function\(y = sin\left( {xcosx} \right)\)is \(cos\left( {xcos\left( x \right)} \right).\left( {cos\left( x \right) - xsin\left( x \right)} \right)\)

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Information

The given function is\(y = sin\left( {xcosx} \right)\). Its derivative is to be found out.

Step 2: Definition of Derivative

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value).

Step 3: Find the derivative

The derivative of the function\(y = sin\left( {xcosx} \right)\)is computed by using the chain rule,

Put \(t = xcosx\)

Now, by the product rule,

\(\begin{array}{c}\frac{{dt}}{{dx}} = \frac{d}{{dt}}\left( {x\cos x} \right)\\ = cos\left( x \right)\frac{d}{{dx}}\left( x \right) + x\frac{d}{{dx}}\left( {cos\left( x \right)} \right)\\ = cos\left( x \right) - xsin\left( x \right)\end{array}\)

Therefore \(y = sin\left( t \right)\)

Now by chain rule:

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}}\\ = \frac{d}{{dt}}\left( {sint} \right).\left( {cos\left( x \right) - xsin\left( x \right)} \right)\\ = cost.\left( {cos\left( x \right) - xsin\left( x \right)} \right)\end{array}\)

Put the value of \(t = xcosx\)

Therefore,

\(\frac{{dy}}{{dx}} = cos\left( {xcos\left( x \right)} \right).\left( {cos\left( x \right) - xsin\left( x \right)} \right)\)

Therefore, the derivative of the function\(y = sin\left( {xcosx} \right)\)is:

\(cos\left( {xcos\left( x \right)} \right).\left( {cos\left( x \right) - xsin\left( x \right)} \right)\)

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