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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

Write the composite function in the form\(f\left( {g\left( x \right)} \right)\). (Identify the inner function\(u = g\left( x \right)\) and the outer function\(y = f\left( u \right)\).) Then find the derivative\(dy/dx\).

\(y = \sqrt {\sin x} \)

The derivative\(dy/dx\)of \(y = \sqrt {\sin x} \)is\(\frac{{\cos x}}{{2\sqrt {\sin x} }}\).

See the step by step solution

Step by Step Solution

Step 1: Given Information.

The given function is\(y = \sqrt {\sin x} \).

Step 2: Definition of Derivative of a Function

The derivative of a function is defined as the rate at which a function changes in relation to a variable.

Step 3: To write the given function in the form of \(f\left( {g\left( x \right)} \right)\)

Consider\(y = \sqrt {\sin x} \)

Let \(y = f\left( x \right)\)

So, \(f\left( x \right) = \sqrt {\sin x} \)

Let\(u = \sin {\rm{x}}\)

So,\(f\left( u \right) = \sqrt u \)

Apply chain rule

\(\begin{array}{l}\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} \times \frac{{dy}}{{du}}\\\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\sin x} \right) \times \frac{d}{{du}}\left( {\sqrt u } \right)\\\frac{{dy}}{{dx}} = \cos x \times \frac{1}{{2\sqrt u }}\end{array}\)

Substitute \(u = \sin x\)in the above equation.

\(\begin{array}{l}\frac{{dy}}{{dx}} = \cos x \times \frac{1}{{2\sqrt {\sin x} }}\\\frac{{dy}}{{dx}} = \frac{{\cos x}}{{2\sqrt {\sin x} }}\end{array}\)

Therefore, the derivative \(dy/dx\)of the function \(y = \sqrt {\sin x} \)is\(\frac{{\cos x}}{{2\sqrt {\sin x} }}\).

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