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Found in: Page 120

### Essential Calculus: Early Transcendentals

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

# 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 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$$\frac{{\cos x}}{{2\sqrt {\sin x} }}$$.