If `y=(sinx)/(x+cosx)`, then find `(dy)/(dx)`.

Options

A. `(xcosx-sinx+1)/((x+cosx)^2)`
B. `(xcosx-sinx+1)/((x-cosx)^2)`
C. `(xcosx-sinx-1)/((x+cosx)^2)`
D. `(xcosx-sinx+1)/((x+cosx))`

Explanation

To find the derivative of the function \( y = \frac{\sin x}{x + \cos x} \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = \sin x \) and \( v = x + \cos x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \sin x \)
- \( v = x + \cos x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we differentiate \( u \) and \( v \):
- \( \frac{du}{dx} = \cos x \)
- \( \frac{dv}{dx} = 1 - \sin x \) (since the derivative of \( x \) is \( 1 \) and the derivative of \( \cos x \) is \( -\sin x \))

### Step 3: Apply the Quotient Rule
Now we apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(x + \cos x)(\cos x) - (\sin x)(1 - \sin x)}{(x + \cos x)^2} \]

### Step 4: Simplify the Numerator
Now we simplify the numerator:
1. Expand the first term:
\[ (x + \cos x)(\cos x) = x \cos x + \cos^2 x \]
2. Expand the second term:
\[ \sin x(1 - \sin x) = \sin x - \sin^2 x \]
3. Combine these:
\[ x \cos x + \cos^2 x - (\sin x - \sin^2 x) = x \cos x + \cos^2 x - \sin x + \sin^2 x \]

### Step 5: Use the Pythagorean Identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \cos^2 x + \sin^2 x = 1 \]
Thus, the numerator simplifies to:
\[ x \cos x + 1 - \sin x \]

### Final Expression
Putting it all together, we have:
\[ \frac{dy}{dx} = \frac{x \cos x + 1 - \sin x}{(x + \cos x)^2} \]

### Summary
The derivative of \( y = \frac{\sin x}{x + \cos x} \) is:
\[ \frac{dy}{dx} = \frac{x \cos x + 1 - \sin x}{(x + \cos x)^2} \]