`y=(x+sin x)/(x+cos x)` find `dy/dx`

Explanation

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

\[ \frac{dy}{dx} = \frac{V \frac{dU}{dx} - U \frac{dV}{dx}}{V^2} \]

Where:
- \( U = x + \sin x \)
- \( V = x + \cos x \)

### Step 1: Differentiate \( U \) and \( V \)

First, we need to find \( \frac{dU}{dx} \) and \( \frac{dV}{dx} \).

1. **Differentiate \( U \)**:
\[ U = x + \sin x \]
\[ \frac{dU}{dx} = 1 + \cos x \]

2. **Differentiate \( V \)**:
\[ V = x + \cos x \]
\[ \frac{dV}{dx} = 1 - \sin x \]

### Step 2: Apply the Quotient Rule

Now we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{(x + \cos x)(1 + \cos x) - (x + \sin x)(1 - \sin x)}{(x + \cos x)^2} \]

### Step 3: Simplify the Expression

Now we will simplify the numerator:

1. **Expand the first term**:
\[ (x + \cos x)(1 + \cos x) = x(1 + \cos x) + \cos x(1 + \cos x) = x + x \cos x + \cos x + \cos^2 x \]

2. **Expand the second term**:
\[ (x + \sin x)(1 - \sin x) = x(1 - \sin x) + \sin x(1 - \sin x) = x - x \sin x + \sin x - \sin^2 x \]

3. **Combine the expansions**:
\[ \text{Numerator} = (x + x \cos x + \cos x + \cos^2 x) - (x - x \sin x + \sin x - \sin^2 x) \]

Distributing the negative sign:
\[ = x + x \cos x + \cos x + \cos^2 x - x + x \sin x - \sin x + \sin^2 x \]

Now, combine like terms:
\[ = x \cos x + x \sin x + \cos x + \cos^2 x - \sin x + \sin^2 x \]

Notice that \( \sin^2 x + \cos^2 x = 1 \):
\[ = x \cos x + x \sin x + \cos x - \sin x + 1 \]

### Step 4: Write the Final Derivative

Thus, we have:
\[ \frac{dy}{dx} = \frac{x \cos x + x \sin x + \cos x - \sin x + 1}{(x + \cos x)^2} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{x(\cos x + \sin x) + \cos x - \sin x + 1}{(x + \cos x)^2} \]