`y=(x+sin x)/(x+cos x) , fi nd dy/dx `

Explanation

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

\[ \frac{dy}{dx} = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2} \]

### Step 1: Identify \( f(x) \) and \( g(x) \)
Here, we identify:
- \( f(x) = x + \sin x \)
- \( g(x) = x + \cos x \)

### Step 2: Differentiate \( f(x) \) and \( g(x) \)
Now we differentiate \( f(x) \) and \( g(x) \):
- \( f'(x) = \frac{d}{dx}(x + \sin x) = 1 + \cos x \)
- \( g'(x) = \frac{d}{dx}(x + \cos x) = 1 - \sin x \)

### Step 3: Apply the Quotient Rule
Now we 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 4: Simplify the Numerator
Now we simplify the numerator:

1. Expand \( (x + \cos x)(1 + \cos x) \):
\[ = x(1 + \cos x) + \cos x(1 + \cos x) = x + x \cos x + \cos x + \cos^2 x \]

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

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

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

5. Combine like terms:
\[ = x \cos x + x \sin x + \cos x - \sin x + \cos^2 x + \sin^2 x \]

6. Use the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):
\[ = x \cos x + x \sin x + \cos x - \sin x + 1 \]

### Step 5: Write the Final Derivative
Thus, the derivative is:

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