Find `(dy)/(dx)` for the functions: `y=(x+sinx)/(x+cosx)`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \frac{x + \sin x}{x + \cos x}\), we will use the quotient rule for differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then:

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

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

### Step-by-step Solution:

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

2. **Differentiate \(u\) and \(v\)**:
- \(\frac{du}{dx} = \frac{d}{dx}(x + \sin x) = 1 + \cos x\)
- \(\frac{dv}{dx} = \frac{d}{dx}(x + \cos x) = 1 - \sin x\)

3. **Apply the Quotient Rule**:
- Using the quotient rule:
\[ \frac{dy}{dx} = \frac{(x + \cos x)(1 + \cos x) - (x + \sin x)(1 - \sin x)}{(x + \cos x)^2} \]

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

5. **Combine the Terms**:
- Now substituting back into the numerator:
\[ \text{Numerator} = (x + x \cos x + \cos x + \cos^2 x) - (x - x \sin x + \sin x - \sin^2 x) \]
- Simplifying this:
\[ = x + x \cos x + \cos x + \cos^2 x - x + x \sin x - \sin x + \sin^2 x \]
- The \(x\) terms cancel out:
\[ = x \cos x + x \sin x + \cos x + \cos^2 x - \sin x + \sin^2 x \]

6. **Combine Like Terms**:
- Noticing that \(\cos^2 x + \sin^2 x = 1\):
\[ = x(\cos x + \sin x) + \cos x - \sin x + 1 \]

7. **Final Expression**:
- Therefore, we have:
\[ \frac{dy}{dx} = \frac{x(\cos x + \sin x) + \cos x - \sin x + 1}{(x + \cos x)^2} \]

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