`((sinx - x cosx)/(x sinx + cos x ))`

Answer

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

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

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

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = \sin x - x \cos x \)
- \( v = x \sin x + \cos x \)

### Step 2: Differentiate \( u \) and \( v \)

Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

**Differentiating \( u \):**
\[ \frac{du}{dx} = \frac{d}{dx}(\sin x) - \frac{d}{dx}(x \cos x) \]
Using the product rule on \( x \cos x \):
\[ \frac{d}{dx}(x \cos x) = \cos x + x(-\sin x) = \cos x - x \sin x \]
Thus,
\[ \frac{du}{dx} = \cos x - (\cos x - x \sin x) = x \sin x \]

**Differentiating \( v \):**
\[ \frac{dv}{dx} = \frac{d}{dx}(x \sin x) + \frac{d}{dx}(\cos x) \]
Using the product rule on \( x \sin x \):
\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x \]
Thus,
\[ \frac{dv}{dx} = (\sin x + x \cos x) - \sin x = x \cos x \]

### Step 3: Apply the Quotient Rule

Now we can apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(x \sin x + \cos x)(x \sin x) - (\sin x - x \cos x)(x \cos x)}{(x \sin x + \cos x)^2} \]

### Step 4: Simplify the Expression

Expanding the numerator:
1. First term: \( (x \sin x + \cos x)(x \sin x) = x^2 \sin^2 x + x \sin x \cos x \)
2. Second term: \( (\sin x - x \cos x)(x \cos x) = x \sin x \cos x - x^2 \cos^2 x \)

Combining these:
\[ \text{Numerator} = x^2 \sin^2 x + x \sin x \cos x - (x \sin x \cos x - x^2 \cos^2 x) \]
This simplifies to:
\[ x^2 \sin^2 x + x^2 \cos^2 x = x^2 (\sin^2 x + \cos^2 x) \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \text{Numerator} = x^2 \]

### Final Result

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