If `y=xsinx`,then

Options

A. `(1)/(y)(dy)/(dx)=(1)/(x)+cotx`
B. `(dy)/(dx)=(1)/(x)+cotx`
C. `(1)/(y)(dy)/(dx)=(1)/(x)-cotx`
D. none of these

Explanation

To solve the problem where \( y = x \sin x \), we need to find the derivative \( \frac{dy}{dx} \) and simplify it to match one of the given options. Here’s the step-by-step solution:

### Step 1: Differentiate \( y = x \sin x \)
We will use the product rule for differentiation, which states that if \( y = u \cdot v \), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Here, let \( u = x \) and \( v = \sin x \).

### Step 2: Apply the product rule
Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = \cos x \)

Now applying the product rule:
\[ \frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 \]
Thus,
\[ \frac{dy}{dx} = x \cos x + \sin x \]

### Step 3: Rearranging the expression
We need to express \( \frac{dy}{dx} \) in a form that can be simplified. We know that \( y = x \sin x \), so we can express \( \sin x \) in terms of \( y \):
\[ \sin x = \frac{y}{x} \]
Substituting this back into our derivative:
\[ \frac{dy}{dx} = x \cos x + \frac{y}{x} \]

### Step 4: Factor out \( y \)
To match the form of the options given, we can multiply and divide \( \sin x \) in the expression:
\[ \frac{dy}{dx} = x \cos x + \frac{y}{x} \]
Now, we can rewrite \( x \cos x \) as \( x \cdot \frac{\cos x}{\sin x} \cdot \sin x \):
\[ \frac{dy}{dx} = x \cdot \cot x \cdot \sin x + \frac{y}{x} \]
This gives us:
\[ \frac{dy}{dx} = y \cdot \cot x + \frac{y}{x} \]

### Step 5: Divide by \( y \)
Now, we divide the entire equation by \( y \):
\[ \frac{1}{y} \frac{dy}{dx} = \cot x + \frac{1}{x} \]

### Final Result
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \cot x + \frac{1}{x} \]

### Conclusion
Now we can check which option matches this result. The answer is:
\[ \frac{1}{y} \frac{dy}{dx} = \cot x + \frac{1}{x} \]
This corresponds to option A.