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 then manipulate the result to match one of the given options. Here’s a step-by-step solution:

### Step 1: Identify the function
We are given:
\[ y = x \sin x \]

### Step 2: Apply the product rule
Since \( y \) is a product of two functions \( u = x \) and \( v = \sin x \), we will use the product rule for differentiation:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
where \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \cos x \).

### Step 3: Differentiate using the product rule
Applying the product rule:
\[ \frac{dy}{dx} = x \cdot \cos x + \sin x \cdot 1 \]
This simplifies to:
\[ \frac{dy}{dx} = x \cos x + \sin x \]

### Step 4: Rearranging the derivative
To match the form of the options provided, we can manipulate the expression. We will express \( \frac{dy}{dx} \) in terms of \( y \):
\[ y = x \sin x \implies \frac{dy}{dx} = x \cos x + \sin x \]

### Step 5: Divide by \( y \)
Now, we divide both sides by \( y \):
\[ \frac{1}{\frac{dy}{dx}} = \frac{y}{x \cos x + \sin x} \]
Substituting \( y = x \sin x \):
\[ \frac{1}{\frac{dy}{dx}} = \frac{x \sin x}{x \cos x + \sin x} \]

### Step 6: Simplifying the expression
Now, we can simplify the right-hand side:
\[ \frac{1}{\frac{dy}{dx}} = \frac{x \sin x}{x \cos x + \sin x} = \frac{\sin x}{\frac{x \cos x}{x \sin x} + 1} = \frac{\sin x}{\cot x + 1} \]
This can be further simplified to:
\[ \frac{1}{\frac{dy}{dx}} = \frac{1}{x} + \cot x \]

### Step 7: Final result
Thus, we find that:
\[ \frac{1}{\frac{dy}{dx}} = \frac{1}{x} + \cot x \]

### Conclusion
The correct option is:
\[ \frac{1}{\frac{dy}{dx}} = \frac{1}{x} + \cot x \]