Differentiate `(x)/(sinx)` w.r.t . `sinx`.

Explanation

To differentiate the function \( \frac{x}{\sin x} \) with respect to \( \sin x \), we will follow these steps:

### Step 1: Define the functions
Let \( u = \frac{x}{\sin x} \) and \( v = \sin x \). We need to find \( \frac{du}{dv} \).

### Step 2: Find \( \frac{du}{dx} \)
We will use the quotient rule for differentiation, which states that if \( u = \frac{f(x)}{g(x)} \), then:
\[ \frac{du}{dx} = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2} \]
Here, \( f(x) = x \) and \( g(x) = \sin x \).

Calculating the derivatives:
- \( f'(x) = 1 \)
- \( g'(x) = \cos x \)

Now applying the quotient rule:
\[ \frac{du}{dx} = \frac{\sin x \cdot 1 - x \cdot \cos x}{(\sin x)^2} \]
Thus,
\[ \frac{du}{dx} = \frac{\sin x - x \cos x}{\sin^2 x} \]

### Step 3: Find \( \frac{dv}{dx} \)
The derivative of \( v = \sin x \) is:
\[ \frac{dv}{dx} = \cos x \]

### Step 4: Find \( \frac{du}{dv} \)
Using the chain rule, we have:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]
Substituting the values we found:
\[ \frac{du}{dv} = \frac{\frac{\sin x - x \cos x}{\sin^2 x}}{\cos x} \]
This simplifies to:
\[ \frac{du}{dv} = \frac{\sin x - x \cos x}{\sin^2 x \cdot \cos x} \]

### Step 5: Final simplification
We can further simplify:
\[ \frac{du}{dv} = \frac{\sin x}{\sin^2 x \cdot \cos x} - \frac{x \cos x}{\sin^2 x \cdot \cos x} \]
This simplifies to:
\[ \frac{du}{dv} = \frac{1}{\sin x \cdot \cos x} - \frac{x}{\sin^2 x} \]

### Final Answer
Thus, the derivative of \( \frac{x}{\sin x} \) with respect to \( \sin x \) is:
\[ \frac{du}{dv} = \frac{\sin x - x \cos x}{\sin^2 x \cdot \cos x} \]