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

Explanation

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

### Step 1: Define the functions
Let:
- \( u = \frac{x}{\sin x} \)
- \( v = \sin x \)

We need to find \( \frac{du}{dv} \).

### Step 2: Differentiate \( u \) with respect to \( x \)
Using the quotient rule, which states that if \( u = \frac{f(x)}{g(x)} \), then \( \frac{du}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \), we have:
- \( f(x) = x \) and \( g(x) = \sin x \)
- \( f'(x) = 1 \) and \( g'(x) = \cos x \)

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

### Step 3: Differentiate \( v \) with respect to \( x \)
Since \( v = \sin x \):
\[ \frac{dv}{dx} = \cos x \]

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

### Step 5: Simplify the expression
This can be simplified:
\[ \frac{du}{dv} = \frac{\sin x - x \cos x}{\sin^2 x \cos x} \]

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