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

Explanation

To differentiate \( \frac{x}{\sin x} \) with respect to \( \sin x \), we will use the chain rule and the quotient rule. Here’s the step-by-step solution:

### Step 1: Define the Variables
Let:
- \( u = x \)
- \( v = \sin x \)

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

### Step 2: Find \( \frac{du}{dx} \)
Since \( u = x \), we differentiate with respect to \( x \):
\[ \frac{du}{dx} = 1 \]

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

### Step 4: Find \( \frac{du}{dv} \)
Using the chain rule, we can express \( \frac{du}{dv} \) as:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{\cos x} \]

### Step 5: Differentiate \( \frac{x}{\sin x} \) using the Quotient Rule
Now, we differentiate \( \frac{x}{\sin x} \) with respect to \( x \) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2} \]
Calculating the derivatives:
- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(\sin x) = \cos x \)

Substituting these into the quotient rule:
\[ \frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x \cdot 1 - x \cdot \cos x}{(\sin x)^2} = \frac{\sin x - x \cos x}{(\sin x)^2} \]

### Step 6: Express \( \frac{du}{dv} \)
Now, we can express \( \frac{du}{dv} \) in terms of \( \frac{d}{dx}\left(\frac{x}{\sin x}\right) \):
\[ \frac{du}{dv} = \frac{\frac{d}{dx}\left(\frac{x}{\sin x}\right)}{\frac{dv}{dx}} = \frac{\frac{\sin x - x \cos x}{(\sin x)^2}}{\cos x} \]

### Step 7: Simplify
This simplifies to:
\[ \frac{du}{dv} = \frac{\sin x - x \cos x}{(\sin x)^2 \cos x} \]

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