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:
- \( u = \frac{x}{\sin x} \)
- \( v = \sin x \)

### Step 2: Differentiate \( u \) with respect to \( x \)
Using the quotient rule, we find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{\sin x \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2} \]
Calculating the derivatives:
- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(\sin x) = \cos x \)

Substituting these into the equation:
\[ \frac{du}{dx} = \frac{\sin x \cdot 1 - x \cdot \cos x}{(\sin x)^2} = \frac{\sin x - x \cos x}{\sin^2 x} \]

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

### Step 4: Use the Chain Rule to Find \( \frac{du}{dv} \)
According to the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]
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
We can simplify this expression:
\[ \frac{du}{dv} = \frac{\sin x - x \cos x}{\sin^2 x \cdot \cos x} \]

### Step 6: Final Simplification
We can separate the terms in the numerator:
\[ \frac{du}{dv} = \frac{\sin x}{\sin^2 x \cdot \cos x} - \frac{x \cos x}{\sin^2 x \cdot \cos x} \]
This simplifies to:
\[ \frac{du}{dv} = \frac{1}{\sin x \cdot \cos x} - \frac{x}{\sin^2 x} \]

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

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