Find the derivatives of the following :
`x/sin^(n)x`

Answer

To find the derivative of the function \( y = \frac{x}{\sin^n x} \), we will apply the quotient rule of differentiation. The quotient rule states that if we have a function in the form \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = x \) and \( v = \sin^n x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = x \)
- \( v = \sin^n x \)

### Step 2: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
Now we need to find the derivatives of \( u \) and \( v \):
- \( \frac{du}{dx} = 1 \)
- To find \( \frac{dv}{dx} \), we will use the chain rule. Since \( v = \sin^n x \), we have:

\[ \frac{dv}{dx} = n \sin^{n-1} x \cdot \cos x \]

### Step 3: Apply the quotient rule
Now we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{\sin^n x \cdot 1 - x \cdot (n \sin^{n-1} x \cdot \cos x)}{(\sin^n x)^2} \]

### Step 4: Simplify the expression
This simplifies to:

\[ \frac{dy}{dx} = \frac{\sin^n x - n x \sin^{n-1} x \cos x}{\sin^{2n} x} \]

### Step 5: Factor out common terms
We can factor out \( \sin^{n-1} x \) from the numerator:

\[ \frac{dy}{dx} = \frac{\sin^{n-1} x \left( \sin x - n x \cos x \right)}{\sin^{2n} x} \]

### Step 6: Final expression
This can be rewritten as:

\[ \frac{dy}{dx} = \frac{\sin x - n x \cos x}{\sin^{n+1} x} \]

### Final Answer
Thus, the derivative of \( y = \frac{x}{\sin^n x} \) is:

\[ \frac{dy}{dx} = \frac{\sin x - n x \cos x}{\sin^{n+1} x} \]