If ` y=log (cosec x -cotx) ,then (dy)/(dx) `

Options

A. ` -cosec x `
B. ` cosec x `
C. ` -(cosec x -cotx ) `
D. ` cosec x -cotx `

Answer

To solve the problem \( y = \log(\csc x - \cot x) \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Differentiate using the chain rule
We start by applying the chain rule for differentiation. The derivative of \( \log(f(x)) \) is given by:

\[ \frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x) \]

In our case, \( f(x) = \csc x - \cot x \).

### Step 2: Find \( f'(x) \)
Now, we need to differentiate \( f(x) = \csc x - \cot x \).

The derivatives are:
- The derivative of \( \csc x \) is \( -\csc x \cot x \).
- The derivative of \( \cot x \) is \( -\csc^2 x \).

Thus, we have:

\[ f'(x) = -\csc x \cot x - (-\csc^2 x) = -\csc x \cot x + \csc^2 x \]

### Step 3: Substitute \( f(x) \) and \( f'(x) \) into the derivative formula
Now we substitute \( f(x) \) and \( f'(x) \) back into our derivative formula:

\[ \frac{dy}{dx} = \frac{1}{\csc x - \cot x} \cdot (-\csc x \cot x + \csc^2 x) \]

### Step 4: Simplify the expression
We can factor out \( \csc x \):

\[ \frac{dy}{dx} = \frac{-\csc x (\cot x - \csc x)}{\csc x - \cot x} \]

Now, notice that \( \cot x - \csc x = -(\csc x - \cot x) \). Thus, we can rewrite the expression:

\[ \frac{dy}{dx} = \frac{-\csc x (-(\csc x - \cot x))}{\csc x - \cot x} \]

This simplifies to:

\[ \frac{dy}{dx} = \csc x \]

### Final Answer
Thus, the derivative is:

\[ \frac{dy}{dx} = -\csc x \]