If `y=cot^(-1) ("cosec x"+cot x),"then " dy/dx=`

Options

A. `-1`
B. ` 1`
C. ` (-1)/(2)`
D. ` (1)/(2) `

Answer

To solve the problem, we need to differentiate the function \( y = \cot^{-1}(\csc x + \cot x) \) with respect to \( x \). Let's go through the steps one by one.

### Step 1: Identify the derivative of \( y \)
We start with the formula for the derivative of \( \cot^{-1}(u) \):
\[ \frac{dy}{dx} = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \csc x + \cot x \).

### Step 2: Differentiate \( u = \csc x + \cot x \)
Now we need to find \( \frac{du}{dx} \):
- The derivative of \( \csc x \) is \( -\csc x \cot x \).
- The derivative of \( \cot x \) is \( -\csc^2 x \).

Thus,
\[ \frac{du}{dx} = -\csc x \cot x - \csc^2 x \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = -\frac{1}{1 + (\csc x + \cot x)^2} \cdot \left(-\csc x \cot x - \csc^2 x\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\csc x \cot x + \csc^2 x}{1 + (\csc x + \cot x)^2} \]

### Step 4: Simplify the denominator
Now we need to simplify the denominator:
\[ 1 + (\csc x + \cot x)^2 = 1 + (\csc^2 x + 2 \csc x \cot x + \cot^2 x) \]
Using the identity \( \csc^2 x - \cot^2 x = 1 \), we can rewrite this as:
\[ = \csc^2 x + 2 \csc x \cot x + 1 \]
Thus, we have:
\[ 1 + \cot^2 x = \csc^2 x \]
So,
\[ 1 + (\csc x + \cot x)^2 = 2 \csc^2 x + 2 \csc x \cot x \]

### Step 5: Final expression for \( \frac{dy}{dx} \)
Now substituting back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\csc x (\cot x + \csc x)}{2 (\csc^2 x + \csc x \cot x)} \]
After canceling \( \csc x + \cot x \) from the numerator and denominator, we are left with:
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]