If `y=cot^(-1)(cotx),` then `(dy)/(dx)` is

Options

A. `1,x"inR`
B. `1,x"inR-{npi}`
C. `{{:(1",",x inR-{npi}),("does not exist", x in{npi},nin "integer" ):}`
D. None of these

Answer

To find the derivative of \( y = \cot^{-1}(\cot x) \), we will follow these steps:

### Step 1: Understand the function
The function \( y = \cot^{-1}(\cot x) \) is not simply equal to \( x \) for all values of \( x \). The cotangent inverse function has a restricted range, which we need to consider.

### Step 2: Differentiate using the chain rule
To differentiate \( y = \cot^{-1}(\cot x) \), we will use the chain rule. The derivative of \( \cot^{-1}(u) \) with respect to \( u \) is given by:
\[ \frac{dy}{du} = -\frac{1}{1 + u^2} \]
where \( u = \cot x \).

### Step 3: Differentiate \( u = \cot x \)
Next, we need to find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = -\csc^2 x \]

### Step 4: Apply the chain rule
Now, we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\frac{1}{1 + \cot^2 x} \cdot (-\csc^2 x) \]

### Step 5: Simplify the expression
Recall that \( 1 + \cot^2 x = \csc^2 x \). Thus, we can simplify:
\[ \frac{dy}{dx} = \frac{\csc^2 x}{\csc^2 x} = 1 \]

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

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