If `y=cos^(-1)((x-x^(-1))/(x+x^(-1))),` then `(dy)/(dx)` is

Options

A. `{:{((2)/(1+x^(2))",",xgt0),(-(2)/(1+x^(2))",",xlt0):}`
B. `{:{((2)/(1+x^(2))",",xgt0),("does not exist"",", x=0),((-2)/(1+x^(2))",",xlt0):}`
C. `{:{((2)/(1+x^(2))",",xlt0),("does not exist"",", x=0),((-2)/(1+x^(2))",",xgt0):}`
D. None of these

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \cos^{-1}\left(\frac{x - \frac{1}{x}}{x + \frac{1}{x}}\right) \), we will follow these steps:

### Step 1: Simplify the argument of the inverse cosine function
We start with the expression inside the inverse cosine:
\[ \frac{x - \frac{1}{x}}{x + \frac{1}{x}} = \frac{x^2 - 1}{x^2 + 1} \]
This is obtained by multiplying the numerator and the denominator by \( x \).

### Step 2: Rewrite the function
Now we can rewrite \( y \):
\[ y = \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) \]

### Step 3: Use the identity for inverse cosine
Using the identity \( \cos^{-1}(-\theta) = \pi - \cos^{-1}(\theta) \), we can express \( y \) as:
\[ y = \pi - \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 4: Use the known formula for inverse cosine
We know that:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2 \tan^{-1}(x) \]
Thus, we can express \( y \) as:
\[ y = \pi - 2 \tan^{-1}(x) \]

### Step 5: Differentiate \( y \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
We get:
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1 + x^2} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = -\frac{2}{1 + x^2} \]

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