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

Options

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

Answer

To solve the problem step by step, we start with the given equation:

**Step 1: Rewrite the function**
Given:
\[ y = \cos^{-1} \left( \frac{x - x^{-1}}{x + x^{-1}} \right) \]

We can simplify the expression inside the cosine inverse.

**Step 2: Simplify the expression**
The expression can be rewritten as:
\[ \frac{x - \frac{1}{x}}{x + \frac{1}{x}} = \frac{x^2 - 1}{x^2 + 1} \]

Thus, we have:
\[ y = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right) \]

**Step 3: Use the identity for cosine inverse**
We know that:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
So we can rewrite:
\[ y = \cos^{-1} \left( -\frac{1 - x^2}{1 + x^2} \right) = \pi - \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]

**Step 4: Use the double angle identity**
We also know that:
\[ \tan^{-1}(x) = \frac{1 - x^2}{1 + x^2} \]
Thus:
\[ y = \pi - 2 \tan^{-1}(x) \]

**Step 5: Differentiate y with respect to x**
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - 2 \cdot \frac{d}{dx} \tan^{-1}(x) \]

Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2} \]

So:
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1 + x^2} \]

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

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