If `y=cos^(-1)((1-x^2)/(1+x^2)) " then at " x=3/2, (dy)/(dx)` is

Options

A. `-8//13`
B. `8//13`
C. `5//12`
D. `-5//12`

Answer

To solve the problem, we need to find the derivative \( \frac{dy}{dx} \) for the function \( y = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) at \( x = \frac{3}{2} \).

### Step-by-Step Solution:

1. **Substitute \( x \) with \( \tan(\theta) \)**:
\[ x = \tan(\theta) \]
This gives us:
\[ y = \cos^{-1}\left(\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}\right) \]

2. **Use the trigonometric identity**:
We know that:
\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \]
Therefore, we can rewrite \( y \) as:
\[ y = \cos^{-1}(\cos(2\theta)) = 2\theta \]
(Note: This holds true when \( 2\theta \) is in the range of \( \cos^{-1} \)).

3. **Express \( \theta \) in terms of \( x \)**:
Since \( x = \tan(\theta) \), we have:
\[ \theta = \tan^{-1}(x) \]
Thus:
\[ y = 2\tan^{-1}(x) \]

4. **Differentiate \( y \) with respect to \( x \)**:
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} \]

5. **Substitute \( x = \frac{3}{2} \)**:
Now we need to evaluate \( \frac{dy}{dx} \) at \( x = \frac{3}{2} \):
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + \left(\frac{3}{2}\right)^2} \]
Calculate \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \):
\[ 1 + \frac{9}{4} = \frac{4}{4} + \frac{9}{4} = \frac{13}{4} \]
Thus:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{\frac{13}{4}} = 2 \cdot \frac{4}{13} = \frac{8}{13} \]

### Final Answer:
\[ \frac{dy}{dx} \text{ at } x = \frac{3}{2} \text{ is } \frac{8}{13} \]