The derivative of `sec^(-1)((1)/(2x^(2)-1))` with respect to `sqrt(1-x^(2))" at "x=1`, is

Options

A. 2
B. -2
C. non-exisent
D. none of these

Explanation

To find the derivative of \( u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( v = \sqrt{1 - x^2} \) at \( x = 1 \), we will follow these steps:

### Step 1: Define the Variables
Let:
- \( u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \)
- \( v = \sqrt{1 - x^2} \)

### Step 2: Change of Variable
We can express \( x \) in terms of \( \theta \) by letting \( x = \cos(\theta) \). Then, \( \theta = \cos^{-1}(x) \).

### Step 3: Substitute \( x \)
Substituting \( x = \cos(\theta) \) into \( u \):
\[ u = \sec^{-1}\left(\frac{1}{2\cos^2(\theta) - 1}\right) \]
Using the identity \( 2\cos^2(\theta) - 1 = \cos(2\theta) \):
\[ u = \sec^{-1}\left(\frac{1}{\cos(2\theta)}\right) = 2\theta \]
Thus, we have:
\[ u = 2\cos^{-1}(x) \]

### Step 4: Differentiate \( u \) with respect to \( x \)
Now, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 2 \cdot \frac{d}{dx}(\cos^{-1}(x)) = 2 \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{2}{\sqrt{1 - x^2}} \]

### Step 5: Differentiate \( v \) with respect to \( x \)
Next, we differentiate \( v \):
\[ v = \sqrt{1 - x^2} \]
Differentiating gives:
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]

### Step 6: Find \( \frac{du}{dv} \)
Now, we can find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{-\frac{2}{\sqrt{1 - x^2}}}{-\frac{x}{\sqrt{1 - x^2}}} = \frac{2}{x} \]

### Step 7: Evaluate at \( x = 1 \)
Finally, we evaluate \( \frac{du}{dv} \) at \( x = 1 \):
\[ \frac{du}{dv} \bigg|_{x=1} = \frac{2}{1} = 2 \]

### Final Answer
Thus, the derivative of \( \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( \sqrt{1 - x^2} \) at \( x = 1 \) is \( \boxed{2} \).

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