The derivative of `sec^(-1)((1)/(2x^(2)-1))` with respect to `sqrt(1-x^(2))" at "x=0`, is

Options

A. 2
B. -2
C. 1
D. none of these

Answer

To find the derivative of \( y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( z = \sqrt{1 - x^2} \) at \( x = 0 \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \]
Using the identity \( \sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right) \), we can rewrite this as:
\[ y = \cos^{-1}(2x^2 - 1) \]

### Step 2: Differentiate \( y \) with respect to \( x \)
To find \( \frac{dy}{dx} \), we differentiate \( y \):
\[ \frac{dy}{dx} = -\frac{d}{dx}\left(2x^2 - 1\right) \cdot \frac{1}{\sqrt{1 - (2x^2 - 1)^2}} \]
Calculating the derivative of \( 2x^2 - 1 \):
\[ \frac{d}{dx}(2x^2 - 1) = 4x \]
Thus, we have:
\[ \frac{dy}{dx} = -\frac{4x}{\sqrt{1 - (2x^2 - 1)^2}} \]

### Step 3: Simplify the expression
Next, we simplify the expression under the square root:
\[ (2x^2 - 1)^2 = 4x^4 - 4x^2 + 1 \]
Thus,
\[ 1 - (2x^2 - 1)^2 = 1 - (4x^4 - 4x^2 + 1) = 4x^2 - 4x^4 \]
So we have:
\[ \frac{dy}{dx} = -\frac{4x}{\sqrt{4x^2 - 4x^4}} = -\frac{4x}{2\sqrt{x^2(1 - x^2)}} = -\frac{2}{\sqrt{1 - x^2}} \quad \text{for } x \neq 0 \]

### Step 4: Find \( \frac{dz}{dx} \)
Now, we differentiate \( z \):
\[ z = \sqrt{1 - x^2} \]
Thus,
\[ \frac{dz}{dx} = -\frac{x}{\sqrt{1 - x^2}} \]

### Step 5: Use the chain rule to find \( \frac{dy}{dz} \)
Using the chain rule:
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = \frac{dy}{dx} \cdot \frac{1}{\frac{dz}{dx}} \]
Substituting the values we found:
\[ \frac{dy}{dz} = \left(-\frac{2}{\sqrt{1 - x^2}}\right) \cdot \left(-\frac{\sqrt{1 - x^2}}{x}\right) = \frac{2}{x} \]

### Step 6: Evaluate at \( x = 0 \)
As \( x \to 0 \), we need to evaluate the limit:
\[ \frac{dy}{dz} = \lim_{x \to 0} \frac{2}{x} \]
This limit approaches \( +\infty \) from the right and \( -\infty \) from the left, indicating a discontinuity.

### Conclusion
The derivative \( \frac{dy}{dz} \) at \( x = 0 \) does not exist due to the discontinuity.