the derivation of ` sec^(-1)"" ((1)/(2x^2-1)) w.r.t sqrt(1-x^2)` at ` x=1//2 ` is

Options

A. 2
B. 4
C. 1
D. `-2`

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 \( x = \frac{1}{2} \), we will use the chain rule and implicit differentiation.

### Step-by-Step Solution:

1. **Identify the Functions**:
- Let \( y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \)
- Let \( z = \sqrt{1 - x^2} \)

2. **Differentiate \( y \) with respect to \( x \)**:
- We will use the formula for the derivative of the inverse secant function:
\[ \frac{dy}{dx} = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx} \]
where \( u = \frac{1}{2x^2 - 1} \).
- First, we need to find \( \frac{du}{dx} \):
\[ u = \frac{1}{2x^2 - 1} \implies \frac{du}{dx} = -\frac{2x}{(2x^2 - 1)^2} \]
- Now, we need to compute \( |u| \) and \( \sqrt{u^2 - 1} \):
\[ |u| = \left|\frac{1}{2x^2 - 1}\right|, \quad u^2 = \left(\frac{1}{2x^2 - 1}\right)^2 \]
\[ u^2 - 1 = \frac{1 - (2x^2 - 1)^2}{(2x^2 - 1)^2} \]

3. **Substituting into the Derivative**:
- Now substitute \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{-\frac{2x}{(2x^2 - 1)^2}}{\left|\frac{1}{2x^2 - 1}\right| \sqrt{\frac{1 - (2x^2 - 1)^2}{(2x^2 - 1)^2}}} \]

4. **Differentiate \( z \) with respect to \( x \)**:
- We differentiate \( z = \sqrt{1 - x^2} \):
\[ \frac{dz}{dx} = \frac{-x}{\sqrt{1 - x^2}} \]

5. **Applying the Chain Rule**:
- Now we apply the chain rule to find \( \frac{dy}{dz} \):
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} \]

6. **Evaluate at \( x = \frac{1}{2} \)**:
- Substitute \( x = \frac{1}{2} \) into the derivatives calculated above to find the value of \( \frac{dy}{dz} \).

### Final Calculation:
1. Calculate \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \) at \( x = \frac{1}{2} \).
2. Substitute these values into \( \frac{dy}{dz} \).