Differentiate `sec^-1""(1)/(2x^2-1)` with respect to `sqrt(1-x^2)`

Options

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

Answer

To differentiate the function \( y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( z = \sqrt{1 - x^2} \), we will use the chain rule and implicit differentiation.

### Step-by-step solution:

1. **Identify the function to differentiate**:
We have \( y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \).

2. **Differentiate \( y \) with respect to \( x \)**:
Using the derivative of the inverse secant function, we know:
\[ \frac{dy}{dx} = \frac{1}{|x| \sqrt{x^2 - 1}} \cdot \frac{d}{dx}\left(\frac{1}{2x^2 - 1}\right) \]
To differentiate \( \frac{1}{2x^2 - 1} \), we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{1}{2x^2 - 1}\right) = -\frac{(0)(2x^2 - 1) - (1)(4x)}{(2x^2 - 1)^2} = -\frac{-4x}{(2x^2 - 1)^2} = \frac{4x}{(2x^2 - 1)^2} \]
Therefore,
\[ \frac{dy}{dx} = \frac{1}{|x| \sqrt{\left(\frac{1}{2x^2 - 1}\right)^2 - 1}} \cdot \frac{4x}{(2x^2 - 1)^2} \]

3. **Differentiate \( z \) with respect to \( x \)**:
We have \( z = \sqrt{1 - x^2} \). The derivative is:
\[ \frac{dz}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]

4. **Use the chain rule to find \( \frac{dy}{dz} \)**:
By the chain rule:
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = \frac{dy}{dx} \cdot \frac{1}{\frac{dz}{dx}} \]
Thus,
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \left(-\frac{\sqrt{1 - x^2}}{x}\right) \]

5. **Substituting the expressions**:
Substitute the expression for \( \frac{dy}{dx} \) into the equation:
\[ \frac{dy}{dz} = \left(\frac{1}{|x| \sqrt{\left(\frac{1}{2x^2 - 1}\right)^2 - 1}} \cdot \frac{4x}{(2x^2 - 1)^2}\right) \cdot \left(-\frac{\sqrt{1 - x^2}}{x}\right) \]

6. **Simplify the expression**:
After simplifying, we get:
\[ \frac{dy}{dz} = -\frac{4\sqrt{1 - x^2}}{|x|(2x^2 - 1)^2 \sqrt{\left(\frac{1}{2x^2 - 1}\right)^2 - 1}} \]

### Final Result:
Thus, the derivative of \( \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( \sqrt{1 - x^2} \) is:
\[ \frac{dy}{dz} = -\frac{4\sqrt{1 - x^2}}{|x|(2x^2 - 1)^2 \sqrt{\left(\frac{1}{2x^2 - 1}\right)^2 - 1}} \]