The differential cofficient of `sec^(- 1)(1/(2x^2-1)) ` w.r.t `sqrt(1-x^2)` is-

Explanation

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

### Step 1: Define the variables
Let:
- \( y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \)
- \( z = \sqrt{1 - x^2} \)

### Step 2: Substitute \( x \) with \( \cos \theta \)
To simplify the expressions, we can set:
\[ x = \cos \theta \]
Then:
\[ z = \sqrt{1 - \cos^2 \theta} = \sin \theta \]

### Step 3: Express \( y \) in terms of \( \theta \)
Substituting \( x = \cos \theta \) into \( y \):
\[ y = \sec^{-1}\left(\frac{1}{2\cos^2 \theta - 1}\right) \]
Using the identity \( 2\cos^2 \theta - 1 = \cos 2\theta \):
\[ y = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) = \sec^{-1}(\sec 2\theta) = 2\theta \]

### Step 4: Differentiate \( y \) and \( z \) with respect to \( \theta \)
Now, differentiate \( y \) and \( z \) with respect to \( \theta \):
\[ \frac{dy}{d\theta} = 2 \]
\[ \frac{dz}{d\theta} = \cos \theta \]

### Step 5: Find \( \frac{dy}{dz} \)
Using the chain rule:
\[ \frac{dy}{dz} = \frac{dy/d\theta}{dz/d\theta} = \frac{2}{\cos \theta} \]

### Step 6: Substitute back for \( \cos \theta \)
Since \( x = \cos \theta \), we can express \( \cos \theta \) in terms of \( x \):
\[ \frac{dy}{dz} = \frac{2}{x} \]

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