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

Explanation

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 can follow these steps:

### Step 1: Substitute \( x = \cos(\theta) \)

Let \( x = \cos(\theta) \). Then, we have:

\[ z = \sqrt{1 - x^2} = \sqrt{1 - \cos^2(\theta)} = \sqrt{\sin^2(\theta)} = \sin(\theta) \]

### Step 2: Rewrite \( y \)

Substituting \( x = \cos(\theta) \) into \( y \):

\[ y = \sec^{-1}\left(\frac{1}{2(\cos^2(\theta)) - 1}\right) = \sec^{-1}(\cos(2\theta)) \]

### Step 3: Simplify \( y \)

Since \( \sec^{-1}(\cos(2\theta)) = 2\theta \), we have:

\[ y = 2\theta \]

### Step 4: Differentiate \( y \) and \( z \)

Now, we need to find \( \frac{dy}{dz} \):

1. Differentiate \( y \) with respect to \( \theta \):

\[ \frac{dy}{d\theta} = 2 \]

2. Differentiate \( z \) with respect to \( \theta \):

\[ \frac{dz}{d\theta} = \cos(\theta) \]

### Step 5: Apply the Chain Rule

Using the chain rule, we find:

\[ \frac{dy}{dz} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dz} = \frac{dy}{d\theta} \cdot \frac{1}{\frac{dz}{d\theta}} = \frac{2}{\cos(\theta)} \]

### Step 6: Find \( \theta \) at \( x = \frac{1}{2} \)

When \( x = \frac{1}{2} \):

\[ \cos(\theta) = \frac{1}{2} \implies \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Step 7: Calculate \( \cos(\theta) \)

At \( \theta = \frac{\pi}{3} \):

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 8: Substitute \( \theta \) back into \( \frac{dy}{dz} \)

Now substitute \( \theta = \frac{\pi}{3} \) into \( \frac{dy}{dz} \):

\[ \frac{dy}{dz} = \frac{2}{\cos\left(\frac{\pi}{3}\right)} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Answer

Thus, the derivative of \( y \) with respect to \( z \) at \( x = \frac{1}{2} \) is:

\[ \frac{dy}{dz} = 4 \]

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