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

Answer

To differentiate \( u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \) with respect to \( v = \sqrt{1 - x^2} \), we will use the chain rule. The steps are as follows:

### Step 1: Define the functions
Let:
- \( u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) \)
- \( v = \sqrt{1 - x^2} \)

### Step 2: Differentiate \( u \) with respect to \( x \)
To find \( \frac{du}{dx} \), we need to differentiate \( u \):
Using the chain rule for \( \sec^{-1}(x) \):
\[ \frac{du}{dx} = \frac{1}{\sqrt{(2x^2 - 1)^2 - 1}} \cdot \frac{d}{dx}\left(\frac{1}{2x^2 - 1}\right) \]
Now, differentiate \( \frac{1}{2x^2 - 1} \):
\[ \frac{d}{dx}\left(\frac{1}{2x^2 - 1}\right) = -\frac{2x}{(2x^2 - 1)^2} \]
Thus,
\[ \frac{du}{dx} = \frac{-2x}{(2x^2 - 1)^2 \sqrt{(2x^2 - 1)^2 - 1}} \]

### Step 3: Differentiate \( v \) with respect to \( x \)
Next, we differentiate \( v \):
\[ v = (1 - x^2)^{1/2} \]
Using the power rule:
\[ \frac{dv}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

### Step 4: Apply the chain rule
Now we use the chain rule to find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{-2x}{(2x^2 - 1)^2 \sqrt{(2x^2 - 1)^2 - 1}}}{\frac{-x}{\sqrt{1 - x^2}}} \]
The \( -x \) cancels out:
\[ \frac{du}{dv} = \frac{2}{(2x^2 - 1)^2 \sqrt{(2x^2 - 1)^2 - 1}} \cdot \sqrt{1 - x^2} \]

### Step 5: Simplify the expression
The expression can be simplified further, but we can stop here as we have found \( \frac{du}{dv} \).

### Final Result
Thus, the derivative \( \frac{du}{dv} \) is:
\[ \frac{du}{dv} = \frac{2\sqrt{1 - x^2}}{(2x^2 - 1)^2 \sqrt{(2x^2 - 1)^2 - 1}} \]