The derivative of `cos^(-1)(2x^(2)-1)` w.r.t. `cos^(-1)x` is

Options

A. `2`
B. `(-1)/(2sqrt(1-x^(2)))`
C. `2/x`
D. `1-x^(2)`

Explanation

To find the derivative of \( \cos^{-1}(2x^2 - 1) \) with respect to \( \cos^{-1}(x) \), we can use the chain rule. Let's denote:

- \( u = \cos^{-1}(2x^2 - 1) \)
- \( v = \cos^{-1}(x) \)

We need to find \( \frac{du}{dv} \).

### Step 1: Find \( \frac{du}{dx} \)

Using the derivative of \( \cos^{-1}(x) \):

\[ \frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1 - x^2}} \]

For \( u = \cos^{-1}(2x^2 - 1) \), we apply the chain rule:

\[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - (2x^2 - 1)^2}} \cdot \frac{d}{dx}(2x^2 - 1) \]

Calculating \( \frac{d}{dx}(2x^2 - 1) \):

\[ \frac{d}{dx}(2x^2 - 1) = 4x \]

Now substituting back:

\[ \frac{du}{dx} = -\frac{4x}{\sqrt{1 - (2x^2 - 1)^2}} \]

### Step 2: Simplify \( 1 - (2x^2 - 1)^2 \)

Calculating \( (2x^2 - 1)^2 \):

\[ (2x^2 - 1)^2 = 4x^4 - 4x^2 + 1 \]

So,

\[ 1 - (2x^2 - 1)^2 = 1 - (4x^4 - 4x^2 + 1) = -4x^4 + 4x^2 = 4x^2(1 - x^2) \]

Thus,

\[ \sqrt{1 - (2x^2 - 1)^2} = \sqrt{4x^2(1 - x^2)} = 2x\sqrt{1 - x^2} \]

### Step 3: Substitute back into \( \frac{du}{dx} \)

Now substituting back into the expression for \( \frac{du}{dx} \):

\[ \frac{du}{dx} = -\frac{4x}{2x\sqrt{1 - x^2}} = -\frac{2}{\sqrt{1 - x^2}} \]

### Step 4: Find \( \frac{dv}{dx} \)

For \( v = \cos^{-1}(x) \):

\[ \frac{dv}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Find \( \frac{du}{dv} \)

Now we can find \( \frac{du}{dv} \):

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-\frac{2}{\sqrt{1 - x^2}}}{-\frac{1}{\sqrt{1 - x^2}}} = 2 \]

### Final Answer

Thus, the derivative of \( \cos^{-1}(2x^2 - 1) \) with respect to \( \cos^{-1}(x) \) is:

\[ \frac{du}{dv} = 2 \]