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

Options

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

Answer

To find the derivative of \( \cos^{-1}(2x^2 - 1) \) with respect to \( \cos^{-1}(x) \), we will use the chain rule and the derivatives of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Define the Functions**:
Let \( u = \cos^{-1}(2x^2 - 1) \) and \( v = \cos^{-1}(x) \).

2. **Find \( \frac{du}{dx} \)**:
The derivative of \( \cos^{-1}(y) \) is given by:
\[ \frac{d}{dy} \cos^{-1}(y) = -\frac{1}{\sqrt{1 - y^2}} \]
Thus, for \( u = \cos^{-1}(2x^2 - 1) \):
\[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - (2x^2 - 1)^2}} \cdot \frac{d}{dx}(2x^2 - 1) \]
The derivative of \( 2x^2 - 1 \) is \( 4x \):
\[ \frac{du}{dx} = -\frac{4x}{\sqrt{1 - (2x^2 - 1)^2}} \]

3. **Simplify \( 1 - (2x^2 - 1)^2 \)**:
We need to simplify \( 1 - (2x^2 - 1)^2 \):
\[ (2x^2 - 1)^2 = 4x^4 - 4x^2 + 1 \]
Therefore:
\[ 1 - (2x^2 - 1)^2 = 1 - (4x^4 - 4x^2 + 1) = -4x^4 + 4x^2 \]
This can be factored as:
\[ 4(x^2 - x^4) = 4x^2(1 - x^2) \]

4. **Substituting Back**:
Now substituting this back into the derivative:
\[ \frac{du}{dx} = -\frac{4x}{\sqrt{-4x^4 + 4x^2}} = -\frac{4x}{\sqrt{4x^2(1 - x^2)}} = -\frac{4x}{2\sqrt{1 - x^2} \cdot |x|} = -\frac{2}{\sqrt{1 - x^2}} \quad (\text{for } x > 0) \]

5. **Find \( \frac{dv}{dx} \)**:
Now, for \( v = \cos^{-1}(x) \):
\[ \frac{dv}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]

6. **Using the Chain Rule**:
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:
The derivative of \( \cos^{-1}(2x^2 - 1) \) with respect to \( \cos^{-1}(x) \) is \( 2 \).