`sin^-1(1-2x^2)`

Answer

To differentiate the function \( y = \sin^{-1}(1 - 2x^2) \), we can follow these steps:

### Step 1: Set up the function
Let:
\[ y = \sin^{-1}(u) \quad \text{where} \quad u = 1 - 2x^2 \]

### Step 2: Differentiate \( y \) with respect to \( u \)
Using the derivative of the inverse sine function, we have:
\[ \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we differentiate \( u \):
\[ u = 1 - 2x^2 \implies \frac{du}{dx} = -4x \]

### Step 4: Apply the chain rule
Using the chain rule, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot (-4x) \]

### Step 5: Substitute \( u \) back into the equation
Now substitute \( u = 1 - 2x^2 \) back into the equation:
\[ \frac{dy}{dx} = \frac{-4x}{\sqrt{1 - (1 - 2x^2)^2}} \]

### Step 6: Simplify the expression
Now we need to simplify the expression under the square root:
\[ 1 - (1 - 2x^2)^2 = 1 - (1 - 4x^2 + 4x^4) = 4x^2 - 4x^4 \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{-4x}{\sqrt{4x^2 - 4x^4}} = \frac{-4x}{\sqrt{4x^2(1 - x^2)}} = \frac{-4x}{2\sqrt{1 - x^2} \cdot \sqrt{4x^2}} = \frac{-2}{\sqrt{1 - x^2}} \]

### Final Result
Therefore, the derivative of \( y = \sin^{-1}(1 - 2x^2) \) is:
\[ \frac{dy}{dx} = \frac{-2}{\sqrt{1 - x^2}} \]