If `y=sin^(-1)(x^(2)-1)"then "dy/dx=`

Options

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

Answer

To find the derivative of the function \( y = \sin^{-1}(x^2 - 1) \), we will use the chain rule and the derivative of the inverse sine function.

### Step-by-step Solution:

1. **Identify the function**:
We have \( y = \sin^{-1}(u) \) where \( u = x^2 - 1 \).

2. **Differentiate using the chain rule**:
The derivative of \( y = \sin^{-1}(u) \) is given by:
\[ \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}} \]
By the chain rule, we also need to multiply by \( \frac{du}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

3. **Calculate \( \frac{du}{dx} \)**:
Since \( u = x^2 - 1 \), we differentiate:
\[ \frac{du}{dx} = 2x \]

4. **Substitute \( u \) back into the derivative**:
Now substituting \( u = x^2 - 1 \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (x^2 - 1)^2}} \cdot 2x \]

5. **Simplify the expression**:
First, simplify \( 1 - (x^2 - 1)^2 \):
\[ (x^2 - 1)^2 = x^4 - 2x^2 + 1 \]
Therefore:
\[ 1 - (x^2 - 1)^2 = 1 - (x^4 - 2x^2 + 1) = -x^4 + 2x^2 \]
Thus:
\[ \frac{dy}{dx} = \frac{2x}{\sqrt{-x^4 + 2x^2}} = \frac{2x}{\sqrt{2x^2 - x^4}} \]

6. **Factor out common terms**:
We can factor out \( x^2 \) from the square root:
\[ \sqrt{2x^2 - x^4} = \sqrt{x^2(2 - x^2)} = x\sqrt{2 - x^2} \]
Therefore:
\[ \frac{dy}{dx} = \frac{2x}{x\sqrt{2 - x^2}} = \frac{2}{\sqrt{2 - x^2}} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2}{\sqrt{2 - x^2}} \]