If `y = sin^(-1) ((2x)/(1 + x^(2))), "then" (dy)/(dx) ` is equal to

Options

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

Explanation

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), we will use implicit differentiation and the chain rule.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \]

2. **Differentiate both sides with respect to \( x \)**:
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2}} \cdot \frac{d}{dx} \left( \frac{2x}{1 + x^2} \right) \]

3. **Differentiate \( \frac{2x}{1 + x^2} \)**:
We will use the quotient rule here. If \( u = 2x \) and \( v = 1 + x^2 \), then:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} = 2x \). Thus:
\[ \frac{d}{dx} \left( \frac{2x}{1 + x^2} \right) = \frac{(1 + x^2)(2) - (2x)(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \]

4. **Substitute back into the derivative**:
Now substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2}} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} \]

5. **Simplify the expression**:
We need to simplify \( 1 - \left( \frac{2x}{1 + x^2} \right)^2 \):
\[ 1 - \left( \frac{2x}{1 + x^2} \right)^2 = 1 - \frac{4x^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1 + x^2)^2} = \frac{1 - 2x^2 + x^4}{(1 + x^2)^2} = \frac{(1 - x^2)^2}{(1 + x^2)^2} \]

Therefore:
\[ \sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2} = \frac{1 - x^2}{1 + x^2} \]

6. **Final expression for \( \frac{dy}{dx} \)**:
Now substituting this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\frac{1 - x^2}{1 + x^2}} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} = \frac{1 + x^2}{1 - x^2} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2(1 + x^2)}{(1 + x^2)^2} = \frac{2}{1 + x^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2}{1 + x^2} \]