`d/(dx) (sin^(-1) "" (2x)/(1+x^(2)))` is equal to

Options

A. `2/(1+x^(2))`
B. ` - 2/( 1+x^(2))`
C. ` (3(1-x^(2)))/(| 1-x^(2)|(1-x^(2))) , x ne 1`
D. ` 2/(1-x^(2))`

Answer

To solve the problem \( \frac{d}{dx} \left( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) \), we will follow these steps:

### Step 1: Recognize the function
We need to differentiate the function \( y = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \).

### Step 2: Use the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) \]

### Step 3: Differentiate the inner function
Now, we need to differentiate \( \frac{2x}{1+x^2} \) using the quotient rule:
\[ \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \]
Simplifying this gives:
\[ = \frac{2(1+x^2) - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

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

### Step 5: Simplify the expression
Next, we need to simplify \( 1 - \left( \frac{2x}{1+x^2} \right)^2 \):
\[ 1 - \left( \frac{2x}{1+x^2} \right)^2 = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} \]

### Step 6: Final expression for the derivative
Thus, we can write:
\[ \frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2} \cdot \frac{1}{\sqrt{\frac{1 - 2x^2 + x^4}{(1+x^2)^2}}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2} \cdot \frac{(1+x^2)}{\sqrt{1 - 2x^2 + x^4}} = \frac{2(1-x^2)}{(1+x^2) \sqrt{(1-x^2)^2}} = \frac{2(1-x^2)}{(1+x^2)(1-x^2)} \]
Thus, the final answer is:
\[ \frac{2}{1+x^2} \]

### Final Answer
\[ \frac{d}{dx} \left( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) = \frac{2}{1+x^2} \]