`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: Let \( y = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \)

We start by defining \( y \) as the inverse sine of the expression \( \frac{2x}{1+x^2} \).

### Step 2: Use the substitution \( x = \tan(\theta) \)

To simplify the differentiation, we can use the substitution \( x = \tan(\theta) \). This gives us:

\[ \frac{2x}{1+x^2} = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]

### Step 3: Rewrite \( y \)

Now we can rewrite \( y \):

\[ y = \sin^{-1}(\sin(2\theta)) = 2\theta \]

### Step 4: Relate \( \theta \) back to \( x \)

Since \( x = \tan(\theta) \), we have:

\[ \theta = \tan^{-1}(x) \]

Thus, we can express \( y \) in terms of \( x \):

\[ y = 2\tan^{-1}(x) \]

### Step 5: Differentiate \( y \) with respect to \( x \)

Now we differentiate \( y \):

\[ \frac{dy}{dx} = 2 \frac{d}{dx} \left( \tan^{-1}(x) \right) \]

Using the derivative of \( \tan^{-1}(x) \):

\[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \]

So we have:

\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

### Step 6: Conclusion

Thus, the final result is:

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