The derivative of `sin^(-1)((x)/(sqrt(1+x^(2))))` w.r.t .x is

Options

A. `(1)/(sqrt(1-x^(2)))`
B. `(1)/((1+x^(2))(3)/(2))`
C. `(1)/(1+x^(2))`
D. none of these

Answer

To find the derivative of \( y = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Differentiate using the chain rule
We start by applying the chain rule to differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{x}{\sqrt{1+x^2}} \).

### Step 2: Find \( u^2 \)
Next, we need to calculate \( u^2 \):
\[ u^2 = \left(\frac{x}{\sqrt{1+x^2}}\right)^2 = \frac{x^2}{1+x^2} \]

### Step 3: Calculate \( 1 - u^2 \)
Now we calculate \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \frac{x^2}{1+x^2} = \frac{(1+x^2) - x^2}{1+x^2} = \frac{1}{1+x^2} \]

### Step 4: Substitute back into the derivative formula
Now we substitute \( 1 - u^2 \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{1}{1+x^2}}} \cdot \frac{du}{dx} = \sqrt{1+x^2} \cdot \frac{du}{dx} \]

### Step 5: Differentiate \( u \)
Next, we need to find \( \frac{du}{dx} \):
\[ u = \frac{x}{\sqrt{1+x^2}} \implies \frac{du}{dx} = \frac{\sqrt{1+x^2} \cdot 1 - x \cdot \frac{1}{2\sqrt{1+x^2}} \cdot 2x}{1+x^2} \]
Simplifying this gives:
\[ \frac{du}{dx} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1+x^2 - x^2}{(1+x^2)\sqrt{1+x^2}} = \frac{1}{(1+x^2)\sqrt{1+x^2}} \]

### Step 6: Substitute \( \frac{du}{dx} \) back into the derivative
Now substituting \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = \sqrt{1+x^2} \cdot \frac{1}{(1+x^2)\sqrt{1+x^2}} = \frac{1}{1+x^2} \]

### Final Answer
Thus, the derivative of \( y = \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{1+x^2} \]