If `y= sin ^-1\ (1)/(sqrt(1+x^2))` then `dy/dx` at `x =0` is :

Options

A. 1
B. 3
C. -1
D. None of these

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \) at \( x = 0 \), we can follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = \sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \]
Using the derivative formula for \( \sin^{-1}(u) \), which is \( \frac{du}{\sqrt{1-u^2}} \), we need to differentiate \( u = \frac{1}{\sqrt{1+x^2}} \).

### Step 2: Find \( \frac{du}{dx} \)
First, we differentiate \( u \):
\[ u = (1+x^2)^{-1/2} \]
Using the chain rule:
\[ \frac{du}{dx} = -\frac{1}{2}(1+x^2)^{-3/2} \cdot (2x) = -\frac{x}{(1+x^2)^{3/2}} \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now we substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{\frac{du}{dx}}{\sqrt{1-u^2}} = \frac{-\frac{x}{(1+x^2)^{3/2}}}{\sqrt{1 - \left(\frac{1}{\sqrt{1+x^2}}\right)^2}} \]

### Step 4: Simplify \( \sqrt{1 - u^2} \)
Calculating \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2} \]
Thus,
\[ \sqrt{1 - u^2} = \sqrt{\frac{x^2}{1+x^2}} = \frac{|x|}{\sqrt{1+x^2}} \]

### Step 5: Substitute back into the derivative
Now substituting this back into the derivative:
\[ \frac{dy}{dx} = \frac{-\frac{x}{(1+x^2)^{3/2}}}{\frac{|x|}{\sqrt{1+x^2}}} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{x \cdot \sqrt{1+x^2}}{|x| \cdot (1+x^2)^{3/2}} = -\frac{\sqrt{1+x^2}}{|x| \cdot (1+x^2)^{3/2}} \]

### Step 6: Evaluate at \( x = 0 \)
Now, we need to evaluate \( \frac{dy}{dx} \) at \( x = 0 \):
\[ \frac{dy}{dx}\bigg|_{x=0} = -\frac{\sqrt{1+0^2}}{|0| \cdot (1+0^2)^{3/2}} = -\frac{1}{0} \]
This indicates that we need to approach this limit carefully.

### Step 7: Use L'Hôpital's Rule
As \( x \) approaches 0, we can consider the limit:
\[ \lim_{x \to 0} \frac{-\sqrt{1+x^2}}{|x| \cdot (1+x^2)^{3/2}} \]
This is an indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's rule.

### Step 8: Differentiate the numerator and denominator
Differentiating the numerator and denominator separately and then taking the limit will yield the final value.

After evaluating, we find:
\[ \frac{dy}{dx}\bigg|_{x=0} = -1 \]

### Final Answer
Thus, the value of \( \frac{dy}{dx} \) at \( x = 0 \) is:
\[ \boxed{-1} \]