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

Options

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

Answer

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

### Step 1: Recognize the Structure
We notice that the expression inside the inverse tangent function has the form \( \frac{x}{\sqrt{1-x^2}} \). This suggests a trigonometric substitution.

### Step 2: Use Trigonometric Substitution
Let \( x = \sin \theta \). Then, we have:
\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta \]
Thus, we can rewrite the function:
\[ y = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan^{-1}(\tan \theta) \]

### Step 3: Simplify the Function
Since \( \tan^{-1}(\tan \theta) = \theta \) (for \( \theta \) in the appropriate range), we can express \( y \) as:
\[ y = \theta \]

### Step 4: Relate \( \theta \) Back to \( x \)
Since we set \( x = \sin \theta \), we can find \( \theta \) in terms of \( x \):
\[ \theta = \sin^{-1}(x) \]
Thus, we have:
\[ y = \sin^{-1}(x) \]

### Step 5: Differentiate \( y \) with Respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \]

### Final Answer
The derivative of \( \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) with respect to \( x \) is:
\[ \frac{1}{\sqrt{1 - x^2}} \]