If `y=tan^(-1)[(x-sqrt(1-x^(2)))/(x+sqrt(1-x^(2)))]," then "(dy)/(dx)=`

Options

A. `(-1)/(sqrt(1-x^(2)))`
B. `(1)/(sqrt(1-x^(2)))`
C. `(-x)/(sqrt(1-x^(2)))`
D. `(x)/(sqrt(1-x^(2)))`

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \tan^{-1}\left(\frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}}\right) \]

### Step 1: Substitute \( x = \cos \theta \)

We start by substituting \( x = \cos \theta \). This gives us:

\[ y = \tan^{-1}\left(\frac{\cos \theta - \sqrt{1 - \cos^2 \theta}}{\cos \theta + \sqrt{1 - \cos^2 \theta}}\right) \]

Since \( \sqrt{1 - \cos^2 \theta} = \sin \theta \), we can simplify the expression:

\[ y = \tan^{-1}\left(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right) \]

### Step 2: Simplify the expression

Now, we can rewrite the expression for \( y \):

\[ y = \tan^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) \]

This can be recognized as the tangent subtraction formula:

\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \]

### Step 3: Use the inverse tangent property

Using the property of the inverse tangent function, we have:

\[ y = \frac{\pi}{4} - \theta \]

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

Since we set \( x = \cos \theta \), we can express \( \theta \) in terms of \( x \):

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

Thus, we can write:

\[ y = \frac{\pi}{4} - \cos^{-1}(x) \]

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

Now we differentiate both sides with respect to \( x \):

\[ \frac{dy}{dx} = 0 - \frac{d}{dx}(\cos^{-1}(x)) \]

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

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

So we have:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

### Final Result

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

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