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

Options

A. `(1)/(sqrt(1-x^4))`
B. `-(1)/( sqrt(1-x^4))`
C. `(x)/( sqrt(1-x^4))`
D. `-(x)/(sqrt(1-x^4))`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

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

we will follow these steps:

### Step 1: Simplify the expression inside the arctangent function

Let

\[ u = \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}. \]

### Step 2: Substitute \( x = \tan(\theta) \)

Using the substitution \( x = \tan(\theta) \), we have:

\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta), \]
\[ \sqrt{1-x^2} = \sqrt{1-\tan^2(\theta)} = \sqrt{\frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta)}} = \frac{\cos(\theta)}{\sqrt{\cos^2(\theta) - \sin^2(\theta)}}. \]

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

Now substituting these into \( u \):

\[ u = \frac{\sec(\theta) + \sqrt{1-\tan^2(\theta)}}{\sec(\theta) - \sqrt{1-\tan^2(\theta)}}. \]

### Step 4: Simplify \( u \)

Using the identities, we can simplify \( u \) further. After simplification, we find that:

\[ u = \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right). \]

### Step 5: Substitute back to find \( y \)

Thus, we have:

\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right) = \frac{\pi}{4} + \frac{\theta}{2}. \]

### Step 6: Express \( \theta \) in terms of \( x \)

From \( x = \tan(\theta) \), we can express \( \theta \) as:

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

### Step 7: Substitute \( \theta \) back into \( y \)

Now substituting back:

\[ y = \frac{\pi}{4} + \frac{1}{2} \tan^{-1}(x). \]

### Step 8: Differentiate \( y \)

Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{1}{1+x^2} \cdot \frac{d}{dx}(x) = \frac{1}{2(1+x^2)}. \]

### Final Answer

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

\[ \frac{dy}{dx} = \frac{1}{2(1+x^2)}. \]

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