If ` y=tan^(-1) "" (x)/(1+ sqrt((1-x^2)))+ sin {2 tan^(-1) "" sqrt(((1-x)/( 1+x)))}` then ` dy // dx=`

Options

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

Answer

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

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

we will solve it step by step.

### Step 1: Simplifying the first part of \( y \)

Let’s first simplify the term \( \tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \).

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

\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta. \]

Thus, we can rewrite the expression:

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

Using the identity \( 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \) and \( \sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \):

\[ \tan^{-1}\left(\frac{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)}{2 \cos^2\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2}. \]

Since \( \theta = \sin^{-1}(x) \):

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

### Step 2: Simplifying the second part of \( y \)

Now, we simplify the second part \( \sin\left(2 \tan^{-1}\left(\sqrt{\frac{1 - x}{1 + x}}\right)\right) \).

Let \( \phi = \tan^{-1}\left(\sqrt{\frac{1 - x}{1 + x}}\right) \). Then:

\[ \sin(2\phi) = 2 \sin(\phi) \cos(\phi). \]

Using the identity \( \sin(\phi) = \frac{\sqrt{1 - x}}{\sqrt{1 - x^2}} \) and \( \cos(\phi) = \frac{\sqrt{1 + x}}{\sqrt{1 - x^2}} \):

\[ \sin(2\phi) = 2 \cdot \frac{\sqrt{1 - x}}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 + x}}{\sqrt{1 - x^2}} = \frac{2\sqrt{(1 - x)(1 + x)}}{1 - x^2} = \frac{2\sqrt{1 - x^2}}{1 - x^2}. \]

Thus, we have:

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

### Step 3: Combining the parts

Now we can combine both parts:

\[ y = \frac{1}{2} \sin^{-1}(x) + \sqrt{1 - x^2}. \]

### Step 4: Differentiating \( y \)

Now we differentiate \( y \):

1. The derivative of \( \frac{1}{2} \sin^{-1}(x) \) is \( \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x^2}} \).
2. The derivative of \( \sqrt{1 - x^2} \) is \( \frac{-x}{\sqrt{1 - x^2}} \).

Combining these, we have:

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

### Final Answer

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

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