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

Options

A. `(1)/( sqrt""(1-x^2))`
B. `-(1)/(sqrt(1-x^2))`
C. `-(1)/(2(sqrt(1-x^2))`
D. None

Answer

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

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

we will follow these steps:

### Step 1: Substitute \(x\) with \(\cos(2\theta)\)

Let \(x = \cos(2\theta)\). Then we can express \(\sqrt{1+x}\) and \(\sqrt{1-x}\) in terms of \(\theta\):

\[ \sqrt{1+x} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2} \cos(\theta), \]

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

### Step 2: Substitute into the function

Now substitute these expressions back into the function:

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

This simplifies to:

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

### Step 3: Simplify the expression

Recognizing that \(\frac{\sqrt{2}}{2} = \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)\), we can rewrite the argument of the \(\sin^{-1}\):

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

Thus, we have:

\[ y = \frac{\pi}{4} + \theta. \]

### Step 4: Find \(\theta\) in terms of \(x\)

Since \(x = \cos(2\theta)\), we can express \(\theta\) as:

\[ \theta = \frac{1}{2} \cos^{-1}(x). \]

### Step 5: Differentiate \(y\)

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

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

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

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

we have:

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

### Final Answer

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

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

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