`(d)/(dx)[sin^(-1)sqrt(((1-x))/(2)]=`

Options

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

Answer

To differentiate the function \( y = \sin^{-1} \left( \sqrt{\frac{1-x}{2}} \right) \), we will use the chain rule and the derivative of the inverse sine function. Here are the steps:

### Step 1: Identify the outer and inner functions
The outer function is \( \sin^{-1}(u) \) where \( u = \sqrt{\frac{1-x}{2}} \).

### Step 2: Differentiate the outer function
The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is:
\[ \frac{d}{du} \left( \sin^{-1}(u) \right) = \frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = \sqrt{\frac{1-x}{2}} \) with respect to \( x \):
\[ u = \left( \frac{1-x}{2} \right)^{1/2} \]
Using the chain rule:
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{1-x}{2} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{1-x}{2} \right) \]
Calculating \( \frac{d}{dx} \left( \frac{1-x}{2} \right) \):
\[ \frac{d}{dx} \left( \frac{1-x}{2} \right) = \frac{-1}{2} \]
Thus,
\[ \frac{du}{dx} = \frac{1}{2} \left( \frac{1-x}{2} \right)^{-1/2} \cdot \left( -\frac{1}{2} \right) = -\frac{1}{4} \left( \frac{1-x}{2} \right)^{-1/2} \]

### Step 4: Combine the derivatives using the chain rule
Now we apply the chain rule:
\[ \frac{dy}{dx} = \frac{d}{du} \left( \sin^{-1}(u) \right) \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \sqrt{\frac{1-x}{2}} \right)^2}} \cdot \left(-\frac{1}{4} \left( \frac{1-x}{2} \right)^{-1/2}\right) \]

### Step 5: Simplify the expression
First simplify \( 1 - \left( \sqrt{\frac{1-x}{2}} \right)^2 \):
\[ 1 - \frac{1-x}{2} = \frac{2 - (1-x)}{2} = \frac{1+x}{2} \]
Now substituting this back:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{1+x}{2}}} \cdot \left(-\frac{1}{4} \left( \frac{1-x}{2} \right)^{-1/2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{4} \cdot \frac{1}{\sqrt{\frac{1+x}{2}}} \cdot \left( \frac{2}{\sqrt{1-x}} \right) \]
\[ = -\frac{1}{4} \cdot \frac{2}{\sqrt{1+x} \cdot \sqrt{1-x}} = -\frac{1}{2} \cdot \frac{1}{\sqrt{(1+x)(1-x)}} \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = -\frac{1}{2 \sqrt{1-x^2}} \]