If ` y = sin^(-1) sqrt(1-x), "then " (dy)/(dx) ` is equal to

Options

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

Explanation

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \sin^{-1}(\sqrt{1 - x}) \), we can use the chain rule and the derivative of the inverse sine function. Here’s a step-by-step solution:

### Step 1: Identify the function
We have:
\[ y = \sin^{-1}(\sqrt{1 - x}) \]

### Step 2: Differentiate using the chain rule
The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is:
\[ \frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \]
Here, \( u = \sqrt{1 - x} \). We will also need to differentiate \( u \) with respect to \( x \).

### Step 3: Differentiate \( u \)
Now, we differentiate \( u = \sqrt{1 - x} \):
\[ \frac{du}{dx} = \frac{1}{2\sqrt{1 - x}} \cdot (-1) = -\frac{1}{2\sqrt{1 - x}} \]

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

### Step 5: Simplify the expression
Now, simplify \( 1 - (\sqrt{1 - x})^2 \):
\[ 1 - (\sqrt{1 - x})^2 = 1 - (1 - x) = x \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}} \cdot \left(-\frac{1}{2\sqrt{1 - x}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x(1 - x)}} \]

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

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