If `y=sin^(-1)x+sin^(-1)sqrt(1-x^(2))`, what is `(dy)/(dx)` equal to?

Options

A. `cos^(-1)x+cos^(-1)sqrt(1-x^(2))`
B. `(1)/(cosx)+(1)/(cos sqrt(1-x^(2)))`
C. `(pi)/(2)`
D. 0

Explanation

To find the derivative of the function \( y = \sin^{-1}(x) + \sin^{-1}(\sqrt{1 - x^2}) \), we will differentiate each term separately.

### Step 1: Differentiate the first term
The first term is \( \sin^{-1}(x) \). The derivative of \( \sin^{-1}(x) \) is given by:
\[ \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 2: Differentiate the second term
The second term is \( \sin^{-1}(\sqrt{1 - x^2}) \). We will use the chain rule to differentiate this term. Let \( u = \sqrt{1 - x^2} \), then:
\[ \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Now, we need to find \( \frac{du}{dx} \):
\[ u = (1 - x^2)^{1/2} \implies \frac{du}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]
Now substituting \( u \) back into the derivative:
\[ \frac{d}{dx} \sin^{-1}(\sqrt{1 - x^2}) = \frac{1}{\sqrt{1 - (\sqrt{1 - x^2})^2}} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - (1 - x^2)}} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{x^2}} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{-1}{\sqrt{1 - x^2}} \quad (\text{since } \sqrt{x^2} = |x| \text{ and we assume } x \geq 0) \]

### Step 3: Combine the derivatives
Now, we combine the derivatives from Step 1 and Step 2:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \left(-\frac{1}{\sqrt{1 - x^2}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} = 0 \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is equal to:
\[ \frac{dy}{dx} = 0 \]

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