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

Options

A. `(1)/(sqrt""(1-x2))`
B. `(x)/(sqrt(1-x^2))`
C. 0
D. None

Answer

To solve the problem, we need to differentiate the function given by:

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

### Step 1: Differentiate the first term
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 using the chain rule
For the second term \( \sin^{-1}(\sqrt{1 - x^2}) \), we apply the chain rule. First, we differentiate the outer function \( \sin^{-1}(u) \) where \( u = \sqrt{1 - x^2} \):

\[ \frac{d}{dx}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]

Now we need to find \( \frac{du}{dx} \):

\[ u = \sqrt{1 - x^2} \implies \frac{du}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

### Step 3: Combine the derivatives
Now substituting \( u \) back into the derivative of \( \sin^{-1}(u) \):

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

This simplifies to:

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

### Step 4: Combine both derivatives
Now we can combine both derivatives to find \( \frac{dy}{dx} \):

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

### Step 5: Simplify the expression
This simplifies to:

\[ \frac{dy}{dx} = \frac{1 - 1}{\sqrt{1 - x^2}} = 0 \]

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

\[ \frac{dy}{dx} = 0 \]

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