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

Options

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

Answer

To solve the problem \(\frac{d}{dx}\left[\sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\right]\), we will differentiate each term separately. Let's go through the steps one by one.

### Step 1: Differentiate \(\sin^{-1}x\)

The derivative of \(\sin^{-1}x\) is given by the formula:
\[ \frac{d}{dx} \sin^{-1}x = \frac{1}{\sqrt{1-x^2}} \]

### Step 2: Differentiate \(\sin^{-1}\sqrt{1-x^2}\)

To differentiate \(\sin^{-1}\sqrt{1-x^2}\), we will use the chain rule. Let \(u = \sqrt{1-x^2}\). Then,
\[ \frac{d}{dx} \sin^{-1}u = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]

First, 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:
\[ \frac{d}{dx} \sin^{-1}\sqrt{1-x^2} = \frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}} \cdot \frac{-x}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-(1-x^2)}} \cdot \frac{-x}{\sqrt{1-x^2}} \]

Simplifying the expression:
\[ 1 - (1-x^2) = x^2 \implies \sqrt{x^2} = |x| \quad \text{(since \(x\) can be positive or negative)} \]

Thus,
\[ \frac{d}{dx} \sin^{-1}\sqrt{1-x^2} = \frac{-x}{|x|\sqrt{1-x^2}} \]

### Step 3: Combine the derivatives

Now we can combine the derivatives from Step 1 and Step 2:
\[ \frac{d}{dx}\left[\sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\right] = \frac{1}{\sqrt{1-x^2}} + \frac{-x}{|x|\sqrt{1-x^2}} \]

### Step 4: Simplify the expression

Combining the two terms:
\[ \frac{1}{\sqrt{1-x^2}} - \frac{x}{|x|\sqrt{1-x^2}} \]
We can express this as:
\[ \frac{1 - \frac{x}{|x|}}{\sqrt{1-x^2}} \]

### Step 5: Analyze the expression

Now, depending on the sign of \(x\):
- If \(x > 0\), \(|x| = x\) and the expression becomes:
\[ \frac{1 - 1}{\sqrt{1-x^2}} = 0 \]

- If \(x < 0\), \(|x| = -x\) and the expression becomes:
\[ \frac{1 + 1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}} \neq 0 \]

### Conclusion

Thus, the derivative \(\frac{d}{dx}\left[\sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\right] = 0\) for \(x > 0\) and is not defined for \(x < 0\). Therefore, the final answer is:
\[ \frac{d}{dx}\left[\sin^{-1}x + \sin^{-1}\sqrt{1-x^2}\right] = 0 \]