Find the second derivative of `sin^(-1)x`.

Answer

To find the second derivative of \( \sin^{-1} x \), we will follow these steps:

### Step 1: Define the function
Let \( y = \sin^{-1} x \).

### Step 2: Find the first derivative
The derivative of \( y \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

### Step 3: Differentiate again to find the second derivative
Now we need to differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \).

Using the quotient rule for differentiation, where \( u = 1 \) and \( v = \sqrt{1 - x^2} \):
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Here, \( \frac{du}{dx} = 0 \) (since \( u = 1 \)) and we need to find \( \frac{dv}{dx} \):
\[ v = (1 - x^2)^{1/2} \]
Using the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]

Now substituting back into the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{\sqrt{1 - x^2} \cdot 0 - 1 \cdot \left(-\frac{x}{\sqrt{1 - x^2}}\right)}{(1 - x^2)} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}} \]

### Final Answer
Thus, the second derivative of \( \sin^{-1} x \) is:
\[ \frac{d^2y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}} \]

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