If `y=(sin^-1x)^2/2,then(1-x^2)y_2-xy_1`=

Options

A. y
B. 2y
C. 1
D. 2

Answer

To solve the problem, we need to compute the expression \( (1-x^2)y_2 - xy_1 \) where \( y = \frac{(\sin^{-1} x)^2}{2} \). We will first find \( y_1 \) (the first derivative of \( y \)) and \( y_2 \) (the second derivative of \( y \)).

### Step 1: Find \( y_1 \)

Given:
\[ y = \frac{(\sin^{-1} x)^2}{2} \]
We differentiate \( y \) with respect to \( x \):
\[ y_1 = \frac{d}{dx}\left(\frac{(\sin^{-1} x)^2}{2}\right) \]
Using the chain rule:
\[ y_1 = \frac{1}{2} \cdot 2(\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) \]
We know that:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \]
Thus,
\[ y_1 = (\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} = \frac{\sin^{-1} x}{\sqrt{1-x^2}} \]

### Step 2: Find \( y_2 \)

Now we differentiate \( y_1 \) to find \( y_2 \):
\[ y_2 = \frac{d}{dx}\left(\frac{\sin^{-1} x}{\sqrt{1-x^2}}\right) \]
Using the quotient rule:
\[ y_2 = \frac{\sqrt{1-x^2} \cdot \frac{d}{dx}(\sin^{-1} x) - \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1-x^2})}{(1-x^2)} \]
Calculating the derivatives:
\[ \frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}} \]
So we have:
\[ y_2 = \frac{\sqrt{1-x^2} \cdot \frac{1}{\sqrt{1-x^2}} - \sin^{-1} x \cdot \left(\frac{-x}{\sqrt{1-x^2}}\right)}{1-x^2} \]
This simplifies to:
\[ y_2 = \frac{1 + x \sin^{-1} x}{(1-x^2)\sqrt{1-x^2}} \]

### Step 3: Substitute \( y_1 \) and \( y_2 \) into the expression

Now we substitute \( y_1 \) and \( y_2 \) into the expression \( (1-x^2)y_2 - xy_1 \):
\[ (1-x^2)y_2 - xy_1 = (1-x^2) \cdot \frac{1 + x \sin^{-1} x}{(1-x^2)\sqrt{1-x^2}} - x \cdot \frac{\sin^{-1} x}{\sqrt{1-x^2}} \]
This simplifies to:
\[ = \frac{1 + x \sin^{-1} x}{\sqrt{1-x^2}} - \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \]
Combining the terms:
\[ = \frac{1}{\sqrt{1-x^2}} \]

### Step 4: Final Result

Thus, the final value of \( (1-x^2)y_2 - xy_1 \) is:
\[ \frac{1}{\sqrt{1-x^2}} \]