If ` y=sin (msin ^(-1) x),then (1-x^(2))(d^(2)y)/(dx^(2))-x( dy)/(dx)=`

Options

A. ` -2m ^(2)y`
B. ` 2m^(2)y`
C. ` -m^(2) y`
D. ` m^(2)y`

Answer

To solve the problem, we start with the given function:

\[ y = \sin(m \sin^{-1}(x)) \]

### Step 1: Differentiate \( y \) with respect to \( x \)

Using the chain rule, we differentiate \( y \):

\[ \frac{dy}{dx} = \cos(m \sin^{-1}(x)) \cdot \frac{d}{dx}(m \sin^{-1}(x)) \]

The derivative of \( \sin^{-1}(x) \) is:

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

Thus, we have:

\[ \frac{dy}{dx} = \cos(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \]

### Step 2: Express \( \cos(m \sin^{-1}(x)) \)

Using the identity \( \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} \), we can express \( \cos(m \sin^{-1}(x)) \):

\[ \cos(m \sin^{-1}(x)) = \sqrt{1 - \sin^2(m \sin^{-1}(x))} = \sqrt{1 - x^2} \]

Thus,

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

### Step 3: Differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \)

Since \( \frac{dy}{dx} = m \) is a constant, we have:

\[ \frac{d^2y}{dx^2} = 0 \]

### Step 4: Substitute into the original expression

Now we substitute \( \frac{d^2y}{dx^2} \) and \( \frac{dy}{dx} \) into the expression:

\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} \]

Substituting the values we found:

\[ (1 - x^2)(0) - x(m) = -mx \]

### Final Result

Thus, the expression simplifies to:

\[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -mx \]