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

Options

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

Answer

To solve the problem, we need to find the expression for \((1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}\) given \(y = e^{m \cos^{-1}(x)}\).

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

Given:
\[ y = e^{m \cos^{-1}(x)} \]

Using the chain rule:
\[ \frac{dy}{dx} = e^{m \cos^{-1}(x)} \cdot \frac{d}{dx}(m \cos^{-1}(x)) \]

We know that:
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1-x^2}} \]

Thus:
\[ \frac{dy}{dx} = e^{m \cos^{-1}(x)} \cdot \left(-\frac{m}{\sqrt{1-x^2}}\right) \]
\[ \frac{dy}{dx} = -\frac{m e^{m \cos^{-1}(x)}}{\sqrt{1-x^2}} \]
Let \(y = e^{m \cos^{-1}(x)}\), then:
\[ \frac{dy}{dx} = -\frac{m y}{\sqrt{1-x^2}} \]

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

Now we differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{m y}{\sqrt{1-x^2}}\right) \]

Using the quotient rule:
\[ \frac{d^2y}{dx^2} = -m \left(\frac{\sqrt{1-x^2} \frac{dy}{dx} + y \frac{d}{dx}(\sqrt{1-x^2})}{1-x^2}\right) \]

Now, \(\frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}}\). Thus:
\[ \frac{d^2y}{dx^2} = -m \left(\frac{\sqrt{1-x^2} \left(-\frac{m y}{\sqrt{1-x^2}}\right) + y \left(\frac{-x}{\sqrt{1-x^2}}\right)}{1-x^2}\right) \]
\[ = -m \left(\frac{-m y + -xy}{1-x^2}\right) \]
\[ = \frac{m^2 y + mx y}{(1-x^2)} \]

### Step 3: Substitute into the expression \((1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx}\)

Now we substitute \(\frac{d^2y}{dx^2}\) and \(\frac{dy}{dx}\) into the expression:
\[ (1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = (1-x^2) \left(\frac{m^2 y + mx y}{1-x^2}\right) - x \left(-\frac{m y}{\sqrt{1-x^2}}\right) \]

Simplifying:
\[ = m^2 y + mx y + \frac{mx y}{\sqrt{1-x^2}} \]

### Final Expression

Thus, the final expression is:
\[ m^2 y + mx y + \frac{mx y}{\sqrt{1-x^2}} \]