The problem asks for the recurrence relation involving the derivatives of the function $y = e^{m \sin^{-1} x}$. We need to find a relation between $y_{n+2}$, $y_{n+1}$, and $y_n$, where $y_n$ denotes the $n$-th derivative of $y$ with respect to $x$. This is typically solved using the Leibniz differentiation theorem.

Deriving the Recurrence Relation

Step 1: Find the first derivative ($y_1$)

Given $y = e^{m \sin^{-1} x}$.

Differentiating with respect to $x$:

Rearranging the terms:

Step 2: Find the second derivative ($y_2$)

Differentiate the equation $\sqrt{1-x^2} y_1 = m y$ with respect to $x$ using the product rule:

Multiply the entire equation by $\sqrt{1-x^2}$:

Substitute $\sqrt{1-x^2} y_1 = m y$ from Step 1:

Rearrange to:

Step 3: Apply Leibniz differentiation theorem

Differentiate the equation $(1-x^2) y_2 - x y_1 - m^2 y = 0$ exactly $n$ times using the Leibniz theorem, which states that $\frac{d^n}{dx^n}(uv) = \sum_{k=0}^{n} \binom{n}{k} (\frac{d^k u}{dx^k}) (\frac{d^{n-k+1} v}{dx^{n-k+1}})$.

Differentiating $(1-x^2) y_2$ $n$ times:

Differentiating $-x y_1$ $n$ times:

Differentiating $-m^2 y$ $n$ times:

Step 4: Combine the results

Summing the results of the differentiation:

Group terms involving $y_{n+2}$, $y_{n+1}$, and $y_n$:

This derived relation matches option 4.