If y = e^{m sin^{-1} x} and (1 - x^2) (frac{d | vedclass

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(C) Given,$y = e^{m \sin^{-1} x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{

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$m^2$

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(C) Given,$y = e^{m \sin^{-1} x}$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{-1} x) = y \cdot \frac{m}{\sqrt{1 - x^2}}$.Squaring both sides,we get:$(\frac{dy}{dx})^2 = \frac{m^2 y^2}{1 - x^2}$.Multiplying both sides by $(1 - x^2)$,we obtain:$(1 - x^2) (\frac{dy}{dx})^2 = m^2 y^2$.Comparing this with the given equation $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,we find that $A = m^2$.

If $y = e^{m \sin^{-1} x}$ and $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,then $A = . . . . . .$

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