If y=e^{sin ^{-1} x},then left(1-x^2right) y_ | vedclass

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(C) Given,$y=e^{\sin ^{-1} x}$.Differentiating with respect to $x$:$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}

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(C) Given,$y=e^{\sin ^{-1} x}$.Differentiating with respect to $x$:$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$$y_1 = \frac{y}{\sqrt{1-x^2}}$$y_1 \sqrt{1-x^2} = y$Differentiating both sides with respect to $x$ using the product rule:$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = y_1$$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = y_1$Multiplying both sides by $\sqrt{1-x^2}$:$y_2 (1-x^2) - x y_1 = y_1 \sqrt{1-x^2}$Since $y_1 \sqrt{1-x^2} = y$,we have:$(1-x^2) y_2 - x y_1 = y$Thus,option $(c)$ is correct.

If $y=e^{\sin ^{-1} x}$,then $\left(1-x^2\right) y_2-x y_1=$

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