If log _e y=3 sin ^{-1} x,then (1-x^2) y^{pri | vedclass

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(D) Given $\ln y = 3 \sin^{-1} x$. Differentiating with respect to $x$: $\frac{1}{y} y' = \frac{3}{\sqrt{1-x^2}} \implies y' = \frac{3y}{\sqrt{1-x^2}}

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$9 e^{\pi / 2}$

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(D) Given $\ln y = 3 \sin^{-1} x$. Differentiating with respect to $x$: $\frac{1}{y} y' = \frac{3}{\sqrt{1-x^2}} \implies y' = \frac{3y}{\sqrt{1-x^2}}$. At $x = \frac{1}{2}$,$y = e^{3 \sin^{-1}(1/2)} = e^{3(\pi/6)} = e^{\pi/2}$. So,$y' = \frac{3 e^{\pi/2}}{\sqrt{1-(1/2)^2}} = \frac{3 e^{\pi/2}}{\sqrt{3}/2} = 2\sqrt{3} e^{\pi/2}$. Now,differentiate $y' \sqrt{1-x^2} = 3y$: $y'' \sqrt{1-x^2} + y' \left( \frac{-2x}{2\sqrt{1-x^2}} \right) = 3y'$. Multiply by $\sqrt{1-x^2}$: $y'' (1-x^2) - xy' = 3y' \sqrt{1-x^2}$. Substitute $y' = \frac{3y}{\sqrt{1-x^2}}$: $y'' (1-x^2) - xy' = 3 \left( \frac{3y}{\sqrt{1-x^2}} \right) \sqrt{1-x^2} = 9y$. At $x = \frac{1}{2}$,$y = e^{\pi/2}$,so the value is $9 e^{\pi/2}$.

If $\log _e y=3 \sin ^{-1} x$,then $(1-x^2) y^{\prime \prime}-x y^{\prime}$ at $x=\frac{1}{2}$ is equal to :

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