If m sin ^{-1} x = log _{e} y,then (1 - x^{2} | vedclass

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(A) Given,$m \sin ^{-1} x = \log _{e} y$ $\Rightarrow y = e^{m \sin ^{-1} x}$ Differentiating with respect to $x$,we get: $y' = e^{m \sin ^{-1} x} 	i

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$m^{2} y$

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(A) Given,$m \sin ^{-1} x = \log _{e} y$ $\Rightarrow y = e^{m \sin ^{-1} x}$ Differentiating with respect to $x$,we get: $y' = e^{m \sin ^{-1} x} 	imes \frac{m}{\sqrt{1 - x^{2}}}$ $\Rightarrow \sqrt{1 - x^{2}} \cdot y' = m y$ Squaring both sides,we get: $(1 - x^{2}) (y')^{2} = m^{2} y^{2}$ Differentiating again with respect to $x$: $(1 - x^{2}) \cdot 2 y' \cdot y'' + (y')^{2} (-2 x) = m^{2} \cdot 2 y y' $ Dividing both sides by $2 y'$,we get: $(1 - x^{2}) y'' - x y' = m^{2} y$

If $m \sin ^{-1} x = \log _{e} y$,then $(1 - x^{2}) y'' - x y'$ is equal to

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