If y=e^{sin ^{-1}(t^{2}-1)} and x=e^{sec ^{-1 | vedclass

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(B) Given that,$y=e^{\sin ^{-1}(t^{2}-1)}$ and $x=e^{\sec ^{-1}(\frac{1}{t^{2}-1})}$.Taking natural logarithm on both sides:$\log y = \sin ^{-1}(t^{2}

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$-\frac{y}{x}$

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(B) Given that,$y=e^{\sin ^{-1}(t^{2}-1)}$ and $x=e^{\sec ^{-1}(\frac{1}{t^{2}-1})}$.Taking natural logarithm on both sides:$\log y = \sin ^{-1}(t^{2}-1)$$\log x = \sec ^{-1}(\frac{1}{t^{2}-1}) = \cos ^{-1}(t^{2}-1)$.Adding the two equations:$\log y + \log x = \sin ^{-1}(t^{2}-1) + \cos ^{-1}(t^{2}-1)$.Using the identity $\sin ^{-1}(	heta) + \cos ^{-1}(	heta) = \frac{\pi}{2}$,we get:$\log(xy) = \frac{\pi}{2}$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(\log y + \log x) = \frac{d}{dx}(\frac{\pi}{2})$$\frac{1}{y} \frac{dy}{dx} + \frac{1}{x} = 0$.Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

If $y=e^{\sin ^{-1}(t^{2}-1)}$ and $x=e^{\sec ^{-1}(\frac{1}{t^{2}-1})}$,then $\frac{dy}{dx}$ is equal to

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