If y = tan^{-1} sqrt{x^{2}-1},then the ratio | vedclass

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(B) Given $y = 	an^{-1} \sqrt{x^{2}-1}$. Let $x = \sec 	heta$,then $\sqrt{x^{2}-1} = 	an 	heta$. So,$y = 	an^{-1}(	an 	heta) = 	heta = \sec^{-

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$\frac{1-2x^{2}}{x(x^{2}-1)}$

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(B) Given $y = 	an^{-1} \sqrt{x^{2}-1}$. Let $x = \sec 	heta$,then $\sqrt{x^{2}-1} = 	an 	heta$. So,$y = 	an^{-1}(	an 	heta) = 	heta = \sec^{-1} x$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x \sqrt{x^{2}-1}}$. Now,find the second derivative $\frac{d^{2}y}{dx^{2}}$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( (x(x^{2}-1)^{1/2})^{-1} ight) = -1 \cdot (x(x^{2}-1)^{1/2})^{-2} \cdot \frac{d}{dx} (x(x^{2}-1)^{1/2})$. Using the product rule: $\frac{d}{dx} (x(x^{2}-1)^{1/2}) = (x^{2}-1)^{1/2} + x \cdot \frac{1}{2}(x^{2}-1)^{-1/2} \cdot 2x = \sqrt{x^{2}-1} + \frac{x^{2}}{\sqrt{x^{2}-1}} = \frac{x^{2}-1+x^{2}}{\sqrt{x^{2}-1}} = \frac{2x^{2}-1}{\sqrt{x^{2}-1}}$. Thus,$\frac{d^{2}y}{dx^{2}} = - \frac{1}{x^{2}(x^{2}-1)} \cdot \frac{2x^{2}-1}{\sqrt{x^{2}-1}} = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}}$. Finally,the ratio is: $\frac{d^{2}y/dx^{2}}{dy/dx} = \left( - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} ight) \div \left( \frac{1}{x \sqrt{x^{2}-1}} ight) = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \cdot x \sqrt{x^{2}-1} = - \frac{2x^{2}-1}{x(x^{2}-1)} = \frac{1-2x^{2}}{x(x^{2}-1)}$.

If $y = \tan^{-1} \sqrt{x^{2}-1}$,then the ratio $\frac{d^{2} y}{dx^{2}} : \frac{dy}{dx}$ is

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