If y = operatorname{Tan}^{-1} sqrt{x^2-1} + o | vedclass

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(B) Let $u = \sqrt{x^2-1}$. Then $y = \operatorname{Tan}^{-1}(u) + \operatorname{Sinh}^{-1}(u)$.Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \c

Vedclass · Accepted Answer

$\frac{x+1}{x \sqrt{x^2-1}}$

Vedclass · Answer

(B) Let $u = \sqrt{x^2-1}$. Then $y = \operatorname{Tan}^{-1}(u) + \operatorname{Sinh}^{-1}(u)$.Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.First,$\frac{dy}{du} = \frac{d}{du}(\operatorname{Tan}^{-1} u) + \frac{d}{du}(\operatorname{Sinh}^{-1} u) = \frac{1}{1+u^2} + \frac{1}{\sqrt{1+u^2}}$.Substitute $u^2 = x^2-1$: $\frac{dy}{du} = \frac{1}{1+(x^2-1)} + \frac{1}{\sqrt{1+(x^2-1)}} = \frac{1}{x^2} + \frac{1}{x} = \frac{1+x}{x^2}$.Next,$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x^2-1}) = \frac{1}{2\sqrt{x^2-1}} \cdot 2x = \frac{x}{\sqrt{x^2-1}}$.Therefore,$\frac{dy}{dx} = \left(\frac{1+x}{x^2}\right) \cdot \left(\frac{x}{\sqrt{x^2-1}}\right) = \frac{1+x}{x \sqrt{x^2-1}}$.Thus,the correct option is $B$.

If $y = \operatorname{Tan}^{-1} \sqrt{x^2-1} + \operatorname{Sinh}^{-1} \sqrt{x^2-1}$,$x > 1$,then $\frac{dy}{dx} = $

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