If y = sqrt{sin^{-1} x + y},then frac{dy}{dx} | vedclass

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Question

(B) Given the equation: $y = \sqrt{\sin^{-1} x + y}$ Squaring both sides,we get: $y^2 = \sin^{-1} x + y$ Differentiating both sides with respect to $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation: $y = \sqrt{\sin^{-1} x + y}$ Squaring both sides,we get: $y^2 = \sin^{-1} x + y$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin^{-1} x + y)$ $2y \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ $(2y - 1) \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ Therefore,$\frac{dy}{dx} = \frac{1}{(2y - 1) \sqrt{1 - x^2}}$ Thus,the correct option is $B$.

If $y = \sqrt{\sin^{-1} x + y}$,then $\frac{dy}{dx} = $ . . . . . . . (where,$x \in (0, 1)$)

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