If x = sin y,then frac{d^2 y}{dx^2} = . . . . | vedclass

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(C) Given $x = \sin y$,we have $y = \arcsin x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-1/2

Vedclass · Accepted Answer

$\frac{x}{(1 - x^2)^{3/2}}$

Vedclass · Answer

(C) Given $x = \sin y$,we have $y = \arcsin x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-1/2}$. Now,differentiating again with respect to $x$ using the chain rule: $\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot \frac{d}{dx}(1 - x^2)$. $\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot (-2x)$. $\frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}}$. Thus,the correct option is $C$.

If $x = \sin y$,then $\frac{d^2 y}{dx^2} = . . . . . .$,$(0 < x < 1)$.

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