If y = (sin^{-1} x)^2,then (1 - x^2) frac{d^2 | vedclass

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Question

(B) Given $y = (\sin^{-1} x)^2$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}}$. Multiply

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $y = (\sin^{-1} x)^2$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}}$. Multiplying both sides by $\sqrt{1 - x^2}$,we have: $\sqrt{1 - x^2} \frac{dy}{dx} = 2 \sin^{-1} x$. Differentiating again with respect to $x$ using the product rule: $\sqrt{1 - x^2} \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot \frac{-2x}{2\sqrt{1 - x^2}} = 2 \cdot \frac{1}{\sqrt{1 - x^2}}$. Multiplying throughout by $\sqrt{1 - x^2}$: $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 2$. Thus,the correct option is $B$.

If $y = (\sin^{-1} x)^2$,then $(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = $

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