If the function y = sin^{-1} x,then left(1-x^ | vedclass

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(C) Given $y = \sin^{-1} x$. On differentiating both sides with respect to $x$,we get: $\frac{d y}{d x} = \frac{1}{\sqrt{1-x^2}}$ ... $(i)$ Squaring b

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$x \frac{d y}{d x}$

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(C) Given $y = \sin^{-1} x$. On differentiating both sides with respect to $x$,we get: $\frac{d y}{d x} = \frac{1}{\sqrt{1-x^2}}$ ... $(i)$ Squaring both sides,we get: $\left(\frac{d y}{d x}ight)^2 = \frac{1}{1-x^2}$ $(1-x^2) \left(\frac{d y}{d x}ight)^2 = 1$ Differentiating both sides with respect to $x$ using the product rule: $(1-x^2) \cdot 2 \left(\frac{d y}{d x}ight) \cdot \frac{d^2 y}{d x^2} + \left(\frac{d y}{d x}ight)^2 \cdot (-2x) = 0$ Dividing by $2 \frac{d y}{d x}$ (assuming $\frac{d y}{d x} 
eq 0$): $(1-x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} = 0$ Therefore,$(1-x^2) \frac{d^2 y}{d x^2} = x \frac{d y}{d x}$.

If the function $y = \sin^{-1} x$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}$ is equal to

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