If the function y = sin^{-1} x, then (1 - x^2 | vedclass

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

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

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(C) Given $y = \sin^{-1} x$. Differentiating with respect to $x$, we get: $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}}$. Squaring both sides: $\left(\frac{d y}{d x}ight)^2 = \frac{1}{1 - x^2}$. $(1 - x^2) \left(\frac{d y}{d x}ight)^2 = 1$. Differentiating again with respect to $x$: $(1 - x^2) \cdot 2 \left(\frac{d y}{d x}ight) \cdot \frac{d^2 y}{d x^2} + \left(\frac{d y}{d x}ight)^2 \cdot (-2x) = 0$. 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 $(1 - x^2) \frac{d^2 y}{d x^2}$ is equal to

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