For x neq -1, y neq -1, if x = frac{1 - sqrt[ | vedclass

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(B) Given $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$.Multiplying both sides by $(1 + \sqrt[3]{y})$, we get $x(1 + \sqrt[3]{y}) = 1 - \sqrt[3]{y}$.$

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$\frac{-(1+x)^4}{6(1-x)^2}$

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(B) Given $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$.Multiplying both sides by $(1 + \sqrt[3]{y})$, we get $x(1 + \sqrt[3]{y}) = 1 - \sqrt[3]{y}$.$x + x\sqrt[3]{y} = 1 - \sqrt[3]{y}$.$\sqrt[3]{y}(x + 1) = 1 - x$.$\sqrt[3]{y} = \frac{1 - x}{1 + x}$.Cubing both sides, $y = \left(\frac{1 - x}{1 + x}\right)^3$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = 3\left(\frac{1 - x}{1 + x}\right)^2 \cdot \frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$.Using the quotient rule, $\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right) = \frac{-(1 + x) - (1 - x)(1)}{(1 + x)^2} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}$.Thus, $\frac{dy}{dx} = 3 \cdot \frac{(1 - x)^2}{(1 + x)^2} \cdot \frac{-2}{(1 + x)^2} = \frac{-6(1 - x)^2}{(1 + x)^4}$.Since $\frac{dx}{dy} = \frac{1}{dy/dx}$, we have $\frac{dx}{dy} = \frac{-(1 + x)^4}{6(1 - x)^2}$.

For $x \neq -1, y \neq -1$, if $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$, then $\frac{dx}{dy} =$

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