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If $x^{3} + y^{3} = 4$ , then $\frac{d y}{d x} =$

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a

\frac{x^{2}}{y^{2}}

b

\frac{- x^{2}}{y^{2}}

c

\frac{y^{2}}{x^{2}}

d

\frac{- y^{2}}{x^{2}}

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detailed solution

Correct option is B

$x^{3} + y^{3} = 4$

Differentiate with respect to $' x' o . b . s$

$\begin{array}{l} \frac{d}{d x} (x^{3} + y^{3}) = \frac{d}{d x} (4) \\ 3 x^{2} + 3 y^{2} \frac{d y}{d x} = 0 \\ 3 y^{2} \frac{d y}{d x} = - 3 x^{2} \\ \frac{d y}{d x} = \frac{- 3 x^{2}}{3 y^{2}} \\ \frac{d y}{d x} = \frac{- x^{2}}{y^{2}} \end{array}$

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