differentiation

Question: If $x = \frac{{3at}}{{1 + {t^3}}},y = \frac{{3a{t^2}}}{{1 + {t^3}}},$then $\frac{{dy}}{{dx}}$=



1) $\frac{{t(2 + {t^3})}}{{1 - 2{t^3}}}$
2) $\frac{{t(2 - {t^3})}}{{1 - 2{t^3}}}$
3) $\frac{{t(2 + {t^3})}}{{1 + 2{t^3}}}$
4) $\frac{{t(2 - {t^3})}}{{1 + 2{t^3}}}$
Solution: Explanation: Explanation
Implicit-function

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