If x = frac{t^2}{1+t^5} and y = frac{2t^3}{1+ | vedclass

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(B) Given $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$.Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v

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$\frac{2t(3-2t^5)}{(2-3t^5)}$

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(B) Given $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$.Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$:$\frac{dx}{dt} = \frac{(1+t^5)(2t) - t^2(5t^4)}{(1+t^5)^2} = \frac{2t + 2t^6 - 5t^6}{(1+t^5)^2} = \frac{2t - 3t^6}{(1+t^5)^2} = \frac{t(2 - 3t^5)}{(1+t^5)^2}$.$\frac{dy}{dt} = \frac{(1+t^5)(6t^2) - 2t^3(5t^4)}{(1+t^5)^2} = \frac{6t^2 + 6t^7 - 10t^7}{(1+t^5)^2} = \frac{6t^2 - 4t^7}{(1+t^5)^2} = \frac{2t^2(3 - 2t^5)}{(1+t^5)^2}$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t^2(3 - 2t^5)}{(1+t^5)^2} 	imes \frac{(1+t^5)^2}{t(2 - 3t^5)} = \frac{2t(3 - 2t^5)}{(2 - 3t^5)}$.

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter,then find $\frac{dy}{dx}$.

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