If y = t^2 + t^3 and x = t - t^4,then the val | vedclass

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(D) Given $y = t^2 + t^3$ and $x = t - t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ No

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$\frac{-4}{3}$

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(D) Given $y = t^2 + t^3$ and $x = t - t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t + 3t^2}{1 - 4t^3}$. To find $\frac{d^2y}{dx^2}$,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) \cdot \frac{dt}{dx} = \frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) \cdot \frac{1}{1 - 4t^3}$. Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$: $\frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) = \frac{(2 + 6t)(1 - 4t^3) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2} = \frac{2 - 8t^3 + 6t - 24t^4 + 24t^3 + 36t^4}{(1 - 4t^3)^2} = \frac{12t^4 + 16t^3 + 6t + 2}{(1 - 4t^3)^2}$. Thus,$\frac{d^2y}{dx^2} = \frac{12t^4 + 16t^3 + 6t + 2}{(1 - 4t^3)^3}$. At $t = 1$: $\frac{d^2y}{dx^2} = \frac{12(1)^4 + 16(1)^3 + 6(1) + 2}{(1 - 4(1)^3)^3} = \frac{12 + 16 + 6 + 2}{(1 - 4)^3} = \frac{36}{(-3)^3} = \frac{36}{-27} = -\frac{4}{3}$.

If $y = t^2 + t^3$ and $x = t - t^4$,then the value of $\frac{d^2 y}{d x^2}$ at $t = 1$ is equal to

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