If y=t^2+t^3 and x=t-t^4,then frac{d^2 y}{d x | vedclass

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(B) Given $y=t^2+t^3$ and $x=t-t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ Then,$\fra

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$-\frac{4}{3}$

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(B) Given $y=t^2+t^3$ and $x=t-t^4$. First,find the derivatives with respect to $t$: $\frac{dy}{dt} = 2t + 3t^2$ $\frac{dx}{dt} = 1 - 4t^3$ Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t + 3t^2}{1 - 4t^3}$. Now,differentiate $\frac{dy}{dx}$ with respect to $x$ using the chain rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) \cdot \frac{1}{1 - 4t^3}$. Using the quotient rule: $\frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) = \frac{(1 - 4t^3)(2 + 6t) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2}$. Thus,$\frac{d^2y}{dx^2} = \frac{(1 - 4t^3)(2 + 6t) + 12t^2(2t + 3t^2)}{(1 - 4t^3)^3}$. At $t=1$: $\left.\frac{d^2y}{dx^2}\right|_{t=1} = \frac{(1 - 4)(2 + 6) + 12(1)^2(2(1) + 3(1)^2)}{(1 - 4)^3} = \frac{(-3)(8) + 12(5)}{(-3)^3} = \frac{-24 + 60}{-27} = \frac{36}{-27} = -\frac{4}{3}$.

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

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