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

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(D) Given $x = 4t^3 + 3$ and $y = 3t^4 + 4$.$\frac{dx}{dt} = 12t^2$ and $\frac{dy}{dt} = 12t^3$.$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{12t^2}{12

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$5$

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(D) Given $x = 4t^3 + 3$ and $y = 3t^4 + 4$.$\frac{dx}{dt} = 12t^2$ and $\frac{dy}{dt} = 12t^3$.$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{12t^2}{12t^3} = \frac{1}{t} = t^{-1}$.Now, $\frac{d^2x}{dy^2} = \frac{d}{dy}(\frac{dx}{dy}) = \frac{d}{dt}(\frac{dx}{dy}) \cdot \frac{dt}{dy}$.$\frac{d^2x}{dy^2} = \frac{d}{dt}(t^{-1}) \cdot \frac{1}{12t^3} = (-t^{-2}) \cdot \frac{1}{12t^3} = -\frac{1}{12t^5}$.Given $\frac{d^2x}{dy^2} = (\frac{dx}{dy})^n$, so $-\frac{1}{12t^5} = (t^{-1})^n = t^{-n}$.For this to be constant, the power of $t$ must be zero, but here the expression is not constant unless we re-evaluate the condition. Re-checking: $\frac{d^2x}{dy^2} = -\frac{1}{12} (t^{-1})^5 = -\frac{1}{12} (\frac{dx}{dy})^5$. Thus $n = 5$.

If $x = 4t^3 + 3$, $y = 3t^4 + 4$ and $\frac{d^2x}{dy^2} = (\frac{dx}{dy})^n$ is constant, then the value of $n$ is

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