If y = at^2 + 2bt + c and t = ax^2 + 2bx + c, | vedclass

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(D) Given $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{d

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$24 a^2 (ax + b)$

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(D) Given $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$. First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$. $\frac{dy}{dt} = 2at + 2b = 2(at + b)$. $\frac{dt}{dx} = 2ax + 2b = 2(ax + b)$. So,$\frac{dy}{dx} = 2(at + b) \cdot 2(ax + b) = 4(at + b)(ax + b)$. Now,differentiate with respect to $x$: $\frac{d^2y}{dx^2} = 4 \left[ \frac{d}{dx}(at + b) \cdot (ax + b) + (at + b) \cdot \frac{d}{dx}(ax + b) \right]$. Since $\frac{dt}{dx} = 2(ax + b)$,we have $\frac{d}{dx}(at + b) = a \cdot \frac{dt}{dx} = 2a(ax + b)$. $\frac{d^2y}{dx^2} = 4 \left[ 2a(ax + b)^2 + (at + b) \cdot a \right] = 8a(ax + b)^2 + 4a(at + b)$. Finally,differentiate again to find $\frac{d^3y}{dx^3}$: $\frac{d^3y}{dx^3} = 8a \cdot 2(ax + b) \cdot a + 4a \cdot \frac{d}{dx}(at + b) = 16a^2(ax + b) + 4a \cdot 2a(ax + b) = 16a^2(ax + b) + 8a^2(ax + b) = 24a^2(ax + b)$.

If $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$,then $\frac{d^3y}{dx^3}$ equals

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