find $\frac{dy}{dt}$. \n$y = 2t(3t^{2}-4)^{3}$\n$\frac{dy}{dt}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u = 2t$ and $v=(3t^{2}-4)^{3}$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. First, find $\frac{du}{dt}$ and $\frac{dv}{dt}$. $\frac{du}{dt}=\frac{d}{dt}(2t) = 2$

Step2: Apply chain - rule to find $\frac{dv}{dt}$

Let $w = 3t^{2}-4$, so $v = w^{3}$. By the chain - rule $\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}$. $\frac{dv}{dw}=\frac{d}{dw}(w^{3}) = 3w^{2}$ and $\frac{dw}{dt}=\frac{d}{dt}(3t^{2}-4)=6t$. Then $\frac{dv}{dt}=3(3t^{2}-4)^{2}\cdot6t = 18t(3t^{2}-4)^{2}$

Step3: Substitute into product - rule formula

$\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}=2t\cdot18t(3t^{2}-4)^{2}+(3t^{2}-4)^{3}\cdot2$ $=36t^{2}(3t^{2}-4)^{2}+2(3t^{2}-4)^{3}$ $=2(3t^{2}-4)^{2}[18t^{2}+(3t^{2}-4)]$ $=2(3t^{2}-4)^{2}(21t^{2}-4)$

Answer:

$2(3t^{2}-4)^{2}(21t^{2}-4)$