Question

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

Explanation:

Step1: Apply product rule

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

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

Let $w = 3t^{3}-4$, so $v = w^{4}$. By the chain - rule $\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}$.
$\frac{dv}{dw}=4w^{3}=4(3t^{3}-4)^{3}$ and $\frac{dw}{dt}=9t^{2}$
So, $\frac{dv}{dt}=4(3t^{3}-4)^{3}\cdot9t^{2}=36t^{2}(3t^{3}-4)^{3}$

Step3: Use product rule to find $\frac{dy}{dt}$

$\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$
$=2t\cdot36t^{2}(3t^{3}-4)^{3}+(3t^{3}-4)^{4}\cdot2$
$=72t^{3}(3t^{3}-4)^{3}+2(3t^{3}-4)^{4}$
$=2(3t^{3}-4)^{3}(36t^{3}+3t^{3}-4)$
$=2(3t^{3}-4)^{3}(39t^{3}-4)$

Answer:

$2(3t^{3}-4)^{3}(39t^{3}-4)$