Question

find $\frac{dy}{dt}$. $y = 2t(3t^{3}-4)^{5}$ $\frac{dy}{dt}=square$

Explanation:

Step1: Apply product rule

The product rule states that if $y = u\cdot v$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Let $u = 2t$ and $v=(3t^{3}-4)^{5}$. First, find $\frac{du}{dt}$ and $\frac{dv}{dt}$. Since $\frac{du}{dt}=\frac{d}{dt}(2t) = 2$.

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

Let $w = 3t^{3}-4$, so $v = w^{5}$. By the chain - rule $\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}$. We know that $\frac{dv}{dw}=5w^{4}$ and $\frac{dw}{dt}=\frac{d}{dt}(3t^{3}-4)=9t^{2}$. Then $\frac{dv}{dt}=5(3t^{3}-4)^{4}\cdot9t^{2}=45t^{2}(3t^{3}-4)^{4}$.

Step3: Use product rule result

Substitute $u$, $v$, $\frac{du}{dt}$ and $\frac{dv}{dt}$ into the product - rule formula: $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}=2t\cdot45t^{2}(3t^{3}-4)^{4}+(3t^{3}-4)^{5}\cdot2$.

Step4: Simplify the expression

$\frac{dy}{dt}=90t^{3}(3t^{3}-4)^{4}+2(3t^{3}-4)^{5}=2(3t^{3}-4)^{4}(45t^{3}+3t^{3}-4)=2(3t^{3}-4)^{4}(48t^{3}-4)=4(3t^{3}-4)^{4}(24t^{3}-2)$.

Answer:

$4(3t^{3}-4)^{4}(24t^{3}-2)$