Question

1. -/1 points find the differential of the function y = t³ sin(6t) dy =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, where $u$ and $v$ are functions of $t$, then $y^\prime=\frac{dy}{dt}=u^\prime v + uv^\prime$. Here, let $u = t^{3}$ and $v=\sin(6t)$.

Step2: Differentiate $u$

Differentiate $u = t^{3}$ with respect to $t$. Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $u^\prime=\frac{d}{dt}(t^{3}) = 3t^{2}$.

Step3: Differentiate $v$

Differentiate $v=\sin(6t)$ with respect to $t$. Using the chain - rule $\frac{d}{dt}(\sin(f(t)))=\cos(f(t))\cdot f^\prime(t)$. Here $f(t)=6t$, so $v^\prime=\frac{d}{dt}(\sin(6t))=\cos(6t)\cdot6 = 6\cos(6t)$.

Step4: Calculate $\frac{dy}{dt}$

By the product - rule $\frac{dy}{dt}=u^\prime v+uv^\prime$. Substitute $u = t^{3}$, $u^\prime = 3t^{2}$, $v=\sin(6t)$ and $v^\prime = 6\cos(6t)$ into the formula:
$\frac{dy}{dt}=3t^{2}\sin(6t)+t^{3}\cdot6\cos(6t)=3t^{2}\sin(6t) + 6t^{3}\cos(6t)$.

Step5: Find the differential $dy$

Since $dy=\frac{dy}{dt}dt$, we have $dy=(3t^{2}\sin(6t)+6t^{3}\cos(6t))dt$.

Answer:

$(3t^{2}\sin(6t)+6t^{3}\cos(6t))dt$