5. 0/1 points details my notes find the differential of the function. y = θ³ sin(6θ) dy =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Here, let $u = \theta^{3}$ and $v=\sin(6\theta)$.

Step2: Find $u^\prime$

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

Step3: Find $v^\prime$

Differentiate $v=\sin(6\theta)$ with respect to $\theta$. Using the chain - rule $\frac{d}{d\theta}(\sin(f(\theta)))=f^\prime(\theta)\cos(f(\theta))$, where $f(\theta)=6\theta$ and $f^\prime(\theta) = 6$. So $v^\prime=6\cos(6\theta)$.

Step4: Calculate $y^\prime$

By the product - rule $y^\prime=u^\prime v+uv^\prime=3\theta^{2}\sin(6\theta)+\theta^{3}\cdot6\cos(6\theta)=3\theta^{2}\sin(6\theta)+6\theta^{3}\cos(6\theta)$.

Step5: Find the differential $dy$

Since $dy = y^\prime d\theta$, then $dy=(3\theta^{2}\sin(6\theta)+6\theta^{3}\cos(6\theta))d\theta$.

Answer:

$(3\theta^{2}\sin(6\theta)+6\theta^{3}\cos(6\theta))d\theta$