find the differential of the function. y = θ³ sin(6θ) dy = (3θ²sin(6θ)+θ³cos(6θ))dθ × need help? read it…

Explanation:

Step1: Recall the chain - rule

If $y = f(u)$ and $u = g(\theta)$, then $dy=\frac{dy}{du}\cdot\frac{du}{d\theta}d\theta$. Let $u = 7+\cos(\theta)$, so $y=\sqrt{u}=u^{\frac{1}{2}}$.

Step2: Differentiate $y$ with respect to $u$

Using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

Step3: Differentiate $u$ with respect to $\theta$

$\frac{du}{d\theta}=-\sin(\theta)$.

Step4: Calculate $dy$

Substitute $u = 7+\cos(\theta)$ into $\frac{dy}{du}$ and use the chain - rule $dy=\frac{dy}{du}\cdot\frac{du}{d\theta}d\theta$. So $dy=\frac{1}{2\sqrt{7 + \cos(\theta)}}\cdot(-\sin(\theta))d\theta=-\frac{\sin(\theta)}{2\sqrt{7+\cos(\theta)}}d\theta$.

Answer:

$-\frac{\sin(\theta)}{2\sqrt{7+\cos(\theta)}}d\theta$