use implicit differentiation to find \\( \\frac { d r } { d \\theta } \\).\n\n\\( \\cos \\left( r \\theta ^…

Explanation:

Step1: Differentiate both sides with respect to $\theta$

Differentiate $\cos(r\theta^{5})$ using the chain rule. Let $u = r\theta^{5}$, then $\frac{d}{d\theta}(\cos(u))=-\sin(u)\cdot\frac{du}{d\theta}$. And $\frac{du}{d\theta}=r\cdot5\theta^{4}+\theta^{5}\cdot\frac{dr}{d\theta}$ (product rule). The right - hand side: $\frac{d}{d\theta}(\frac{1}{6}) = 0$. So, $-\sin(r\theta^{5})\left(5r\theta^{4}+\theta^{5}\frac{dr}{d\theta}\right)=0$.

Step2: Solve for $\frac{dr}{d\theta}$

Since $-\sin(r\theta^{5})\neq0$ (otherwise the original equation $\cos(r\theta^{5})=\frac{1}{6}$ would not hold), we can divide both sides by $-\sin(r\theta^{5})$. We get $5r\theta^{4}+\theta^{5}\frac{dr}{d\theta}=0$. Then $\theta^{5}\frac{dr}{d\theta}=- 5r\theta^{4}$.

Step3: Isolate $\frac{dr}{d\theta}$

Divide both sides by $\theta^{5}$ (assuming $\theta\neq0$). So, $\frac{dr}{d\theta}=-\frac{5r}{\theta}$.

Answer:

$-\frac{5r}{\theta}$