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

Explanation:

Step1: Differentiate both sides with respect to (\theta)

Using the chain rule, (\frac{d}{d\theta}(\sin(r\theta^{4}))=\frac{d}{d\theta}(\frac{1}{6})). The left - hand side: (\cos(r\theta^{4})\cdot\frac{d}{d\theta}(r\theta^{4})), and the right - hand side is (0).

Step2: Apply the product rule on (\frac{d}{d\theta}(r\theta^{4}))

By the product rule ((uv)^\prime = u^\prime v+uv^\prime) (where (u = r) and (v=\theta^{4})), (\frac{d}{d\theta}(r\theta^{4})=\frac{dr}{d\theta}\cdot\theta^{4}+r\cdot4\theta^{3}). So we have (\cos(r\theta^{4})\left(\theta^{4}\frac{dr}{d\theta}+4r\theta^{3}\right)=0). Since (\cos(r\theta^{4})\neq0) (because (\sin(r\theta^{4})=\frac{1}{6}), and (\cos^{2}x = 1-\sin^{2}x), so (\cos(r\theta^{4})\neq0)), then (\theta^{4}\frac{dr}{d\theta}+4r\theta^{3}=0).

Step3: Solve for (\frac{dr}{d\theta})

Isolate (\frac{dr}{d\theta}): [ \begin{align*} \theta^{4}\frac{dr}{d\theta}&=- 4r\theta^{3}\ \frac{dr}{d\theta}&=-\frac{4r}{\theta} \end{align*} ]

Answer:

(\frac{dr}{d\theta}=-\frac{4r}{\theta})