Implicit Differentiation: cot(r(Θ)^4) = 1/6 ????

Question

cot(r(Θ)^4) = 1/6 ---- Solving for dr/dΘ

So I differentiate, and I get -csc^2(r(Θ)^4)(4r(Θ)^3 + (dr/dΘ)(Θ^4)) = 0

farthest point I got to is (-4r(Θ)^3-(dr/dΘ)(Θ^4))/sin^2(r(Θ^4)) = 0

how are you supposed to put anything else on the right hand side when all is being multiplied?

Answer 1

$$\cot(r\theta^4)=\frac{1}{6}$$ $$-\csc^2(r\theta^4)\big(4r\theta^3+\frac{dr}{d\theta}\theta^4\big)=0$$ Notice that $\csc^2(x)\ne0$ for all $x\in \mathbb R$, so we can divide both sides by it: $$-(r\theta^4)\big(4r\theta^3+\frac{dr}{d\theta}\theta^4\big)=0$$ Now notice that since $$\cot(r\theta^4)=\frac{1}{6}$$ it is impossible for $\theta$ to equal $0$. Thus we can divide both sides by $\theta$ as much as we want, and the same goes for $r$: $$\big(4r\theta^3+\frac{dr}{d\theta}\theta^4\big)=0$$ $$4r+\frac{dr}{d\theta}\theta=0$$ $$\color{green}{\frac{dr}{d\theta}=-\frac{4r}{\theta}}$$

Answer 2

Consider the implicit function $$f=\cot(r\theta^4)-\frac{1}{6}=0$$ and compute the partial derivatives $$\frac{\partial f}{\partial r}=-\theta ^4 \csc ^2\left(r\theta ^4 \right)$$ $$\frac{\partial f}{\partial \theta}=-4 \theta ^3 r \csc ^2\left(r\theta ^4 \right)$$ and, from the implicit function theorem $$\frac{dr}{d \theta}=-\frac{\frac{\partial f}{\partial \theta} } {\frac{\partial f}{\partial r} }=-\frac{4 r}{\theta }$$