find $\frac{dr}{d\theta}$ for $\theta^{2/9}-r^{2/9}=3$. $\frac{dr}{d\theta}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $\theta^{\frac{2}{9}}-r^{\frac{2}{9}} = 3$ with respect to $\theta$. The derivative of a constant 3 is 0. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $\frac{2}{9}\theta^{\frac{2}{9}-1}-\frac{2}{9}r^{\frac{2}{9}-1}\frac{dr}{d\theta}=0$. $$\frac{2}{9}\theta^{-\frac{7}{9}}-\frac{2}{9}r^{-\frac{7}{9}}\frac{dr}{d\theta}=0$$

Step2: Isolate $\frac{dr}{d\theta}$

First, move the term with $\frac{dr}{d\theta}$ to one side: $\frac{2}{9}r^{-\frac{7}{9}}\frac{dr}{d\theta}=\frac{2}{9}\theta^{-\frac{7}{9}}$. Then, divide both sides by $\frac{2}{9}r^{-\frac{7}{9}}$. $$\frac{dr}{d\theta}=\frac{\theta^{-\frac{7}{9}}}{r^{-\frac{7}{9}}}=\left(\frac{\theta}{r}\right)^{-\frac{7}{9}}=\left(\frac{r}{\theta}\right)^{\frac{7}{9}}$$

Answer:

$\left(\frac{r}{\theta}\right)^{\frac{7}{9}}$