find $\frac{dr}{d\theta}$ for $\theta^{4/9}-r^{4/9}=-7$. $\frac{dr}{d\theta}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $\theta^{4/9}-r^{4/9}=-7$ with respect to $\theta$. The derivative of a constant $(-7)$ is 0. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $\frac{d}{d\theta}(\theta^{4/9})-\frac{d}{d\theta}(r^{4/9})=\frac{d}{d\theta}(-7)$. $\frac{4}{9}\theta^{\frac{4}{9}-1}-\frac{4}{9}r^{\frac{4}{9}-1}\frac{dr}{d\theta}=0$

Step2: Simplify the exponents

Simplify the exponents: $\frac{4}{9}\theta^{-\frac{5}{9}}-\frac{4}{9}r^{-\frac{5}{9}}\frac{dr}{d\theta}=0$

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

First, move $\frac{4}{9}\theta^{-\frac{5}{9}}$ to the other side: $-\frac{4}{9}r^{-\frac{5}{9}}\frac{dr}{d\theta}=-\frac{4}{9}\theta^{-\frac{5}{9}}$. Then divide both sides by $-\frac{4}{9}r^{-\frac{5}{9}}$ (assuming $r\neq0$). $\frac{dr}{d\theta}=\frac{\theta^{-\frac{5}{9}}}{r^{-\frac{5}{9}}}=\left(\frac{\theta}{r}\right)^{-\frac{5}{9}}=\left(\frac{r}{\theta}\right)^{\frac{5}{9}}$

Answer:

$\left(\frac{r}{\theta}\right)^{\frac{5}{9}}$