How to evaluate this partial derivative in terms of polar coordinates

Question

How to evaluate this partial derivative in terms of polar coordinates?

How to solve this question?

Answer 1

The chain rule says that $\frac{\partial z}{\partial r}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$.

Here, $z= ln(x+ 2y)$ so $frac{1}{x+2y}$ and $\frac{\partial z}{\partial y}= \frac{2}{x+ 2y}$. $x= r cos(\theta)$ so $\frac{\partial x}{\partial r}= cos(\theta)$ and $y= r sin(\theta)$ so $\frac{\partial y}{\partial r}= -r sin(\theta)$.

$\frac{\partial z}{\partial r}= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}= \frac{cos(\theta)}{x+ 2y}+ \frac{2 sin(\theta)}{x+ 2y}$.

If you want that entirely in terms of r and $\theta$, replace x with $r cos(\theta)$ and replace y with $r sin(\theta)$: $\frac{\partial z}{\partial r}= \frac{cos(\theta)}{r(cos(theta)+ 2sin(\theta))}+ \frac{sin(\theta)}{r(cos(theta)+ 2sin(\theta))}$.