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

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How to solve this question?

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closed as off-topic by Eevee Trainer, Kavi Rama Murthy, Cesareo, user3658307, José Carlos Santos Apr 28 at 16:01

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  • $\begingroup$ I think I have the solution but I can't find the solution for the problem so I would like someone to give me the answer $\endgroup$ – Kiran Brown Apr 22 at 9:57
  • $\begingroup$ Welcome to MSE. It will be better for you in the long run if you post your (attempted) solutions. Then we know how to help or how to explain misconceptions you have. $\endgroup$ – Ted Shifrin Apr 22 at 23:19
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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 $r sin(\theta)$: $\frac{\partial z}{\partial r}= \frac{cos(\theta)}{r(cos(theta)+ 2sin(\theta))}+ \frac{sin(\theta)}{r(cos(theta)+ 2sin(\theta))}$.

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