# How to evaluate this partial derivative in terms of polar coordinates [closed]

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

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

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• 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 – Kiran Brown Apr 22 at 9:57
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## 1 Answer

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