# Partial derivative with unknown function f(x,y)

I got a task from my lecture, i have tried to do that and this is my solution for the task. I was confused because the function f(x,y), g(r,$$\theta$$), and h(r,$$\theta$$) were not defined, so i just can use the value of that functions.

[Click on this link to see my solution] (https://i.sstatic.net/MpxJN.png)

Is it my solution correct? Thanks for any help.

• No, it's not correct because it's a common trap instructors set. Hint: the partials of $f$ w.r.t. $x$ and $y$ are not functions of $r$ and $\theta$. Sep 3, 2019 at 4:12
• @NinadMunshi , oh i see. So, to calculate $\frac{\partial f}{\partial r}(r,\theta)$, i must calculate the value of $\frac{\partial f}{\partial x} (x,y)* \frac{\partial x}{\partial r} (r,\theta) + \frac{\partial f}{\partial y} (x,y)* \frac{\partial y}{\partial r} (r,\theta)$? Sep 3, 2019 at 4:27
• That is correct Sep 3, 2019 at 4:36
• Ok, thanks @NinadMunshi Sep 3, 2019 at 6:29

## 1 Answer

$$\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} = 2*6 + 4*8 = 44$$

Here the $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial x}$$ should be evaluated at $$(-1,1)$$