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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.

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  • $\begingroup$ 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 $\theta$. $\endgroup$ Sep 3, 2019 at 4:12
  • $\begingroup$ @NinadMunshi , oh i see. So, to calculate $\frac{\partial f}{\partial r}(r,\theta)$, i must calculate the value of $\frac{\partial f}{\partial x} (x,y)* \frac{\partial x}{\partial r} (r,\theta) + \frac{\partial f}{\partial y} (x,y)* \frac{\partial y}{\partial r} (r,\theta)$? $\endgroup$
    – user136524
    Sep 3, 2019 at 4:27
  • $\begingroup$ That is correct $\endgroup$ Sep 3, 2019 at 4:36
  • $\begingroup$ Ok, thanks @NinadMunshi $\endgroup$
    – user136524
    Sep 3, 2019 at 6:29

1 Answer 1

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

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

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