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I'm having trouble understanding the chain rule for partial derivatives. If I'm given that $\omega=f(x,y)$ where $x$ and $y$ are functions of both $t$ and $r$, then by chain rule I can write that: $$\frac{\partial \omega}{\partial t}=\frac{\partial \omega}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial \omega}{\partial y}\frac{\partial y}{\partial t}$$

But if I'm asked to find out what $\frac{\partial f}{\partial x}$ is equal to then can I write that it's equal to $\frac{\partial \omega}{\partial x}?$ If I'm wrong then what is $\frac{\partial f}{\partial x}$ equal to?

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Since $\omega$ is defined as $f$, then $\partial f/\partial x = \partial \omega/\partial x$. In general: $$ \frac{\partial \omega}{\partial x} = \frac{\partial f}{\partial x},\quad \frac{\partial \omega}{\partial y} = \frac{\partial f}{\partial y} $$ $$ \frac{\partial \omega}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t},\quad \frac{\partial \omega}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}. $$

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  • $\begingroup$ If I consider an example, $w=sin(xy)$, then $\frac{\partial \omega}{\partial x}=ycos(xy)$. But $\frac{\partial f}{\partial x}=cosxy$. Which means they're not equal? $\endgroup$ Oct 19, 2018 at 13:37
  • $\begingroup$ Your steps are not correct, because $f(x,y) = \sin xy$, so $\partial f/\partial x = y\cos xy$. $\endgroup$
    – Gibbs
    Oct 19, 2018 at 13:39
  • $\begingroup$ But $f$ corresponds to $sin$ only isn't it? Not the variables passed to the function $\endgroup$ Oct 19, 2018 at 13:42
  • $\begingroup$ No, $f \colon \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto \sin xy$. $\endgroup$
    – Gibbs
    Oct 19, 2018 at 13:42

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