Problem in understanding Chain rule for partial derivatives

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?

Since $$\omega$$ is defined as $$f$$, then $$\partial f/\partial x = \partial \omega/\partial x$$. In general: $$\frac{\partial \omega}{\partial x} = \frac{\partial f}{\partial x},\quad \frac{\partial \omega}{\partial y} = \frac{\partial f}{\partial y}$$ $$\frac{\partial \omega}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t},\quad \frac{\partial \omega}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}.$$
• 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? Oct 19, 2018 at 13:37
• Your steps are not correct, because $f(x,y) = \sin xy$, so $\partial f/\partial x = y\cos xy$. Oct 19, 2018 at 13:39
• But $f$ corresponds to $sin$ only isn't it? Not the variables passed to the function Oct 19, 2018 at 13:42
• No, $f \colon \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto \sin xy$. Oct 19, 2018 at 13:42