# Another one on chain rule

Hi,
here is what I'm trying to do:

Find

$$\frac{\partial}{\partial x} f(2x, 3y)$$

First of all, I'm confused by the

$$f(2x, 3y)$$

How does the function look like? I imagine that it is for example

$$f(x,y) = cos(xy) - sin(3xy^2}$$

and that therefore

$$f(2x, 3y) = cos(6xy) - sin(54xy^2)$$

I'm confused, I don't have any good picture of how it might look in real.

$$\frac{\partial}{\partial x} f(2x, 3y) = \frac{\partial f}{\partial x}\frac{d(2x)}{dx} + \frac{\partial f}{\partial y}\frac{d(3y)}{dx} = 2\frac{\partial f}{\partial x}$$

Is it ok? If yes, could you please give an example of this scenario? I mean, how could f(2x, 3y) and f(x,y), respectively, look like so that I could see it with the particular functions?
Thank you very much.

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$$f(x,y)$$ is just a function mapping of your x and y coordinates.

As for the derivative, when you take the derivative with respect to x, the y values are treated as if they were constants, same for x when you derive with respect for y.

ComputerGeek said:
$$f(x,y)$$ is just a function mapping of your x and y coordinates.

As for the derivative, when you take the derivative with respect to x, the y values are treated as if they were constants, same for x when you derive with respect for y.
So, is my approach ok? And how could the functions look like?

Thank you.

Just imagine that

v(x)=2x, and w(y)=6y

Then

f(2x,6y)=f(v(x), w(y)) for arbitrary f.

for example
if
f(x)=cos(x)
then f(v)= cos(v)=cos(2x)