Another one on chain rule

  • Thread starter twoflower
  • Start date
  • #1
368
0
Hi,
here is what I'm trying to do:

Find

[tex]
\frac{\partial}{\partial x} f(2x, 3y)
[/tex]

First of all, I'm confused by the

[tex]
f(2x, 3y)
[/tex]

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

[tex]
f(x,y) = cos(xy) - sin(3xy^2}
[/tex]

and that therefore

[tex]
f(2x, 3y) = cos(6xy) - sin(54xy^2)
[/tex]

I'm confused, I don't have any good picture of how it might look in real.
Anyway, to accomplish the task:

[tex]
\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}
[/tex]

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.
 

Answers and Replies

  • #2
379
0
[tex]f(x,y)[/tex] 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.
 
  • #3
368
0
ComputerGeek said:
[tex]f(x,y)[/tex] 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.
 
  • #4
84
0
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)
 

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