- Thread starter
- Banned
- #1

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

- Admin
- #2

- Jan 26, 2012

- 4,197

You can do

$$df= \frac{ \partial f}{ \partial x}dx+ \frac{ \partial f}{ \partial y}dy+ \frac{ \partial f}{ \partial y}dy,$$

where each sum is vector addition. That is, $df$ is a vector made up of

$$df= \begin{bmatrix} df_{1} \\ df_{2}\end{bmatrix},$$

and you follow the usual rules for finding the differential of each component.

- Thread starter
- Banned
- #3

- Thread starter
- Banned
- #4

- Admin
- #5

- Jan 26, 2012

- 4,197

I get

$$df(1,0,-1)=\begin{bmatrix}2\,dx-2 \, dy \\ -dy \end{bmatrix}.$$

As a slight correction to Jameson's post, I think I would write it as

$$df= \begin{bmatrix} f_{1x} & f_{1y} & f_{1z} \\ f_{2x} &f_{2y} &f_{2z}\end{bmatrix}

\begin{bmatrix} dx \\ dy \\ dz \end{bmatrix}.$$

- Admin
- #6

- Jan 26, 2012

- 4,052

Yep, that's exactly it. The dimensions of my matrix didn't work with the function mapping R^3 to R^2. You're also right that the (x, y, z) coordinate should be evaluated for each partial derivative in the matrix. Thanks for clearing this up.