Derivative of $f(x, y)=x+y$ with respect to $z=(x, y)$

Solution 1:

If $ z $ is the pair $ ( x , y ) $, then a derivative with respect to $ z $ is essentially the gradient $$ \frac { \mathrm d f ( z ) } { \mathrm d z } = \nabla f ( z ) = \nabla f ( x , y ) = \left \langle \frac { \partial f ( x , y ) } { \partial x } , \frac { \partial f ( x , y ) } { \partial y } \right \rangle \text , $$ although it's more proper to write this as a matrix $$ \left [ \matrix { \displaystyle \frac { \partial f ( x , y ) } { \partial x } & \displaystyle \frac { \partial f ( x , y ) } { \partial y } } \right ] \text . $$

You should be careful with your Chain Rule, since this really involves some partial derivatives. You can write $$ \frac { \mathrm d f ( z ) } { \mathrm d z } = \frac { \partial f ( z ) } { \partial x } \frac { \mathrm d x } { \mathrm d z } + \frac { \partial f ( z ) } { \partial y } \frac { \mathrm d y } { \mathrm d z } \text , $$ and then since $ \mathrm d x / \mathrm d z = \langle \partial x / \partial x , \partial x / \partial y \rangle = \langle 1 , 0 \rangle $ and $ \mathrm d y / \mathrm d z = \langle \partial y / \partial x , \partial y / \partial y \rangle = \langle 0 , 1 \rangle $ (where I'm writing vectors rather than matrices for convenience), this ends up giving the answer that I said. But I don't think that this is helpful here.

Solution 2:

Not sure what you notion of derivative is here, but $f$ maps two real variables to one and hence, the derivative should be a $2\times 1$-matrix consisting of the partial derivatives of $f$, namely $$Df(x,y) = \left[\frac{df(x,y)}{dx}\quad \frac{df(x,y)}{dy}\right].$$