This is probably a really bad question with some major oversights, but I don't seem to see them right now.
If I define a function
$$ z = f(x,y) = x^2 + y^2$$
and took the partial derivative respect to $z$ of $z$ and $f(x,y)$ is it correct to say:
\begin{align}
\frac{\partial z}{\partial z} &= \frac{\partial f(x,y)}{\partial z} \\
1 &= 2x\frac{\partial x}{\partial z} + 2y\frac{\partial y}{\partial z} \ ?
\end{align}
I tried Wolfram Alpha but it gave me this instead.
Following the chain rule I get the same answer.
Best Answer
Woflram outputs this result because it assumes x and y don't depend on z, which means it finds that:
$\frac{\partial x}{\partial z} = 0, \frac{\partial y}{\partial z} = 0$
But yes, you can technically say:
$1=2x \frac{\partial x}{\partial z}+2y \frac{\partial y}{\partial z}$
Although it's rather meaningless. You're expecting the partials of x and y with respect to z to yield "something else", but x and y DO depend on z by:
$z=x^2 + y^2$
We can explicitly show it by:
$x=\sqrt{z-y^2}$ and $y=\sqrt{z-x^2}$
Then using the Chain Rule:
$\frac{\partial x}{\partial z} =\frac{1}{2 \sqrt{z-y^2}}(1-2y\frac{\partial y}{\partial z})=\frac{1}{2x}(1-2y\frac{\partial y}{\partial z})$ and $\frac{\partial y}{\partial z} =\frac{1}{2 \sqrt{z-x^2}}(1-2x\frac{\partial x}{\partial z})=\frac{1}{2y}(1-2x\frac{\partial x}{\partial z})$
Plugging them into each other:
$\frac{\partial x}{\partial z} =\frac{1}{2x}(1-2y(\frac{1}{2y}(1-2x\frac{\partial x}{\partial z})))=\frac{1}{2x}(2x\frac{\partial x}{\partial z})=\frac{\partial x}{\partial z}$
$\frac{\partial y}{\partial z} =\frac{1}{2y}(1-2x(\frac{1}{2x}(1-2y\frac{\partial y}{\partial z})))=\frac{1}{2y}(2y\frac{\partial y}{\partial z})=\frac{\partial y}{\partial z}$
And we come to a roundabout answer.