Derivative: $f_x, f_y, f_{xy}$ of function - $f(x,y)$

Question

Let's say $f(x,y) = x^2 + 2xy +y^2$

$f'_x = 2x + 2y$

$f'_y = 2y + 2x$

$f'_{xy} = 2x + 2y$ ?

Am I right about the third?

Answer 1

$$f_x = 2x + 2y$$

$$f_y = 2y + 2x$$

$$f_{xy} = \frac{\partial{f_x}}{\partial{y}}=\frac{\partial{(2x + 2y)}}{\partial{y}}=\frac{\partial{(2x)}}{\partial{y}}+\frac{\partial{(2y)}}{\partial{y}}=2\frac{\partial{x}}{\partial{y}}+2\frac{\partial{y}}{\partial{y}}\overset{(*)}{=} 2 \cdot 0+2 \cdot 1=2$$

$(*)$ When we differentiate $x$ with respect to $y$, $x$ is like a constant, so $\displaystyle{\frac{\partial{x}}{\partial{y}}=0}$.

$$\text{ OR }$$

$$f_{xy} = \frac{\partial{f_y}}{\partial{x}}=\frac{\partial{(2y + 2x)}}{\partial{x}}=\frac{\partial{(2y)}}{\partial{x}}+\frac{\partial{(2x)}}{\partial{x}}=2\frac{\partial{y}}{\partial{x}}+2\frac{\partial{x}}{\partial{x}}\overset{(**)}{=} 2 \cdot 1+2 \cdot 0=2$$

$(**)$ When we differentiate $y$ with respect to $x$, $y$ is like a constant, so $\displaystyle{\frac{\partial{y}}{\partial{x}}=0}$.