Combined Partial Derivatives Suppose $f : \mathbb{R}^2 \to \mathbb{R}$ is in $C^1(\mathbb{R}^2)$. Does
$$
\frac{f(x_1+h, x_2+h)-f(x_1, x_2)}{h} \to \frac{\partial f}{\partial x_1} + \frac{\partial f}{\partial x_2}
$$
as $h \to 0$ ? I am led to believe this is true, but I can't quite see how to prove it. I would like to "break" $f(x_1+h, x_2+h)$ in two, but in general I don't see how this can be done. 

The function $g:\mathbb{R}\to \mathbb{R},\ t\mapsto g(t)=f(x_1+t,x_2+t)$ is $C^1$, and we have
$$
g(t)=g(0)+tg'(0)+o(t)=f(x_1,x_2)+t\frac{\partial f}{\partial x_1}(x_1,x_2)+t\frac{\partial f}{\partial x_2}(x_1,x_2)+o(t).
$$
Hence
$$
\lim_{h\to0}\frac{f(x_1+h,x_2+h)-f(x_1,x_2)}{h}=\frac{\partial f}{\partial x_1}(x_1,x_2)+\frac{\partial f}{\partial x_2}(x_1,x_2).
$$
