Deriving $\Delta z=\frac{\partial y}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y+\alpha\sqrt{\Delta x^2+\Delta y^2}$ I was reading a math book, which contained.  
"Let us consider a function $$z=f(x,y)$$ 
of two variables. If it has continuous partial derivatives, we can prove that its increment
$$\Delta z=f(x+\Delta x, y+\Delta y)-f(x,y)$$
corresponding to the increments $\Delta x$ and $\Delta y$ of its arguments, may be put in the form
$$\Delta z=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y+\alpha\sqrt{\Delta x^2+\Delta y^2}$$
And the differential will be the first 2 summands." 
Now how did they get the last expression?
 A: should $\frac{\partial y}{\partial x}$ be $\frac{\partial f}{\partial x}$ I would say this is a linear approximation, i.e. like $$f(x+h)=f(x)+hf'(x)+O(h^2)$$ but in two dimensions.
The $\sqrt{\Delta x^2+\Delta y^2}$ is a radius of point $(\Delta x, \Delta y)$, i.e. the distance from original point $(x,y)$
which plays the role of $h$ in this case. The $\alpha$ comes from the definition of $f=O(h)$ which means there is exists an $\alpha$ such that $f\le \alpha h$ and as you can see the equality in your expression should be changed by $\approx$.
A: As the question is written now, the second equation for $\Delta z$ is simply wrong.  The first order terms are correct (assuming $\partial y/\partial x$ is really $\partial f/\partial x$), but the "higher order terms" should depend on $f$.
Perhaps the equation was supposed to say:
\begin{equation}
\Delta z = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + o\left(\sqrt{\Delta x^2 + \Delta y^2}\right).
\end{equation}
(In this corrected equation we're using little-o notation.)
