TITLE: Why $h\phi (x)$ is analog to $\frac{\partial f}{\partial x}h+ \frac{\partial f}{\partial y}k $?
QUESTION [0 upvotes]: I'm reading Lang's: Calculus of Several Variables. Here:


And comparing the two formulas:
$$f(x+h)-f(x)=h\phi (x) - hg(x)$$
$$f(x+h,y+k)-f(x,y)=\frac{\partial f}{\partial x}h+ \frac{\partial f}{\partial y}k - |H|g(H)$$
He does an analogy with the purpose of how to build the derivative for $2$ variables. I can understand/especulate why $hg(x)$ and $|H|g(H)$ are analog: If I take one coordinate vectors, then I guess $|(h)|=h$ and hence, it's just a norm in disguise. My problem is that I have no clue on why $h\phi (x)$ is analog to $\frac{\partial f}{\partial x}h+ \frac{\partial f}{\partial y}k $, specially why there are $k,h$ in there?

REPLY [0 votes]: I discovered it! It was really silly, I just took an example and it cleared things a bit. I took $f(x,y)=xy$ and wrote:
$$f(x+h,y+k)-f(x,y)=(x+h)(y+k)-xy=xk+hy+hk$$
And hence:
$$\left[ \frac{\partial f}{dx}h+\frac{\partial f}{dy}k=yh+xk\right]\sim [xk+hy+hk]$$
Now it feels more natural to accept that.
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