if x,y are sufficiently close to 0 and f is differentiable, show that $$f(x,y)\approx f(0,0)+xf_x(0,0)+yf_y(0,0)$$
I tried messing around with the definition of partial derivatives and Leibniz notation etc. and failed to make any progress.
So I decided to solve it inductively by example,
For example;
LHS: $f(x,y)=x^2+y^2+2$,
$\,f_x(0,0)=(2x)_0^0=0,\,f_y(0,0)=(2y)_0^0=0,\,f(0,0)=2$,
RHS: $\,xf_x(0,0)+yf_y(0,0)+f(0,0)=2$,
To me this example seems completely redundant. So for a function that has partial derivatives at $(x,y)=(0,0)$ equal to $0$, the function is approximately equal to $f(0,0)$ as x and y approach 0.
I'm essentially telling myself that as x and y approach $0$ the function will be approximately be equal to the evaluation of the function.. lol
However not every reasonable functions have null partial derivatives evaluated at the origin. for example consider the function $x^2+2x+y$
LHS: $f(x,y)=x^2+2x+y$
RHS: $\,xf_x(0,0)+yf_y(0,0)+f(0,0)=x(2x+2)^0_0+y(1)^0_0+0=2x+y$,
when $x$ and $y$ are sufficiently close to $0$ $f$ will also take a value extremely close to zero. therefore this relation makes absolutely no sense.
I think it will even be less true for functions of products of x and y and I will provide examples of this if you guys think this will help me understand.
Regardless, could the community give me an example, or definition that I could use to address this situation, please don't solve this for me just give me a piece of information to make my life easier.