Why $\Delta f \approx f_x(x_0,y_0) \Delta x+f_y(x_0,y_0) \Delta y$

Question

Why $\Delta f \approx f_x(x_0,y_0) \Delta x+f_y(x_0,y_0) \Delta y$

The part that I don't get it is why the sum of the two differential is approximately equal to $\Delta f$?

Answer 1

It follows from the definition of Fréchet derivative: We call $f: \mathbb{R}^n \to \mathbb{R}^m$ differentiable at a point $a \in \text{Int}(\text{dom(f))}$, if there exists a linear transformation $A: \mathbb{R}^n \to \mathbb{R}^m$ and a function $r:\text{dom}(f) \to \mathbb{R}^m$ with $\lim\limits_{x \to a} \frac{r(x)}{||x-a||}=0$, so that $f$ has the following form: $$f(x)=f(a)+A(x-a)+r(x)$$ And of course $A$ is called as the derivative of $f$ at $a$. So the best linear approximation around $a$ is: $$f(x) \approx f(a)+A(x-a)$$ In the case of $\mathbb{R}^2 \to \mathbb{R}$ functions, $A=(\nabla f)(a)$, so: $$f(x) \approx f(a) +(\nabla f)(a) \cdot (x-a)$$ Or if you want to use the $\Delta$ notation: $$f(x) -f(a) \approx (\nabla f)(a) \cdot (x-a)$$ $$\Delta f \approx (\nabla f)(a) \cdot \Delta x $$

Answer 2

Here's one way to look at it:

\begin{align} \underbrace{f(x + \Delta x, y + \Delta y) - f(x,y)}_{\Delta f} &= \underbrace{f(x+ \Delta x,y+ \Delta y) - f(x + \Delta x, y)}_{\approx D_2 f(x + \Delta x,y) \Delta y} + \underbrace{f(x + \Delta x, y) - f(x,y)}_{D_1 f(x,y) \Delta x}. \end{align} If we make the further approximation that $$ D_2 f(x + \Delta x, y) \approx D_2 f(x,y), $$ then we get the approximation $$ \Delta f \approx D_1 f(x,y) \Delta x + D_2 f(x,y) \Delta y. $$

Comment: Are the approximations above accurate? It seems plausible that if $D_1 f$ and $D_2 f$ exist in a neighborhood of $(x,y)$, and are continuous at $(x,y)$, then the approximations are good when $\Delta x$ and $\Delta y$ are small. Analysis textbooks show how to convert the above argument into a rigorous proof that having continuous partial derivatives at $(x,y)$ is sufficient to guarantee that $f$ is Frechet differentiable at $(x,y)$.