Taylor's theorem implies $f(x,y) - f(a,b) = (x-a)\frac{\partial}{\partial x}f(\bar{x},y) + (y-b)\frac{\partial}{\partial y}(fx,\bar{y})$?

Question

Let $f: B_{r}(a,b) \subset \mathbb{R}^{2} \to \mathbb{R}$ be a differentiable function. Prove that for all $(x,y) \in B_{r}(a,b)$ there is $\bar{x} \in [a,x]$ and $\bar{y} \in [b,y]$ such that $$f(x,y) - f(a,b) = (x-a)\frac{\partial}{\partial x}f(\bar{x},y) + (y-b)\frac{\partial}{\partial y}f(\color{red}{a},\bar{y})$$

I dont know if there is any typo in the question (where is red), but it seems like Taylor expansion.

We can write $f(x,y)$ as $$f(x,y) = f(a,b) + (x,a)\frac{\partial}{\partial x}f(a,b) + (y-b)\frac{\partial}{\partial y}f(a,b) + \frac{1}{2!}\left(\cdots\right) \cdots$$

I cannot see which way to complete the question using Taylor.

Answer 1

Note that $ f(x, y)-f(a, b)=[f(x, y)-f(a, y)]+[f(a, y)-f(a, b)] $.

Then use the mean value theorem, we have $$ f(x, y)-f(a, y)=(x-a)\frac{\partial}{\partial x}f(\bar{x}, y) $$ and $$ f(a, y)-f(a, b)=(y-b)\frac{\partial}{\partial y}f(a, \bar{y}) $$ for some $ \bar{x}\in [a, x] $ and $ \bar{y}\in [b, y] $.

Hence

$$ f(x, y)-f(a, b)=(x-a)\frac{\partial}{\partial x}f(\bar{x}, y)+(y-b)\frac{\partial}{\partial y}f(a, \bar{y}) .$$