If $f\in C^2$ is a real function of two variables, we know that $f(x+h,y+k)-f(x,y)= hf_1(x+\theta h,y+\theta k)+kf_2(x+\theta h, y+\theta k)$ for some $\theta \in (0,1)$. I know how to derive this identity (using taylor series). However, in order to save time is there an exact analog for functions of more than two varibles?
For example, is it true that if $f\in C^2$ is a real function of three variables then $f(x+h,y+k,z+l)-f(x,y,z)= hf_1(x+\theta h,y+\theta k,z+\theta l)+kf_2(x+\theta h, y+\theta k, z+\theta l)+lf_3(x+\theta h,y+\theta k, z+\theta l)$ for a $\theta\in (0,1)$? As is clear I haven’t yet proved a similar taylor series anolog for real functions of more than two variables that’s why I wonder if my guess is correct.
You can define $$g:\mathbb{R}\rightarrow \mathbb{R}, g(t)=f(x_1+th_1, x_2+th_2, \dots, x_n +th_n)$$ Then everything reduces to the mean value theorem of one variable. Because you get $$ g(1)-g(0)= g'(\Theta)$$ for some $\Theta\in (0;1)$ by the mean value theorem. However, by the Leibniz formula and the chain rule you obtain for the derivate $$ g'(t)= \sum_{j=1}^n f_j(x_1 + t h_1, \dots, x_n + th_n)\cdot (t h_j)' = \sum_{j=1}^n h_n \cdot f_j(x_1 + th_1, \dots, x_n + t h_n) $$ which is exactly your expression.