Answer to Question #207662 in Calculus for Nikhil rawat

Question

Apply Young's theorem to justify that

fxy(1,1)= fyx(1,1)

for the function f: R^2 to R , defined by

f(x,y) = | x+ y|

Answer 1

The function f(x, y) is defined in a neighborhood of a point $(1,1)$

$f(x, y)=|x+y|=x+y, x>0, y>0.$

The partial derivatives $f_x, f_y$ are defined in a neighborhood of $(1,1)$ and are differentiable at $(1, 1).$

$f_x=1, f_y=1$ in a neighborhood of $(1,1)$

$f_{xy}(1,1)=0, f_{yx}(1,1)=0$

We see that $f_{xy}(1,1)=0=f_{yx}(1,1)$ in accordance with the Young's theorem.