Mathematica Asked on January 25, 2021
I have this function:
f = Sqrt(Abs(x^2 - x*y))
So, how can I prove that it is derivable and/or continue in a point, like (0,0)?
A good question for learning purpose. Let us somewhat refine the definition of the function under consideration:
f[x_, y_] := Sqrt[RealAbs[x^2 - x*y]]
We start from the verification of its continuity at the origin through
Limit[f[x, y], {x, y} -> {0, 0}] == f[0, 0]
(*True*)
Next, in order to test its differentiability at the origin, we find its partial derivatives at the origin by
Limit[(f[x, y] /. y -> 0 - f[0, 0])/x, x -> 0]
(*Indeterminate*)
and the analogous limit for the partial derivative by $y$. Since the partial derivative by $x$ does not exist at the origin, the function under consideration is not differentiable at the origin.
Try another point on your own.
Addition. On the OP's request, let
f[x_, y_] := Sqrt[x^2 + y^2]*Cos[x*y]
Then
D[f[x, y], x] /. {x -> 0, y -> 1}
(*0*)
and
D[f[x, y], y] /. {x -> 0, y -> 1}
(*1*)
Up to the definition of the differentiability , the condition
Limit[(f[x, y] - f[0, 1] + 0*(x - 0) - 1*(y - 1))/Sqrt[x^2 + y^2],{x, y}-> {0, 1}] == 0
(*True*)
should be tested in this case.
Edit. It should be
Limit[(f[x, y] - f[0, 1] + 0*(x - 0) - 1*(y - 1))/Sqrt[x^2 + (y - 1)^2], {x, y} -> {0, 1}] == 0
(*0*)
in the above. Sorry for the typo.
Correct answer by user64494 on January 25, 2021
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