TITLE: Differentiability of the remainder in Taylor's theorem (1D vs 2D)
QUESTION [5 upvotes]: I am interested in the Lagrange's form of the remainder for $C^1$ function $f$ defined in a convex neighborhood of the origin.
One dimension case. Let $f:\Omega\to\mathbb{R},$ where $0\in\Omega\subset\mathbb{R}.$ Then
$$f(x)=f(0)+\overbrace{x\int_0^1\frac{\partial f}{\partial x}(tx)dt}^{R(x)}.$$
Two dimensions case. Let $f:\Omega\to\mathbb{R},$ where $(0,0)\in\Omega\subset\mathbb{R}^2.$ Then
$$f(x,y)=f(0,0)+\overbrace{x\int_0^1\frac{\partial f}{\partial x}(tx,ty)dt}^{R_x(x,y)}+\overbrace{y\int_0^1\frac{\partial f}{\partial y}(tx,ty)dt}^{R_y(x,y)}.$$
Some people may call above formulas Hadamard's lemma, but I will stick to Taylor's formula.
Clearly $R$ is $C^1$ cause $R(x)=f(x)-f(0).$ However we cannot use the same argument for multivariable case.
Question. Are $R_x,R_y$ also $C^1$?

REPLY [2 votes]: This is the same answer I posted over at MO (normally a comment with a link might be sufficient, but since there's a bounty here now, I might as well try to collect it rather than let it expire).
No, not in general. Consider $f(x,y)=(x+y)|x+y|$. This is $C^1$ with partial derivatives $f_x=f_y=2|x+y|$, but you lose one derivative when you form
$$
R= x \int_0^1 2|tx+ty|\, dt = x|x+y| .
$$