Give $D$ is an open set in $\mathbb{R}^n$, $f: D\to \mathbb{R}^p$ is differentiable on $D$. Supposing that $f$ has second derivative at $x_0\in D$.
For every $u\in \mathbb{R}^n$, give $g: D\to \mathbb{R}^p$ define by $$g(x)=f'(x)(u), \forall x\in D.$$ Prove that $g$ is differentiable at $x_0$ and $g'(x_0)(v)=f^{(2)}(x_0)(u,v)$.
My attempt:
Give $h\in \mathbb{R}^n$ that $x+h \in D$.
Firstly, we have $$g(x_0+h)-g(x_0)=f'(x_0+h)(u)-f'(x_0)(u)=\left[f'(x_0+h)-f'(x_0)\right](u) \quad (1)$$ Secondly, since $f$ have second dervative at $x_0$ that exist linear mapping $A:\mathbb{R}^n\to L(\mathbb{R}^n,\mathbb{R}^p)$ that $$f'(x_0+h)-f'(x_0)=A(h)+\vert h \vert_2\varphi(h) \quad (2)$$ with $\lim\limits_{h \to 0_{\mathbb{R}^n}}\varphi(h)=0_{L(\mathbb{R}^n,\mathbb{R}^p)}$.
From (1) and (2), we infer that $$g(x_0+h)-g(x_0)=A(h)(u)+\vert h \vert_2\varphi(h)(u)$$
I stuck here now.
I wonder that there exists linear mapping $A_1: \mathbb{R}^n \to \mathbb{R}^p$ and function $\varphi_1:\mathbb{R}^n \to \mathbb{R}^p$ that $A_1(u)(h)=A(h)(u)$ and $\varphi_1(u)(h)=\varphi(h)(u)$?