Partial differentiation of a function

I found this definition of function $\phi_{p}(x)$: $\phi_{p}(x)=\sum_{i}(x^{i}-x^{i}(p))\frac{\partial \phi}{\partial x^{i}}(x)-(\phi(x)-\phi(p))$. Conclusion is that $\phi_{p}(p)=0$. Should we then put $x^{i}(x)$ instead of $x^{i}$ in the first part of the right hand side of the equation above. And if we do that, how can we get that $\frac{\partial^{2}\phi_{p}}{\partial x^{i}\partial x^{j}}(p)=\frac{\partial^{2}\phi}{\partial x^{i}\partial x^{j}}(p)$ (the other conclusion). I am sure that this is a very trivial thing but at this moment i can't see why.

-

Well for the first conclusion yes, and for the the second:

$$\frac{\partial \phi_p}{\partial x^k} = \sum_i \frac{\partial x^i}{\partial x^k} \frac{\partial \phi}{\partial x^i} + \sum_i(x^i -x^i (p)) \frac{\partial ^2 \phi}{\partial x^i \partial x^k} - \frac{\partial \phi}{\partial x^k}$$

$$= \sum_i \delta_{ki} \frac{\partial \phi}{\partial x^i} +\sum_i (x^i -x^i(p)) \frac{\partial ^2 \phi}{\partial x^i \partial x^k} -\frac{\partial \phi}{ \partial x^k}=\sum_i (x^i-x^i(p))\frac{\partial^2 \phi}{\partial x^i \partial x^k }$$

Differentiate once more to get:

$$\frac{\partial \phi_p}{\partial x^j \partial x^k} = \sum_i \delta_{ji}\frac{\partial^2 \phi}{\partial x^i \partial x^k } + (x^i -x^i(p)) \frac{\partial^3 \phi}{\partial x^j \partial x^i \partial x^k }$$

Plug in and get your answer (btw, the $\delta$ is Kronecker's delta.

Hope I helped.

-

A priori you are given a function $$\phi:\quad {\mathbb R}^n\to{\mathbb R},\qquad x\mapsto \phi(x)$$ of one vector variable, and you use this $\phi$ to construct a new function $\Phi$ of two vector variables, namely $$\Phi(p,x):=\sum_k (x_k-p_k)\phi_{.k}(x) \ -\ \bigl(\phi(x)-\phi(p)\bigr)\ ,$$ where differentiation with respect to $x_k$ is denoted by ${}_{.k}\$. Obviously $\Phi(p,p)=0$. It follows that $$\Phi_{.i}(p,x)=\phi_{.i}(x)+\sum_k (x_k-p_k)\phi_{.ki}(x)-\phi_{.i}(x)=\sum_k (x_k-p_k)\phi_{.ki}(x)\ ,$$ and at the next step $$\Phi_{.ij}(p,x)=\phi_{.ji}(x)+\sum_k(x_k-p_k)\phi_{.kij}(x)\ .$$ In particular we have $$\Phi_{.ij}(p,p)=\phi_{.ij}(p,p)$$ for all $i$, $j\in[n]$.

-