Please give me an example $d:C[\mathbb{R}]\times‎ C[\mathbb{R}]\longrightarrow\mathbb{R}$ Such that: I need an example $d$ such that:
$$d:C[\mathbb{R}]\times‎ C[\mathbb{R}]\longrightarrow\mathbb{R}$$ 
$$C[\mathbb{R}]=\lbrace f:\mathbb{R}\longrightarrow\mathbb{R}\ | \ f ‎‎\text{is differentiable on }‎\mathbb{R}\rbrace$$
And $\forall f,g \in C[\mathbb{R}]$
$$1.\ \ d(f,g)=-d(g,f)$$
$$2.\ \ d(f,g)=0 \Longrightarrow f=g$$
Thanks to all of you.
 A: The construction I present below is a little bit contrived, but I think it works. In fact, no differentiability or even continuity is needed; it works on the space of all real-valued functions.
Define a function $s:C[\mathbb R]\times C[\mathbb R]\to\mathbb R$ as follows. If $f\in C[\mathbb R]$, $g\in C[\mathbb R]$, and $f=g$, then $s(f,g)\equiv 0$. If $f\neq g$, then define $s(f,g)$ to be any one point $x\in\mathbb R$ such that $f(x)\neq g(x)$. Specify also that $s(f,g)=s(g,f)$, so that one avoids considering the same pair of functions twice and assigning two different points at which they do not agree. (Note that I need to rely on the axiom of choice to construct such a function $s$.)
Define another function $t:C[\mathbb R]\times C[\mathbb R]\to\mathbb R$ as follows:
\begin{align*}
t(f,g)\equiv\begin{cases}\phantom{-}1&\text{if $f(s(f,g))>g(f(s,g))$,}\\\phantom{-}0&\text{if $f(s(f,g))=g(f(s,g))$,}\\-1&\text{if $f(s(f,g))<g(f(s,g))$}
\end{cases}
\end{align*}
for each $(f,g)\in C[\mathbb R]\times C[\mathbb R]$. Intuitively, $t$ “orders” each pair $(f,g)$ according as which one has a greater value at the distinguished point $s(f,g)$ at which they do not agree. [If $f(s(f,g))=g(s(f,g))$, then $f=g$ because of the way $s$ was constructed, but this doesn't really matter.]
Now define the desired function $d:C[\mathbb R]\times C[\mathbb R]\to\mathbb R$ as follows for each $(f,g)\in C[\mathbb R]\times C[\mathbb R]$:
\begin{align*}
d(f,g)\equiv t(f,g)\times\sup_{x\in\mathbb R}\left\{\frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}\right\}.
\end{align*}
Note that the supremum is always finite, given that the range of the function $y\mapsto y/(1+y)$ is $[0,1)$ on the domain $[0,\infty)$. Also, the supremum is zero if and only if $f(x)=g(x)$ for all $x\in\mathbb R$. Finally, the way $t$ is defined ensures that $d$ changes sign when you swap the roles of $f$ and $g$ (but the quantity defined by the supremum remains unchanged).

ADDED #1: Actually, ignore the last paragraph. One can just take the desired function to be $t$.

ADDED #2: Some details about how to construct the function $s$. It is well-known that the axiom of choice is logically equivalent to the well-ordering theorem. According to it, $\mathbb R$ can be endowed with such a linear order $\succsim$ that every non-empty subset of $\mathbb R$ has a well-defined least element according to this ordering. If $X\subseteq\mathbb R$ is not empty, let $\min_{\succsim} X$ be that least element. (The purpose of having this ordering $\succsim$ at hand is for one to be able to choose one and only one well-defined element from any non-empty subset of $\mathbb R$.)  One can then define $s$ for each $(f,g)\in C[\mathbb R]\times C[\mathbb R]$ as follows:
\begin{align*}
s(f,g)\equiv\begin{cases}0&\text{if $f=g$,}\\\min_{\succsim}\{x\in\mathbb R\,|\,f(x)\neq g(x)\}&\text{if $f\neq g$.}
\end{cases}
\end{align*}
With this definition, it is not difficult to see that $s(f,g)=s(g,f)$. This implies that the function $t$ as defined above satisfies $t(f,g)=-t(g,f)$ and $t(f,g)=0$ if and only if $f=g$.
A: Pick an enumeration of the rationals $r_1,r_2,r_3,...$.
Let $d(f,g) = 0$ if $f = g$.
Else let $r_n$ be the first rational in the enumeration for which $f(r_n) \neq g(r_n)$.  Define $d(f,g) = f(r_n) - g(r_n)$.
This is well defined since if $f \neq g$, there must be some rational number for which $f(r) \neq g(r)$, since $f$ and $g$ are continuous, and the rationals are dense.
