Derivative of $p\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}du} $? I found a result,
I have,
$p\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_x^\infty  {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}du} $
$y =  - p\left( x \right)\log p\left( x \right) - (1 - p\left( x \right))\log (1 - p\left( x \right))$
I found a solution for the derivative of $y$ at $x=0$ is,
$\frac{{dy\left( 0 \right)}}{{dx}} = \frac{2}{\pi }$
But I could not able to show that. It comes zero for me every time.

 A: Since you edited the question, I edited my answer.

FIRST PART
We will show that

$$Q'(x) = -\frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x^2}{2}} $$

The main tool here is the Fundamental theorem of Calculus.
First method
$$
\begin{split}
\frac{\mathrm d}{\mathrm d x} \left[ \frac{1}{\sqrt{ 2 \pi}} \int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du} \right] &= \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du}\right]\\
&\overset{t=-u}{=}\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_{-\infty}^{-x} {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{t^2}}}{2}}}\,\mathrm dt}\right]\\
&=\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_{-\infty}^{-x} {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{t^2}}}{2}}}\,\mathrm dt}\right]\\
&=-\frac{1}{\sqrt{ 2 \pi}} {{\mathop{\rm e}\nolimits} ^{ - \frac{{{x^2}}}{2}}}\\
\end{split}
$$
Second method (essentially the same)
Let $G(x)$ a primitive of $\mathrm{e}^{-\frac{x^2}{2}}$, i. e. $G'(x) = \mathrm{e}^{-\frac{x^2}{2}}$; therefore
$$
\begin{split}
\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[ \int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du} \right] &= \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x}\left[G(+\infty)-G(x)\right] \\
&=  \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} G(-x)\\
&= -\frac{1}{\sqrt{ 2 \pi}}  G'(-x)  \\
&= -\frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x^2}{2}}
\end{split}
$$
Third method
Remembering that $\int_{-\infty}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \sqrt{2\pi}$ we can write
$$
\int_{x}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \int_{-\infty}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du -\int_{-\infty}^x \operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \sqrt{2\pi} - \int_{-\infty}^x \operatorname e^{-\frac{u^2}{2}}\, \mathrm du
$$
And then apply FTC.

SECOND PART
According to your notation we have

$$C(x) := 2 \log 2 -2h\left(Q\left(\sqrt x\right)\right) $$

If we evaluate $\dot C$, thanks to the chain rule we have
$$
\begin{split}
\dot C (x) &= -2h'\left(Q\left(\sqrt x\right)\right) \cdot Q'\left(\sqrt x\right) \cdot \frac{1}{2 \sqrt x} \\
&= -\log \left[ \frac{Q\left(\sqrt x\right)}{1-Q\left(\sqrt x\right)} \right]\cdot \frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x}{2}} \cdot \frac{1}{\sqrt x}\\
&= \frac{\operatorname{e}^{-\frac{x}{2}}}{\sqrt{ 2 \pi}} \cdot \frac{\log \left( 1 - Q\left(\sqrt x\right)\right) -\log Q\left(\sqrt x\right) }{\sqrt x}
\end{split}
$$
Since $Q(0)=\frac{1}{2}$, we obtain an indeterminate form $\frac{0}{0}$. Can you handle from here?

THIRD PART
We have to evaluate

$$\lim_{x \to 0}\frac{\log \left( 1 - Q\left(\sqrt x\right)\right) -\log Q\left(\sqrt x\right) }{\sqrt x} \overset{y=\sqrt x}{=} \lim_{y \to 0}\frac{\log \left( 1 - Q(y) \right) -\log Q(y) }{y}$$

to be continued...
