pi-th derivative of x^pi
pi-th derivative of x^pi

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…

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