Find $\frac{{dy(0)}}{{dx}}$ for $x \ge 0$, where, $y = Q(\sqrt x )$?

Question

$Q(\sqrt x )$ is a decreasing function for $x\ge0$ where,

$Q(x) = \frac{1}{{\sqrt {2\pi } }}\int\limits_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}du}$

The derivative of the function is

$\frac{{dy}}{{dx}} = - \frac{{{e^{ - \frac{x}{4}}}}}{{4\sqrt {\pi x} }}$.

Is is possible to find $\frac{{dy(0)}}{{dx}}$? I think this should exist ( it exists for $x \to {0_ + }$ but I dont know how to show that). Is there any way to do that?

Answer 1

Write, for $h>0$, $$ \frac{y(h)-y(0)}{h}=-\frac{1}{h\sqrt{2\pi}}\int_0^\sqrt{h} e^{-\frac{u^2}{2}}du=-\frac{e^{-\frac{c^2}{2}}}{\sqrt{2h\pi}} $$ where $0<c<h$ by the Mean Value Theorem for integrals.

Letting $h\to 0^+$, we get that the limit does not exist.

Answer 2

Let me denote $$f(x) = \int_{x}^{+ \infty}e^{-t^2}dt = \left(\int_{0}^{+ \infty}-\int_{0}^{x} \right)e^{-t^2}dt$$ For derivative we have $f^{'}(x) = -e^{-x^2} \rightarrow -1$, when $x \rightarrow 0$.

Then, let's consider definition of derivative in $x=0$: $$\dfrac{1}{x} \cdot (f(x)-f(0))=-\dfrac{1}{x}\int_{0}^{x}e^{-t^2}dt = - e^{-c(x)^2} $$ where $0 \leqslant c(x)\leqslant x $. This gives $f^{'}(0)$ = -1 and coincides with limit of $f^{'}(x)$.

$$Q(x) =\frac{1}{\sqrt {2 \pi }} \int_{x}^{+ \infty}e^{-\frac{u^2}{2}}du =\frac{1}{\sqrt {2 \pi }} \int_{\sqrt{2}x}^{+ \infty}e^{-u^2}du = \frac{1}{\sqrt {2 \pi }} f(\sqrt{2}x)$$