1. ## Stationary points

Hi i am having issues with a question from a textbook.
The normal distribution with a mean 0 and standard deviation Q(i dont know how to do the correct sign) is given by y= e^ -(x/Q)^2 . Find and classify all stationary points and points of inflection for this function Q>0.

The main issue i have occurred is finding the first and second derivatives of this function.

My first derivative is coming out to be dy/dx = - 2xy/Q using logarithmic and implicit differentiation.
However this does not appear to be right. I dont see how i could find when f'(x)=0 to find my stationary points.

Thank you.

2. Originally Posted by agent josh
Hi i am having issues with a question from a textbook.
The normal distribution with a mean 0 and standard deviation Q(i dont know how to do the correct sign) is given by y= e^ -(x/Q)^2 . Find and classify all stationary points and points of inflection for this function Q>0.

The main issue i have occurred is finding the first and second derivatives of this function.

My first derivative is coming out to be dy/dx = - 2xy/Q using logarithmic and implicit differentiation.
However this does not appear to be right. I dont see how i could find when f'(x)=0 to find my stationary points.

Thank you.
$
y = e^{-\left(\frac{x}{\sigma}\right)^2}
$

$\ln{y} = -\left(\frac{x}{\sigma}\right)^2$

$
\frac{1}{y} \cdot \frac{dy}{dx} = -\frac{2x}{\sigma^2}
$

$
\frac{dy}{dx} = -\frac{2x}{\sigma^2} \cdot e^{-\left(\frac{x}{\sigma}\right)^2}
$

$\frac{dy}{dx} = 0$ at $x = 0$

3. To find the second derivative, start from $\frac{1}{y}\frac{dy}{dx}= y'/y= \frac{-2x}{\sigma^2}$

Differentiating both sides, $\frac{yy''- y'^2}{y^2}= -\frac{2}{\sigma^2}$. Since the problem is to find the points of inflection, just put y"= 0 and solve $\frac{y'^2}{y^2}= \left(\frac{-2x}{\sigma^2}\right)^2= \frac{2}{\sigma^2}$