The image you uploaded contains the probability density function (PDF) of a normal distribution:

$p(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2}$

To find the derivative of this function $p(x)$:

We'll differentiate the function with respect to $x$. Using the chain rule and recognizing that this is a Gaussian function, we differentiate it step by step:

$\frac{d}{dx} \left[ \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2} \right]$

First, treat the constant factor $\frac{1}{\sigma \sqrt{2\pi}}$ as a constant, and then differentiate the exponential term. The derivative of the exponential function $e^{-u(x)}$ is $e^{-u(x)} \cdot \frac{du}{dx}$, where $u(x) = \frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2$.

The result will be:

$\frac{d}{dx} p(x) = -\frac{1}{\sigma \sqrt{2\pi}} \left( \frac{x - \mu}{\sigma^2} \right) e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2}$

Graph of $p(x)$:

The graph of this function is a normal distribution (Gaussian curve), which is symmetric about the mean $\mu$ with standard deviation $\sigma$.

Do you need further explanation of the steps, or would you like me to generate the graph of the PDF or its derivative?