Partial derivative of a 2D integration?

26 Views Asked by At

I am currently trying to retrace steps in a process, and I am stuck a bit between two equations:

$$Eq.1: \frac{d}{d\theta_x}[\theta_I * \int_{A_n}\frac{1}{2\pi\sigma^2}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}}dxdy + \theta_b]$$

$$Eq.2: (\frac{\theta_I}{2\sigma^2})\int_{A_n}(\theta_x-x)\frac{1}{2\pi\sigma^2}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}}dxdy$$

From what I understand, $\theta_b$ obviously is lost upon derivation, you keep $\frac{1}{2\pi\sigma^2}$ as a coefficient, then take the derivative of $e^u$, which is just $u'e^u$ with respect to $\theta_x$. But for that I am coming up with:

$$\frac{d}{d\theta_x}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}} = -(\frac{1}{2\sigma^2})2(x-\theta_x)[\frac{d}{d\theta_x}(x-\theta_x)]e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}} = \frac{(x-\theta_x)(-1)}{\sigma^2}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}} = \frac{\theta_x-x}{\sigma^2}e^{\frac{-(x-\theta_x)^2-(y-\theta_y)^2}{2\sigma^2}}$$

...Which means I'm missing a 2 in the denominator?

I have repeatedly checked the source material, and either I'm missing something with the derivation steps somewhere, or somehow I overlooked something in the original reference despite repeatedly checking to ensure I didn't.

Any thoughts would be appreciated!