<s>[INST] Derivative of this function (quadratic over quadratic) $$K = \frac{-(\sigma^2 + 3\sigma + 2)}{\sigma^2 - 8\sigma +15}$$
How is this differentiated with respect to $\sigma$?
The answer is
$$\dfrac{dK}{d\sigma} = \frac{11\sigma^2 - 26\sigma -61}{(\sigma^2 - 8\sigma + 15)^2}$$
And the quotient rule is a very long process, I get the feeling there was a short method.
 [/INST] A fast way to find the Derivative of Quadratic over Quadratic:
$$  \bbox[20px,border:1px solid red]
{
y = \frac{ax^2+bx+c}{a'x^2+b'x+c'}\quad\Rightarrow\quad y'= \frac{\left| \begin{array}\\ a&b\\ a'&b'\\ \end{array}\right| \times x^2 + 2 \times \left|\begin{array} \\ a & c \\ a' & c' \end{array}\right|\times x + \left|\begin{array}\\ b & c \\ b' & c' \end{array}\right|}{V^2}
}
$$
For example: 
$$
y=\frac{x^2-2x+1}{x^2+2x+3}\,\Rightarrow\,y'=\frac{(2+2)x^2+2\times(3-1)x+(-6-2)}{V^2}=\frac{4x^2+4x-8}{V^2}
$$
Now your question: 
$$
y=\frac{-x^2-3x-2}{x^2-8x+15}\,\Rightarrow\,y'=\frac{(8+3)x^2+2\times(-15+2)x+(-45-16)}{(x^2-8x+15)^2}=\frac{11x^2-26x-61}{(x^2-8x+15)^2}
$$
It is faster than the regular method. Even you can do it in your head. It is just calculating a determinant of a matrix. $\left| \begin{array}\\ a&b\\ a'&b'\\ \end{array}\right| = (a\times b')-(a'\times b)$
</s>