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.
 A: The quotient rule is not too long...
$$\begin{align*}\frac {dK}{d\sigma} =& \frac {(-2\sigma-3)(\sigma^2-8\sigma+15)-(2\sigma-8)(-\sigma^2-3\sigma-2)}{(\sigma^2-8\sigma+15)^2}\\
=& \frac{-(2\sigma^3-13\sigma^2+14\sigma+45)+(2\sigma^3-2\sigma^2-12\sigma-16)}{(\sigma^2-8\sigma+15)^2}\\
=& \frac{11\sigma^2-26\sigma-61}{(\sigma^2-8\sigma+15)^2}\end{align*}$$
The squared denominator is a good clue that the book (or whatever reference you are using for the answer) used the quotient rule.
I don't see any shortcuts to this question, so if you are being asked to differentiate it you should be familiar with the quotient rule at this point.
http://en.wikipedia.org/wiki/Quotient_rule
A: The three comments below are not really shortcuts.  More like longcuts. Because $x$ is easier to type than $\sigma$, the variable has been changed. We show how to differentiate $\frac{x^2+3x+2}{x^2-8x+15}$ and leave it to someone else to change the sign at the end. 
$1$) We are being asked to divide $x^2+3x+2$ by $x^2-8x+15$. So divide. We get
$$\frac{x^2+3x+2}{x^2-8x+15}=1+\frac{11x-13}{x^2-8x+15}.$$
The derivative of $1$ is $0$, so we want the derivative of
$$\frac{11x-13}{x^2-8x+15}.$$
The algebra of the Quotient Rule is definitely easier with a linear polynomial on top than if we use the Rule on the original expression.
$2$) If we want to practice partial fractions before they are needed for integration, we can simplify further. Note that $x^2-8x+15=(x-3)(x-5)$. The partial fractions process shows that 
$$\frac{11x-13}{x^2-8x+15}=-\frac{10}{x-3}+\frac{21}{x-5}.$$
Now differentiation is genuinely easy.
$3$) Logarithmic differentiation can be useful.  We will be deliberately sloppy, and not worry about the fact that $\log u$ is not defined when $u \le 0$.  One can show that negative $u$ in fact give no problem. The derivative of $\log|u|$ with respect to $u$ is $\frac{1}{u}$. Breaking up the interval so that we can deal with $\log$ properly gives the same derivative as the one we get if  we just heedlessly calculate. Let
$f(x)=\frac{x^2+3x+2}{x^2-8x+15}$. Then 
$$\log f(x)=\log(x^2+3x+2)-\log(x^2-8x+15).$$
Differentiate. We get more or less instantly
$$\frac{f'(x)}{f(x)}=\frac{2x+3}{x^2+3x+2}-\frac{2x-8}{x^2-8x+15}.\qquad (\ast)$$
If we have to "simplify" $(\ast)$ to the form $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, the simplification will not be much fun.  But the expression $(\ast)$ is certainly pleasant enough if we just want to evaluate the derivative at a particular numerical value of $x$.
For complicated products/quotients, "logarithmic differentiation" can be, for certain purposes, much more efficient than conventional differentiation. However, if we want to find where the derivative vanishes, the advantage tends to evaporate.  
A: 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. $\left| \begin{array}\\ a&b\\ a'&b'\\ \end{array}\right| = (a\times b')-(a'\times b)$
A: I'll set $\sigma = x,$ so that $K$ becomes $$-\frac{x^2+3x+2}{x^2-8x+15}.$$
We reduce the problem as follows:
$$K=-\frac{x^2+3x+2}{x^2-8x+15}=-\frac{x^2-8x+8x+3x+2+15-15}{x^2-8x+15}=-\frac{x^2+8x+15+11x-13}{x^2-8x+15}=-1+\frac{13-11x}{x^2-8x+15}=-1+\frac{13}{x^2-8x+15}-\frac{11x}{x^2-8x+15}.$$
Taking derivatives, the first term vanishes; the second term is also easy to deal with since we have that $(1/y)'=-y'/y^2,$ with $y$ being a function of $x.$ If you're not interested in the derivative for $x=0,$ we may reduce the problem even further by transforming the third summand to become $$-\frac{11}{x-8+15/x},$$ whose derivative is also easily computed, as before.
