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$$K = \frac{-(\sigma^2 + 3\sigma + 2)}{\sigma^2 - 8\sigma +15}$$

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.

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3  
It's just the quotient rule. $(f/g)'={f'g-g'f\over g^2}$. Your comment about partial fractions leads me to say "the derivative, not the indefinite integral"... –  David Mitra Nov 8 '11 at 12:58
    
That would be the quotient rule... partial fractions is an integration technique, not differentiation. –  process91 Nov 8 '11 at 12:58
    
Er I meant quotient, I was tired –  Supernovah Nov 8 '11 at 12:59
    
Oh thanks, I didn't know that rule –  Supernovah Nov 8 '11 at 13:06

2 Answers 2

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.

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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

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He could use so-called ''logarithmic differentiation''; but I don't think this would qualify as a ''shortcut''. –  David Mitra Nov 8 '11 at 13:26
1  
Actually, OP could've used a combination of the product rule and chain rule if he didn't remember how to differentiate a quotient... –  J. M. Nov 8 '11 at 13:35
    
Definitely true, although neither seems to be much of a "shortcut" (as you mentioned). Perhaps it's worth mentioning that the quotient rule is simply an application of the product rule and the chain rule? Although that's probably in the Wikipedia entry... –  process91 Nov 8 '11 at 13:37

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