0

I need to find the third order derivative of $\frac{1}{s+3}$. I know that I need to calculate the derivative of the derivative till I get to the desired order.

So, here are my answers:
First-order derivative: $\frac{-1}{(s+3)^2}$
Second-order derivative: $\frac{2(s+3)}{(s+3)^3}$
and Third-order derivative: $\frac{4}{(s+3)}$

Can someone please confirm my calculations ? :)

How I calculated the second-order derivative:

$\frac{(s+3).0 + (1)2(s+3)(1)}{(s+3)^2}$

An SO User
  • 937

1 Answers1

2

Your first derivative is fine, but after that you went astray. The easiest way to do the computations is to write $f(s)=(s+3)^{-1}$ and apply the power rule three times: $f\,'(s)=-(s+3)^{-2}$, $f''(s)=2(s+3)^{-3}$, and $f'''(s)=-6(s+3)^{-4}$ or, if you prefer,

$$f'''(s)=\frac{-6}{(s+3)^4}\;.$$

Added: I see from your edit that you actually used the quotient rule and, as I kind of suspected, forgot to square the denominator. As a general rule, when the numerator is a constant it’s easier to rewrite the fraction as a constant multiple of a negative power and use the power rule.

Brian M. Scott
  • 616,228
  • It should perhaps be mentioned that there is an "invisible Chain Rule factor", in that the first derivative in full would be $ \ (-1) \cdot (s+3)^{-2} \cdot \frac{d}{ds}(s+3) = \ (-1) \cdot (s+3)^{-2} \cdot 1 \ , $ and similarly for the higher derivatives. Were the function $ \ \frac{1}{2s+3} \ $ , you'd have $ \ f'(s) = -2(2s+3)^{-2} \ , \ f''(s) = 4(2s+3)^{-3} \ , \ $ and so on... When you have a rational function which is simply a constant divided by some polynomial, using the "Power Rule" and Chain Rule is definitely to be preferred over applying the Quotient Rule. – colormegone Aug 29 '13 at 02:44
  • @RecklessReckoner Please see my edit. :) – An SO User Aug 29 '13 at 02:58
  • You want a minus sign between the terms in the numerator, no? Also, the $ \ s+3 \ $ shouldn't be in the second term there... – colormegone Aug 29 '13 at 03:00
  • @RecklessReckoner It becomes $-(-1)$ right ? – An SO User Aug 29 '13 at 03:02
  • You may want to do yourself the favor, in making these calculations using the Quotient Rule, to do less of it in your head, and to write out more of the steps -- I suspect that is where your errors arise from... (see the other comments as to why I am saying this) – colormegone Aug 29 '13 at 03:06