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Need help deriving multiple quantities

  1. Nov 19, 2012 #1
    Need help differentiating multiple quantities

    1. The problem statement, all variables and given/known data

    [itex]f′(x) = (x − 1)(x − 2)^2(x − 3)^3(x − 4)^4(x − 5)^5[/itex]

    I need help in trying to differentiate this equation. I know could use a combination of the chain and product rule to figure it out, but my teacher said that doing so would take a while and the resulting equation before simplifying would be very long and convoluted and didn't want us to do it that way.

    My teacher told me she already taught us a method of doing derivatives of long equations a while ago, but I can't remember that method. Any help in either solving it or just giving me the setup of this "shorter" method is appreciated. (prefer the latter since then it gives me a chance to practice it)
     
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    Mark44

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    What are you asking? f'(x) already is a derivative. Do you need to find f''(x)? In English we don't say we're "deriving an equation" if the goal is to differentiate a function.
     
  4. Nov 19, 2012 #3
    Yeah sorry I need second derivative because the main question asked for the critical points for f(x) (not known) which I already know are 1, 2, 3, 4, 5, but it also asked what kind of extrema the points are and to prove which extrema they are and I need the second derivative to prove it.
     
  5. Nov 19, 2012 #4

    haruspex

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    I think what you're after is a different way of expressing the product rule:

    If y = f(x)g(x)h(x)... then y' = f'(x)g(x)h(x)...+f(x)g'(x)h(x)...+f(x)g(x)h'(x)...
    = (f'(x)/f(x))f(x)g(x)h(x)...+f(x)(g'(x)/g(x))g(x)h(x)...+f(x)g(x)(h'(x)/h(x))h(x)...
    = y(x){f'(x)/f(x)+g'(x)/g(x)+h'(x)/h(x)...}
    This is even easier using logs:
    y = ∏fi
    ln(y) = Ʃln(fi)
    y'/y = Ʃf'i/fi
    And if fi(x) = (x+ai)bi then
    ln(fi(x)) = biln(x+ai)
    f'i/fi = bi/(x+ai)
     
  6. Nov 19, 2012 #5
    Yeah my teacher wanted me to avoid using the product rule and chain rule combo to differentiate it, which was your first example.

    Though your second example with the logarithmic differentiation is the one I needed for my problem thanks for reminding me and the help.
     
  7. Nov 20, 2012 #6

    HallsofIvy

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    The simplest way to do this is logarithmic differentiation.
    log(f'(x))= log(x-1)+ 2log(x- 2)+ 3log(x-3)+ 4log(x- 4)+ 5log(x- 5)
     
  8. Nov 20, 2012 #7
    yeah I already got to that part.

    so that simplified it is:

    [itex]f"(x) = (\frac {1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}+\frac{4}{x-4}+\frac{5}{x-5})((x-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5)[/itex]

    thing is now how would I find out what kind of extrema the roots of f'(x)/critical points of f(x) with such a long function?

    I could just plug-in the critical values and look for sign changes, but I don't know what to do it with the fractions going to end up being [itex] \frac{v}{0} [/itex] with v being any of the numerator of my fractions.


    Edit: Never mind forgot that for sign change test I don't use the critical values as plug-in values

    Edit 2: Ok I did my sign change test and my results were +1+2-3-4+5+ if put on a number line, but I'm not sure if that's right
     
    Last edited: Nov 20, 2012
  9. Nov 20, 2012 #8

    Ray Vickson

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    I'm not quite sure what you are saying here, but for x < 1 all factors x-j are < 0, and three of them are taken to odd powers while two are taken to even powers, so f'(x) < 0 for x < 1. As x passes through 1, f'(x) changes sign, so f'(x) > 0 for 1 < x < 2. Look carefully what happens to the sign of f'(x) as x passes through 2, 3, 4 and 5.

    RGV
     
    Last edited: Nov 20, 2012
  10. Nov 20, 2012 #9
    Yeah somehow my brain started to die out on me when doing my sign change test.

    So would the correct order of sign change be:

    - 1 + 2 + 3 - 4 - 5 +

    (+'s and -'s are the sign in between the intervals) so that the local extrema are: 1, 3, 5 with points 2 and 4 where the f(x) just goes flat.

    If that is the case I could have just done it with the original f'(x) instead of having to use the f"(x) to find them.
     
  11. Nov 20, 2012 #10

    Ray Vickson

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    Another way is to note that near x = 1, f'(x) is essentially the same as c1*(x-1), where c1 = (1-2)^2*(1-3)^3*(1-4)^4*(1-5)^5 is a constant; this is essentially because the other factors of f' do not vary much as x varies near 1. You can figure out that c1 > 0. For x near 2, f'(x) looks like c2*(x-2)^2, where c2 =(2-1)*(2-3)^3*(2-4)^4*(2-5)^5 > 0 is a constant. Similarly near x = 3, 4 or 5. Note that these ways of looking at f' allow you to figure out f''(x) easily at x = 1,2,3,4,5.

    RGV
     
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