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Homework Help: X^x and xsin(x)

  1. Nov 20, 2004 #1
    x^x and sin(x)/x

    The derivative of x^x is:

    Why is this not a possible way of solving it:
    I see it's not the same answer but I wonder why

    The reason is if the above was correct it would be easy to integrate x^x (which is not possible):
    why is this not correct?

    Ok I just realized that the inner derivative (do you also call it this?) of ln x is 1 thus
    (e^(xlnx))'=e^(xlnx)(lnx + (inner derivative of lnx)) which is one thus x^x(lnx+1)
    How does this affect the integral?

    Secondly how would you go about drawing sin(x)/x by hand without a calculator
    What I realize is the following:
    Though not defined x=0 would seem to be at 1 mainly since for small sin(x)/x=x thus 1/x*sin(x)=1/x*x)=1
    Next the function would reach the first zero (at the positive side) at pi/4 so how would the line look between 0 and pi/4? I have seen it but I don't know how to realize it myself.
    Next I understand for x=pi/4->pi/2 since:
    it would look something like
    Code (Text):

    \    /
    as opposed to
    \   /
    slightly bulged to the left due to decreasing 1/x but where between pi/4 and pi/2 would the maximum be?
    Following this the amplitude of the waves would just decrease which can be done by appreciation...

    Any ideas?
    Last edited: Nov 20, 2004
  2. jcsd
  3. Nov 20, 2004 #2


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    The last step quoted is wrong. For the chain rule, you need the derivative of xlnx. Think it through again.
  4. Nov 20, 2004 #3
    i dont think you did the derivative of the second one correct
    y= e^(xlnx) the derivative of e^x is e^x dx
    y' = e^(xlnx) (1)lnx + (x)(1/x) from the derivative of the exponent you go into the product rule
    simplified this is
    y' = e^(xlnx) (lnx +1)
    y' = x^x (lnx + 1)

    sometimes it's easy to forget simple thing like product and quotient rules when doing harder problems like this. you just have to think things though and not try to jump ahead of yourself.
  5. Nov 20, 2004 #4


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    "Secondly how would you go about drawing x*sin(x) by hand without a calculator"
    Draw the two lines [tex]y=\pm{x}[/tex]
    xsin(x) always lies in-between these two lines; you should be able to figure out how..
  6. Nov 20, 2004 #5
    Let u = xlinx

    du/dx= ?
  7. Nov 20, 2004 #6
    my god, i meant sin(x)/x sorry =)
  8. Nov 20, 2004 #7
    Originally posted by Ponjavic:
  9. Nov 20, 2004 #8
    It's pretty much the same concept for [tex]\sin{x}/x[/tex], except instead you draw the lines [tex]y=\pm\frac{1}{x}[/tex].

    Hope that helps! :)
  10. Nov 20, 2004 #9


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    I would say curves, rather than lines, but whatever..
    In any case, you should also utilize the fact that sin(x)/x is an even function, so you only have to draw for positive x's.
    Also, scale your horizontal axis so that [tex]x=n\pi[/tex] for integer n's are easy to place (at these points, your function is zero).
  11. Nov 20, 2004 #10
    yeah that would probably give me the amplitude for the coming waves but for the first 0 to pi/2 it will go from 1->0 what will be the shape of this line and why? Also as I mentioned before the waves will be bent to one side. So +-1/x helps me get some sort of evaluation of the amplitude but not more than that...
  12. Nov 20, 2004 #11
    Doh, I was thinking of [tex]x[/tex] as constant! *feels stupid* You're right, it's a curve.
  13. Nov 20, 2004 #12


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    There's a complicated way to prove that
    [tex]\lim_{x\rightarrow0}\frac{sin x}{x}=1[/tex]
    but early on, even if it's proven, it's just something you remember (you can also find this with L'Hopital's Rule).

    You can also evaluate the opposite limit:
    [tex]\lim_{x\rightarrow\infty}\frac{sin x}{x}=0[/tex]
    (sin x) can have a max value of 1 and a min value of -1. (x) just keeps getting bigger as you approach infinity. 1 (or -1) over infinity is almost zero, so you know you're getting closer and closer to zero. The fact that the value of sin x can range from -1 to 1 means you have to cross zero.

    You just have to figure out where the zero crossings are (every [tex]\pi[/tex] ). Then you ought to get a feel for how the max values between each zero vary.
    Test [tex]\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}[/tex] etc.
    (You might notice that the max values are just the reciprical of x with just the sign changing).

    If you wanted to, you could also use your first derivative test and second derivative test to find your local maximums and minimums and your inflection points.
  14. Nov 20, 2004 #13
    I would like to disagree with
    According to my calculator the minimum between pi/2 and pi/4 is: x=4.49...
    while 3pi/2=4.71...
    This is due to the descending 1/x multiplayer which displaces the minimum point/maximum point and gives the curve another shape
    And the derivatives would give cos(x)-1/x^2=0 which is clearly not solvable with paper/pencil so that doesn't help me I think
  15. Nov 20, 2004 #14


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    Ignore first.

    Edit: I misread your post.

    [tex]dy(\frac{sin x}{x})=\frac{x cos x-sin x}{x^2}[/tex]

    Finding 0= x cos x - sin x is harder, but it is the only way (I forgot - you can ignore the denominator when you're looking for the zero points, but they push you max/min points since the number you're dividing by keeps getting bigger)
    I have to think about this one.
    Last edited: Nov 20, 2004
  16. Nov 20, 2004 #15
    again mistake by me i though sin(x)+1/x
  17. Nov 20, 2004 #16


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    You could also use the Newton-Raphson method, but that's more fun on a spread sheet than manually (in this case, it takes about 4 or 5 iterations to converge).

    You also know the max gets pushed low, but only by a small amount (in fact, close enough one might not even notice unless they looked close). And the max gets closer and closer to being right on one of the 90 degree points (diff goes from .22 rad to .13 rad to .09 rad to .07 rad).
  18. Nov 21, 2004 #17
    yea... just make sure that u chain rule ALL of the steps, especially if u are dealing with variable to variable or function to function. also, the "inner derivative" is just called the last term of the chain rule.
  19. Nov 22, 2004 #18


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    Answer for X^X

  20. Nov 22, 2004 #19
    again... I realized that, thanks for the help but please read the whole post =)
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