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Homework Help: Differentiation Question Help

  1. Feb 28, 2004 #1
    Hey
    I'd really appreciate some help with the following differntiation problems:

    Here are my workings for the first part:
    b^2 - 4ac = 0 (from formula)
    4b^2 - 4(3a)(c) = 0
    4b^2 = 12ac
    b^2 = 3ac

    For the second part, finding the coordinates, im not sure what i should do, any advice?

    (ii) Using logarithms, or otherwise, differentiate:

    y=(e^x^2\sqrt{SinX})/(2x + 1)^3

    Ok, so I've been trying to do this, but I'm not sure where the logarithms part comes into it, or where to start, maybe a quotient rule?
    Any help is appreciated.
     
  2. jcsd
  3. Feb 28, 2004 #2

    FZ+

    User Avatar

    Whaa? Where did you get this formula from?

    I would do it by differentiating:

    f'(x) = 3ax^2 + 2bx + c = 0

    This gives one turning point, given repeated roots in x.

    ie. (2b)^2 - 4*3*a*x = 0

    This can be solved easily.

    Coordinates? Quadratic formulae, again.
    x = -2b/6a (since the square root bit equals 0 for repeated roots)

    Substitute, and viola!

    That does look nasty, doesn't it?
    Can you clarify with brackets etc what (e^x^2\sqrt{SinX}) is? Do you mean:

    e^((x^2)/(sqrt(sinx)))?

    As for logs, perhaps you are meant to log both sides, and differentiate implicitly?
     
  4. Feb 28, 2004 #3

    HallsofIvy

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    Science Advisor

    "Here are my workings for the first part:
    b^2 - 4ac = 0 (from formula)"

    My first reaction was the same as FZ+: "where did you get that formula"? Then I realized that you are using the same letters to mean different things.

    Given a cubic ax3+ bx2+ cx+ d, a turning point can only occur where the derivative: 3ax2+ 2bx+ c= 0

    That can be solved using the quadratic formula and, in order that there be exactly 1 turning point, there must be exactly one solution so "b2- 4ac= 0" where a,b,c are now the coefficients of ax2+ bx+ c. Putting in the values 3a and 2b for "a","b" gives you the rest:

    "4b^2 - 4(3a)(c) = 0
    4b^2 = 12ac
    b^2 = 3ac"

    BAD IDEA! Never use the same letters to represent different things! Or at least tell us you are doing that!

    As for finding the turning point: 3ax2+ 2bx+ c= 0 and
    b2= 3ac. Okay, then, again, from the quadratic formula,
    since the discriminant is 0 x= -2b/(6a). You can plug that into the original cubic to find y (exactly what FZ+ suggested)

    Like FZ+, I have no idea what "y=(e^x^2\sqrt{SinX})/(2x + 1)^3" is, in part because of that "\". My guess, like his, is that you meant
    y=(e^{x^(x^2/sqrt(sin(x)))})/(2x+1)^3.

    log(y)= x^(x^(2/sqrt(sin(x)))))-3log(2x+1) which should be easier to differentiate.
     
  5. Feb 28, 2004 #4
    Ok, thanks for the help with the first part. I should probably have explained that the formula was taken from the quadratic formula x = [-b +/- squrt(b^2-4ac)]/2a, using the b^2-4ac = 0 to show that it has one real root.
    I think the reason the question was asked using a,b,c,d was to confuse people trying to do it.

    As for the log question, i didnt make a great job of typing it out, so ill try again:
    Y = [{e^(x^2)}.sqrt{SinX}]/[(2x + 1)^3

    Y = e to the power of x squared by the square root of sinX, divided by (2x + 1) to be cubed.
    Thanks again for the help.
     
  6. Feb 29, 2004 #5
    Any ideas on solving the second problem?
     
  7. Feb 29, 2004 #6
    is it

    [tex]y=\frac{e^{x^2} \sqrt{\sin x}}{(2x+1)^3}[/tex]
     
  8. Feb 29, 2004 #7
    Yeah, thats it, my latex skills arent really up to scratch yet. Thanks, any clues on solving?
     
  9. Feb 29, 2004 #8
    take log u will get

    [tex]logy= x^2+\frac{log sinx}{2} - 3log(2x+3)[/tex]

    and now differentiate u will get the desired result
     
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