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Derivative of e^x[(3x+2)/3x^(1/3)]

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data

    original equation: f(x) = x[itex]^{2/3}[/itex]e[itex]^{x}[/itex]

    first derivative: f'(x) = e^x[(3x+2)/3x^(-1/3)]

    second derivative f"(x) = e^x[(9x^2 + 12x - 2)/9x^(4/3)]


    2. Relevant equations


    So far, I'm right here in the second derivative:

    f"(x) e^x[(9x^2/3+12x^(-1/3)-2x^(-4/3)]

    But I don't know how to get it into this final form:

    f"(x) = e^x[(9x^2 + 12x - 2)/9x^(4/3)]
    9x4=3

    I am thinking I need to factor out an x to some power, but I don't know which power.

    I know I'm supposed to factor out the smallest exponent, but does that mean the absolute smallest exponent, or the smallest in terms of the number furthest to the left on the number line?

    In otherwords, do I factor out x^(-1/3) or x^(-4/3) ? And by factoring out one of these terms, am I heading in the right direction to that final second derivative form?



    EDIT:

    I should specify that my question is in regard to getting from the first derivative to the second derivative. That's where I'm getting stuck. I can get the derivative, but not into the simplified I posted.
     
    Last edited: Nov 19, 2011
  2. jcsd
  3. Nov 19, 2011 #2

    eumyang

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    If you mean this:
    [tex]f'(x) = e^x \left( \frac{3x+2}{3x^{-1/3}} \right)[/tex]
    ... then you have a small error. There shouldn't be a negative in the exponent in the denominator.

    EDIT: I see that there is no negative in the thread title.
    I wouldn't change the 1st derivative into fraction form. I would go from this step (which is the 1st derivative):
    [itex]x^{2/3} e^x + \frac{2}{3}x^{-1/3} e^x[/itex]
    and use the product rule twice. Then rearrange into fraction form.
     
    Last edited: Nov 19, 2011
  4. Nov 19, 2011 #3

    eumyang

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    Disregard my previous post. I shouldn't be posting when I'm a little tipsy. :rofl:

    Can you show all steps to get to the above? What I have is a little different. (The coefficients I have are fractions.)
     
  5. Nov 19, 2011 #4
    Aw dang man, I completely wrote that exponent incorrectly in the message. Should be as written in the title. My bad on that mistake.

    These are the steps I used to to get where I got:

    [tex]f'(x) = e^x \left( \frac{3x+2}{3x^{1/3}} \right)[/tex]

    [itex]e^{x}[/itex][itex]\left(3x+2\right)[/itex][itex]\left(3x^{-1/3}\right)[/itex]

    Then I used Foil Method

    [itex]e^{x}[/itex][itex]\left(9x^{2/3}+6x^{-1/3}\right)[/itex]

    Then product rule

    [itex]e^{x}[/itex][itex]\left(9x^{2/3}+6x^{-1/3}\right)[/itex]+[itex]\left(e^{x}\right)[/itex]d/dx[itex]\left(9x^{2/3}+6x^{-1/3}\right)[/itex]

    Then got

    [itex]e^{x}[/itex][itex]\left(9x^{2/3}+12x^{-1/3}-2x^{-4/3}\right)[/itex]


    I think its just a matter of a few more steps to get:

    [tex]f'(x) = e^x \left( \frac{9x^{2}+12x-2}{9x^{4/3}} \right)[/tex]


    I'm trying to figure out if its a matter of factoring out another term, but no terms seem like they would give me a 9 for the denominator, so I must have done something wrong maybe
     
    Last edited: Nov 19, 2011
  6. Nov 19, 2011 #5
    hold on..I think i know what i did wrong here...

    From the very first step, I brought the 3 from the denominator to the numerator when maybe I should have only brought the x^(1/3) to the numerator!?!


    Could I have instead written:


    [itex]e^{x}[/itex][itex]\left(3x+2\right)[/itex][itex]\left(3^{-1}x^{-1/3}\right)[/itex]



    EDIT:

    Okay so now that I took the derivative after the correction, I get:

    [itex]e^{x}[/itex][itex]\left(x^{2/3}+\frac{4}{3}x^{-1/3}-\frac{2}{3}x^{-4/3}\right)[/itex]


    EDIT 2:

    Okay and now I multiplied everything in the parenthesis by 9 to cancel out the denominators and I get:

    [itex]e^{x}[/itex][itex]\left(9x^{2/3}+12x^{-1/3}-2x^{-4/3}\right)[/itex]

    But since I mutiplied everything inside the parenthesis by nine, I also have to multiply everything outside the parenthesis by nine also, right?

    So it should look like this?

    [itex]9e^{x}[/itex][itex]\left(9x^{2/3}+12x^{-1/3}-2x^{-4/3}\right)[/itex]

    Then to factor out an x to some power... "conveniently" back to my original point of confusion lol...isn't math beautiful lol...

    Now, I am trying to figure out what exact concept/definition of smallest exponent I need to think about when factoring out negative exponents. Should it be the smallest, as in absolute value smallest, which would be x^(-1/3) or smallest, as in furthest in the negative direction on a number line, which would be x^(-4/3).

    Well, since I see the end answer has an x^(4/3) at the bottom, I am gonna say B) the furthest to the left on the number line is the exponent to be factored out.

    So factoring that out, I end up with

    the final form!

    If you hadn't told me you had gotten a different form and to rewrite my steps, I never would have noticed my mistake.
     
    Last edited: Nov 19, 2011
  7. Nov 20, 2011 #6

    eumyang

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    No! It's 1/9!
    [itex]\frac{1}{9}e^{x}[/itex][itex]\left(9x^{2/3}+12x^{-1/3}-2x^{-4/3}\right)[/itex]

    Not sure how! You would have a 9 in the numerator instead of the denominator.
     
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