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Differentiation of (2x)^(2x)

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{d}{dx}\, (y = (2x)^{(2x)})[/tex]

    2. Relevant equations

    Chain Rule: d/dx (g * x) = g'x + x'g

    3. The attempt at a solution

    [tex]y = (2x)^{(2x)}[/tex]

    1. Take natural log of both sides.

    [tex]ln(y) = ln(2x)·(2x)[/tex]

    2. Differentiate both sides

    [tex]\frac{dy}{dx}\,·\frac{1}{y}\, = 2·ln(2x)+2[/tex]

    3. Substitute y in and distribute

    [tex]\frac{dy}{dx}\, = 2x^{2x}(2·ln(2x)+2)[/tex]

    This solution is incorrect. Any help would be appreciated.
     
  2. jcsd
  3. Jun 8, 2013 #2
    This may just be a technicality but there are no parentheses around the (2x)^(2x) in step 3. That would be technically wrong. How do you know the solution is incorrect? Is a computer grading it or are you looking at the correct answer in a text book.
     
  4. Jun 8, 2013 #3
    Hi, cpi255. It's a computer graded problem. I made sure to add parentheses in my answer when I submitted it to their system.

    I'm looking at wolframalpha's solution right now, and it seems to be different than the one I have arrived at.
     
  5. Jun 8, 2013 #4
    Enter this expression into Wolfram Alpha. You will see that your answer is correct.
    (d/dx (2x)^(2x)) = ((2x)^(2x)(2*ln(2x) + 2)

    Also are you sure in the problem the that the base was in the parentheses?
     
  6. Jun 8, 2013 #5

    Dick

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    Science Advisor
    Homework Helper

    It looks pretty ok to me. Except you meant ##\frac{dy}{dx}\, = (2x)^{2x}(2·ln(2x)+2)## in 3. right? You just missed a parentheses.
     
  7. Jun 8, 2013 #6
    Huh, I guess that's what it was! I was adding parenthesis to the top exponent and not the bottom. Damn!

    Thanks a lot, cp255 and Dick!
     
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