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Finding the derivatives of functions

  1. Sep 29, 2008 #1
    Hi, i'm working with finding the derivatives of functions, which i'm not very comfortable with; if someone could please check my answers to the following questions i would be VERY grateful! Thank you! :)

    find the derivative of the following function:

    Question 1:

    y = [tex]\frac{ 2^{x} }{ e^{x} }[/tex]

    My Answer

    y1 = [tex]\frac{ e^{x} . ln2 . 2^{x} + 2^{x} . e^{x} }{ e^{x}^{2} }[/tex]
    = [tex]\frac{ 2^{x} (ln2 + 1) }{ e^{x} }[/tex]

    Question 2:

    f(x) = 2x ln(x[tex]^{2}[/tex] + 5)

    My answer

    f [tex]^{1}[/tex] (x) = 2ln(x[tex]^{2}[/tex] + 5) + (2x) . [tex]\frac{1}{x^{2} + 5}[/tex] . (2x)

    = 2 ln(x[tex]^{2}[/tex] + 5) + [tex]\frac{4x^{2}}{x^{2} + 5}[/tex]

    Question 3:

    g(x) = [tex]\frac{ln x}{e^{x}^{2} + 2}[/tex]

    My answer:

    g[tex]^{1}[/tex](x) = [tex]\frac{(e^{x}^{2} + 2) . (1/x) - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

    = [tex]\frac{\frac{e^{x}^{2} + 2}{x} - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

    for the last two questions i'm not sure if i simplified enough...if anyone could guide me in the right direction where needed i'd really appreciate it! thanks in advance!
     
  2. jcsd
  3. Sep 29, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Almost right you need "-" , not "+" in the numerator- quotient rule: (u/v)'= (u'v- uv')/v^2.

    Yes, that looks good.

    The third problem also looks good to me.
     
  4. Sep 29, 2008 #3
    Great! Thank you HallsofIvy! :D
     
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