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Natural logs and derivatives

  1. Oct 14, 2009 #1
    Q1)

    If (x^5)-x = 8
    then
    x = e^(solve for this)

    not sure, x*x*x*x*x-x = 8
    help please

    -

    q2)
    ln(3x)/x^2
    dy/dx = [ ]-[ ]ln(3x) / x^3
    or solve for dy/dx

    I get

    (x^2)*(dy/dx(ln(3x))) - ln(3x)*(dy/dx(x^2)) / x^2*x^2
    =(x^2)*(1/x) - ln(3x)*2x / x^4
    = [x^2]-[2x]*(ln(3x)) / x^3
    (is this right?)


    -

    q3)

    if dy/dx = 3e^x
    and y = 5 when x=0

    then y = [ ]

    I get
    I think i have to integrate 3e^x
    and it's just 3e^x
    y = 3e^0 + c
    e^0 = 1
    so 3e^x + 2 = 5

    y = 3e^x + 2
    and I got it wrong???


    -

    q4)

    evaluate
    the integral of (1/3 * x-1)dx between 6 and 3

    integrate it = 1/3*x^2/2 - x
    = 1/6x^2 - x
    (1/6 * 6^2 - 6 ) - (1/6 * 3^2 - 3) = 1/6*36-6 - 1/6*9-3
    = 0 - 1.5-3
    = -4.5
    and I got it wrong
    although I think during the test I may have made the mistake of doing 1.5 - 3 instead of -1.5 - 3...


    thanks please mark
     
  2. jcsd
  3. Oct 14, 2009 #2

    Mentallic

    User Avatar
    Homework Helper

    Q1) I don't see any simple solution to the problem. You can try approximate the solution, but I wouldn't know how to get any nice analytical solution if there is one.

    Q2) You were right with the derivatives but didn't simplify correctly.

    [tex]\frac{x^2*\frac{1}{x} - ln(3x)*2x}{x^4}[/tex]

    [tex]=\frac{x-2xln(3x)}{x^4}[/tex]

    [tex]=\frac{1-2ln(3x)}{x^3}[/tex]

    Q3) Uh it looks right...

    Q4) Again, it was the algebra that got you.

    [tex]0 - (9/6 - 3)= -9/6 + 3 = 1.5[/tex]
     
  4. Oct 14, 2009 #3
    thanks
     
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