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Differentiate y= sin(3x)/x^2

  1. Jun 2, 2009 #1
    hiya, do you mind checking my awnsers please? i think i may have gotten them all wrong.

    a: y= sin(3x)/x^2

    b: (x^3) * (cos(4x))

    c: If dy/dx =5/x and y=6 when x=0, then y=

    d: if ln(x^3) - ln(x) = 4, then x=e^something

    e: if dy/dx = 3e^x, and y = 7 when x =0, then y =

    my attempts

    a: y= sin(3x)/x^2
    lo*dhi - hi*dlo/lo^2
    ((x^2)*(3cos(3x))) - ( (2x)*(sin(3x)) )/((x^2)^2)
     
    Last edited: Jun 2, 2009
  2. jcsd
  3. Jun 2, 2009 #2
    Re: Differentiate

    b: (x^3) * (cos(4x))

    ( (3x^2) * cos(4x) ) + ( (x^3) * 4(-sin(4x)) )
     
  4. Jun 2, 2009 #3
    Re: Differentiate

    c: If dy/dx =5/x and y=6 when x=0, then y=

    integral of 5/x= 5ln(x) + c
    5ln(x) + c = 6
    ln(0)=1 so
    5*1 + c = 6
    c=1, 5ln(x) + 1 = y
     
  5. Jun 2, 2009 #4
    Re: Differentiate

    d: if ln(x^3) - ln(x) = 4, then x=e^something

    ln(x3)-ln(x) = 4

    eln(x^3) - eln(x) = e4
    = x3 - x = e4

    not sure how to solve from here, mind helpin or showing me an example?
     
  6. Jun 2, 2009 #5

    rock.freak667

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    Re: Differentiate

    This looks good

    Good.


    Are you sure those are the initial conditions? Since ln(0) does not exist.

    Use this logarithm rule [itex]log_ax - log_a y=log_a (\frac{x}{y})[/itex]
     
  7. Jun 2, 2009 #6
    Re: Differentiate

    x=e not 0, sorry
     
  8. Jun 2, 2009 #7

    rock.freak667

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    Re: Differentiate

    Then part c would be correct.
     
  9. Jun 2, 2009 #8
    Re: Differentiate

    so ln(e) = 1?
     
  10. Jun 2, 2009 #9
    Re: Differentiate

    d: if ln(x^3) - ln(x) = 4, then x=e^something

    i'm not sure how to use that equation you gave me to solve this
    [itex]log_ax - log_a y=log_a (\frac{x}{y})[/itex]
     
  11. Jun 2, 2009 #10

    rock.freak667

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    Re: Differentiate

    Yes

    If I tell you that lnX is the same as logeX, would that help you more?
     
  12. Jun 2, 2009 #11
    Re: Differentiate

    d: if ln(x^3) - ln(x) = 4, then x=e^something
    logex^3 - logex

    which is loge(x^3/x)
    do i use the quotient rule or something there?
     
    Last edited: Jun 2, 2009
  13. Jun 2, 2009 #12
    Re: Differentiate


    integral of 3e^x dx = 3e^x + c
    so 3e^x + c = 7
    e^0 = 1,
    3*1 + c = 7
    y=3e^x + 4
     
  14. Jun 2, 2009 #13

    rock.freak667

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    Re: Differentiate

    using this [itex]log_aX - log_a Y=log_a (\frac{X}{Y}) [/itex]

    if X=x3 and Y=x, can you find one log term?
     
  15. Jun 2, 2009 #14
    Re: Differentiate

    is it x^2? x^3/x^1 = X^3-1

    ln(x^2) = 4
    x^2 = e^4
    x = e^4/nothing^2
    x = e^4-2

    x=e^2?
     
  16. Jun 2, 2009 #15

    rock.freak667

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    Re: Differentiate


    Yes, the final answer is correct, but it should be something like this

    [tex]x^2=e^4[/tex]
    [tex]x=\pm (e^4)\frac{1}{2}[/tex]

    [tex]x= \pm e^2[/tex]

    But x can't be negative since ln(-ve) does not exist so

    x=e2 only.
     
  17. Jun 2, 2009 #16
    Re: Differentiate


    OHO sweet! cheers mate, thanks for the help =]
     
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