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Math Help - Calculus Derivatives HW help

  1. #1
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    Calculus Derivatives HW help

    1) Find an eq. of the line tangent to y=(e^2cos4x)+1 at x=Pi/6.
    2) Find the nth derivative of f(x) = xe^x.

    Any help will do. Thanks a lot.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post
    1) Find an eq. of the line tangent to y=(e^2cos4x)+1 at x=Pi/6.
    Any help will do. Thanks a lot.
    the derivative gives the slope of the tangent line at any value of x, so let's start by finding that.

    y=(e^2cos4x)+1 ..........i suppose 2cos4x is the power of e
    => y ' = -8sin4x*e^(2cos4x)
    when x = pi/6
    y' = -8sin(2pi/3)*e^(2cos(2pi/3)) = -4sqrt(3)*e^-1 = -4sqrt(3)/e ...this is the slope at x=pi/6

    now when x = pi/6
    y = e^(2cos(2pi/3)) + 1 = e^-1 + 1

    using the point slope form:
    y-y1 = m(x - x1)
    => y - e^-1 - 1 = [-4sqrt(3)/e]x + [-4sqrt(3)/e](pi/6)
    => y = [-4sqrt(3)/e]x - (2sqrt(3)pi/3e - e^-1 - 1) ...........equation of the tangent line
    and you can simplify this if you wish

    ...wow, this is a weird one, you might want to wait till someone checks this
    Last edited by Jhevon; April 2nd 2007 at 07:52 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post
    2) Find the nth derivative of f(x) = xe^x.
    Let's find a few derivatives to see if we get a pattern

    f(x) = xe^x
    => f ' (x) = e^x + xe^x = (1 + x)e^x
    => f '' (x) = e^x + + e^x + xe^x = 2e^x + xe^x = (2 + x)e^x
    => f ''' (x) = 2e^x + e^x + xe^x = 3e^x + xe^x = (3 + x)e^x
    => f '''' (x) = 3e^x + e^x + xe^x = 4e^x + xe^x = (4 + x)e^x ......starting to see the pattern?
    .
    .
    .
    => f ^(n) (x) = (n + x)e^x ................the nth derivative
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    Thank you so much for your help Jhevon. I have one more question:

    An avian disease is decimating the bird population of a South Pacific island. The population at any time (t) is roughly approximated by the equation
    y=A*e^(cos Pi/2 t), where t is in years, 0<t<2, and A is the population at t=0. Find the rate at which the population is declining at t=1 (in terms of A).

    Again, any help will do. Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cupcakelova87 View Post
    Thank you so much for your help Jhevon. I have one more question:

    An avian disease is decimating the bird population of a South Pacific island. The population at any time (t) is roughly approximated by the equation
    y=A*e^(cos Pi/2 t), where t is in years, 0<t<2, and A is the population at t=0. Find the rate at which the population is declining at t=1 (in terms of A).

    Again, any help will do. Thanks!
    I assume you mean y=A*e^(cos(Pi*t/2))

    We begin by finding the derivative:

    y=A*e^(cos(Pi*t/2))
    => y' = A*[-(pi/2)sin(pi*t/2)]e^(cos(Pi*t/2))
    when t = 1
    y' = A*(-pi/2)e^0 = -(pi/2)A ........the rate at which the population is declining in terms of A
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