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.
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
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?
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=> f ^(n) (x) = (n + x)e^x ................the nth derivative
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!