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WhiteBuffa...

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posted by WhiteBuffalo on 7/24/2008 12:56:57 AM | status: Closed

derivatives of exponetial

Course	Textbook	Chapter	Problem
Calculus	N/A	N/A	N/A

Question Details:

y = (e^-x)cos(x)

Need the first and second deriv and when the equal zero.
Thanks

Tags: Math, Calculus

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Answers:

zsm28

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posted by zsm28 on 7/24/2008 3:40:22 AM | status: Live

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Response Details:

y = (e^-x)cos(x)
y' = - (e^-x)cos(x) - (e^-x)sin(x) = - (e^-x)[cos(x) + sin(x)]
when y' = 0, cos(x) + sin(x) = 0, 1 + tan(x) = 0, tan(x) = -1, x = -π/4, 3π/4
y" = (e^-x)[cos(x) + sin(x)] - (e^-x)[-sin(x) + cos(x)] = 2(e^-x)sin(x)
when y" = 0, sin(x) = 0, x = 0, π

Tags: Math, Calculus

I, Mathema...

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posted by I, Mathemagician on 7/24/2008 3:46:10 AM | status: Live

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Response Details:

y=(e^-x)cos(x)
y'=-(e^-x)cos(x)-(e^-x)sin(x)=-((e^-x)cos(x)+(e^-x)sin(x))
y''=-(-(e^-x)cos(x)-(e^-x)sin(x)-(e^-x)sin(x)+(e^-x)cos(x))
=(e^-x)cos(x)+(e^-x)sin(x)+(e^-x)sin(x)-(e^-x)cos(x).

y'=0=-((e^-x)cos(x)+(e^-x)sin(x)).
0=cos(x)+sin(x).
-sin(x)=cos(x)
sin(-x)=cos(-x)
x=-π/4+C(2π), where C is an integer.

y''=0=(e^-x)cos(x)+(e^-x)sin(x)+(e^-x)sin(x)-(e^-x)cos(x).
0=2(e^-x)cos(x)
cos(x)=0
x=π/2+C(2π), where C is an integer.

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Tags: Math, Calculus

Achilles

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posted by Achilles on 7/24/2008 3:48:09 AM | status: Live

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Response Details:

y = (e^-x)cos(x)...............................>product rule
Let u=e^-x...................................>du=-e^-xdx
v=cosx.......................................>dv=-sinx

∴uv'+vu'----------------------->product rule

∴dy/dx=-e^-xcosx-e^-xsinx................................>using same rule as above find double derivative
when the equal zero..............................>