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Re: using product to find stationary points (49 Views)
Posted by: joejoe (IP Logged)
Date: October 18, 2009 05:57AM
y=xe^(-2x)
Using the product rule
u=x, v=e^(-2x)
du/dx=1, dv/dx=(-2)e^(-2x)
Product rule
du/dx=v*du/dx + u*dv/dx
=(x)(-2)e^(-2x) + (1)e^(-2x)
=(-2x+1)e^(-2x)
=(1-2x)e^(-2x)
Staionary point occurs when dy/dx=0 therefore
Staionary point is
0=(1-2x)e^(-2x)
x=0.5 dy/dx=0.
Substitute x=0.5 back into y=xe^(-2x) to get y-co-ordinate.
Using the product rule
u=x, v=e^(-2x)
du/dx=1, dv/dx=(-2)e^(-2x)
Product rule
du/dx=v*du/dx + u*dv/dx
=(x)(-2)e^(-2x) + (1)e^(-2x)
=(-2x+1)e^(-2x)
=(1-2x)e^(-2x)
Staionary point occurs when dy/dx=0 therefore
Staionary point is
0=(1-2x)e^(-2x)
x=0.5 dy/dx=0.
Substitute x=0.5 back into y=xe^(-2x) to get y-co-ordinate.
| Subject | Written By | Posted |
|---|---|---|
 using product to find stationary points (82 Views)
|
buggy | 10/17/2009 05:50PM |
| Re: using product to find stationary points (49 Views) |
joejoe | 10/18/2009 05:57AM |
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