MathsRevision.net Forum : MathsRevision.net Forum
A discussion board where people can post problems or questions they are having about their course.
Re: find stationary points (69 Views)
Posted by: joejoe (IP Logged)
Date: September 29, 2009 11:55AM
1>
y=e^(x^2(x^2-18))
Use the chain rule to determine the derivative.
The chain rule is dy/dx= dy/dt * dt/dx.
Let t=x2(x2-18))= x^4-18x^2
So dt/dx=4x^3-36x
y=e^t
dy/dt=e^t
So putting all this into the chain rule equation dy/dx= dy/dt * dt/dx
dy/dx = e^t * 4x^3-36x
Remember we put t=x^4-18x^2
dy/dx = e^(x4-18x2) * 4x^3-36x
dy/dx = e^(x4-18x2) * 4x(x^2-9)
dy/dx = e^(x4-18x2) * 4x(x+3)(x-3)
Stationary point is when dy/dx=0 so by looking at the right hand side of the equation we can determine when dy/dx=0. Determine when 4x(x+3)(x-3)=0
when x=0, 4x(x+3)(x-3)=0
when x=3, 4x(x+3)(x-3)=0
when x=-3, 4x(x+3)(x-3)=0
Please work out the y co-ordinates by subsituting x=0,3,-3 into the equation y=e^(x^2(x^2-18)).
y=e^(x^2(x^2-18))
Use the chain rule to determine the derivative.
The chain rule is dy/dx= dy/dt * dt/dx.
Let t=x2(x2-18))= x^4-18x^2
So dt/dx=4x^3-36x
y=e^t
dy/dt=e^t
So putting all this into the chain rule equation dy/dx= dy/dt * dt/dx
dy/dx = e^t * 4x^3-36x
Remember we put t=x^4-18x^2
dy/dx = e^(x4-18x2) * 4x^3-36x
dy/dx = e^(x4-18x2) * 4x(x^2-9)
dy/dx = e^(x4-18x2) * 4x(x+3)(x-3)
Stationary point is when dy/dx=0 so by looking at the right hand side of the equation we can determine when dy/dx=0. Determine when 4x(x+3)(x-3)=0
when x=0, 4x(x+3)(x-3)=0
when x=3, 4x(x+3)(x-3)=0
when x=-3, 4x(x+3)(x-3)=0
Please work out the y co-ordinates by subsituting x=0,3,-3 into the equation y=e^(x^2(x^2-18)).
| Subject | Written By | Posted |
|---|---|---|
 find stationary points (233 Views)
|
sammy | 08/12/2009 05:38PM |
| Re: find stationary points (69 Views) |
joejoe | 09/29/2009 11:55AM |
Sorry, only registered users may post in this forum.