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4th derivative of cos(2x)

  • Thread starter Cacophony
  • Start date
  • #1
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Homework Statement



see title.

Homework Equations


no


The Attempt at a Solution



Ok so the solution is 16cos(2x) but I'm not sure how it is derived to that. I've tried the product rule but it's not working for me. What rule or rules do I use to get this solution?
 

Answers and Replies

  • #2
584
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Well if you calculate the first derivative properly you should be yielded to -2sin(2x).



Edit: Take into consideration that the derivative of cos(Ux)= -u'(x)*sin(x)
 
  • #3
41
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Ok cool but what rule did you use there?
 
  • #4
22,097
3,282
The chain rule.
 
  • #5
HallsofIvy
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You don't need the product rule because do not have a product of two functions of x. You need the chain rule because you have f(y)= 2cos(y) and y= 2x:
[tex]\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}[/tex]
With f(y)= cos(y), what is df/dy? With y= 2x, what is dy/dx?
 
  • #6
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You don't need the product rule because do not have a product of two functions of x. You need the chain rule because you have f(y)= 2cos(y) and y= 2x:
Make that f(y)= cos(y)
[tex]\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}[/tex]
With f(y)= cos(y), what is df/dy? With y= 2x, what is dy/dx?
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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Make that f(y)= cos(y)
Right. Thanks for the correction.
 

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