July 10th, 2017, 03:41 PM  #1 
Newbie Joined: Jul 2017 From: United Kingdom Posts: 4 Thanks: 0 Math Focus: General  Second order derivative
So the question asks first to find y' of sin(x)cos(2x) which I worked out to be cos(x)cos(2x)  2sin(x)sin(2x) Then question asks for y'' for which the answer is supposed to be 5sin(x)cos(2x)  4cos(x)sin(2x) I am having trouble getting to that point, my answer somehow ends up a lot longer than that. Would anybody be able to help me with this problem? Thanks! Last edited by skipjack; July 11th, 2017 at 12:00 AM. 
July 10th, 2017, 04:10 PM  #2 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus 
Please see the attachment. Be mindful of the double product rule and the 2 that must be distributed.

July 10th, 2017, 05:40 PM  #3 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,322 Thanks: 453 Math Focus: Yet to find out.  
July 10th, 2017, 06:09 PM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,657 Thanks: 681  Quote:
sin(x)cos(2x)  2cos(x)sin(2x)  2cos(x)sin(2x)  4 sin(x)cos(2x) Now, sin(x)cos(2x)  4 sin(x)cos(2x) = 5 sin(x)cos(2x) and 2cos(x)cos(2x)  2 cos(x)sin(2x) = 4 sin(x)cos(2x). Last edited by skipjack; July 11th, 2017 at 12:03 AM.  
July 10th, 2017, 07:00 PM  #5 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  

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derivative, order 
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