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May 9th, 2017, 11:54 AM  #1 
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  I dont understand how this is the answer deriative
tan^2(2+3x) I have no clue how they get (6tan)(2+3x)(cos^2)(2+3x) I used chain and what I got is 6sec^2(2+3x) 
May 9th, 2017, 12:05 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
$$(\tan^2(2+3x))'=2\cdot\tan(2+3x)\cdot\sec^2(2+3x )\cdot3$$ Does that help? 
May 9th, 2017, 12:55 PM  #3 
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  
May 9th, 2017, 01:01 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
Do you know what $\sec(x)$ is in terms of $\cos(x)$?


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