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May 9th, 2017, 11:54 AM   #1
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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)
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May 9th, 2017, 12:05 PM   #2
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$$(\tan^2(2+3x))'=2\cdot\tan(2+3x)\cdot\sec^2(2+3x )\cdot3$$

Does that help?
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May 9th, 2017, 12:55 PM   #3
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Originally Posted by greg1313 View Post
$$(\tan^2(2+3x))'=2\cdot\tan(2+3x)\cdot\sec^2(2+3x )\cdot3$$

Does that help?
how do they get cos?
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May 9th, 2017, 01:01 PM   #4
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Do you know what $\sec(x)$ is in terms of $\cos(x)$?
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