How do you find the derivative g(x)=arcsec(2x)?

1 Answer
Dec 29, 2017

ddx(sec1(2x))=1|x|4x21

Explanation:

We know that ddx(sec1(x))=1|x|x21
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In this case, ddx(sec1(2x))=1|2x|(2x)212.

We can rewrite this a bit, since |2x|=2|x|, the 2 from here will cancel with the 2 we got from the chain rule. Also, (2x)2=4x2. So we end up with:

ddx(sec1(2x))=1|x|4x21.