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

1 Answer
turksvids
Dec 29, 2017

#d/dx(sec^(-1)(2x)) = 1/(|x|sqrt(4x^2-1)#

Explanation:

We know that #d/dx(sec^-1(x)) = 1/(|x|sqrt(x^2-1))#
See this video, for reference.

In this case, #d/dx(sec^(-1)(2x)) = 1/(|2x|sqrt((2x)^2)-1)*2#.

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=4x^2#. So we end up with:

#d/dx(sec^(-1)(2x)) = 1/(|x|sqrt(4x^2-1)#.

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