Question

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

Answer 1

`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)`.