I wanted to find the derivative of $$g(t) = \operatorname{arcsec} 2x$$
I know I can plug it into an equation for the derivative, but I want to learn how to solve it
So the way I am working on is to solve for $a^2 + b^2 = c^2$ to find the values given the arcsin angle. So I get tha $ a = \sqrt{1-4t^2}$, the second side is $b = 1$ and $c = 2t$
Then I solve to find the derivative of $f(t)$ using the equation for the Derivatives of Inverse function. So $$f'(t)= 2(sec2t)(tan2t)$$ I plug this into the equation $$g'(t) = \frac{\frac{du}{dx}}{f'(\operatorname{arcsec}2t)} $$
Which gives me the final answer of $$g'(t) = \frac{1}{2t \sqrt{1-4t^2}} $$ This answer is incorrect. Can someone help me figure out where I am going wrong. Thanks!
$g(x) = \text{arcsec}(2x)$, assuming $x$ and $t$ are the same.
I'll re-write this as $y = \sec^{-1} (2x)$. With a bit of work... $$\sec y = 2x$$ $$y'\sec y \tan y = 2$$ A really helpful identity here is $\tan^2 x + 1 = \sec^2 x$
This is equivalent to $\tan x = \sqrt{\sec^2 x -1}$. Let's plug this in now.
$$y' = \frac2{\sec y \tan y}$$ $$=\frac2{(2x)(\sqrt{4x^2-1})}$$ $$=\frac1{x\sqrt{4x^2-1}}$$
BTW, the book may be wrong.
http://www.math.com/tables/derivatives/tableof.htm
https://en.wikipedia.org/wiki/List_of_trigonometric_identities
https://www.desmos.com/calculator/yxpojm1etv
Edit: It seems the correct answer is $=\frac1{|x|\sqrt{4x^2-1}}$. Could someone please explain why an absolute value is necessary? It may have to do with plus-or-minus square roots...