Shown are two different methods for finding $du$ when $u=\frac{1}{2}sec(v)$. I'm failing to see how we are able to justify dropping the absolute value sign in the second method. Thanks!
Justify dropping the absolute value when taking the derivative of secant
absolute valuederivativessecanttrigonometry
Best Answer
The first method gives you the correct derivative for all $v$ where $\sec v$ is defined.
The second method, as written, is only applicable to $0\le v\le\pi$ (excluding $v=\pi/2$), but in this interval, $\sec v\tan v\ge 0$ anyways, and so the absolute value can be dropped.
However, in other intervals for $v$, the formula $v=\text{arcsec}(2u)$ will be incorrect, and a more sophisticated formula will be valid. E.g. for $\pi\le v\le 2\pi$, you would instead have this formula: $v=2\pi-\text{arcsec}(2u)$, and that minus sign (in front of $\text{arcsec}$) will cancel the absolute value, because, between $\pi$ and $2\pi$, we will have $\sec v\tan v\le 0$.
I would leave it to you to look at a few further intervals and convince yourself that the second method, when the $\text{arcsec}$ formula is corrected, always yields the same result as the first method.