How to justify dropping the absolute value when taking the derivative of secant?

Question

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!

Answer 1

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.