Question

find the derivative
y=sec^-1 (2s+1)

Answer 1

Couple of ways to go about this. You COULD just remember the silly thing.

\(\dfrac{d}{dx}\sec^{-1}(x) = \dfrac{1}{|x|\sqrt{x^{2} - 1}}\)

Isn't that beautiful? (I had to look it up.)

On the other hand, one might want to break out some implicit differentiation.

\(y = \sec^{-1}(2s+1)\)

Then, after noting that \(\pi/2 \le y \le \pi/2\)

\(\sec(y) = 2s + 1\)

And we have rather easily,

\(\sec(y)\tan(y)\dfrac{dy}{ds} = 2\;or\;\dfrac{dy}{ds} = \dfrac{2}{\sec(y)\tan(y)}\)

You can draw a right triangle and play around with that a little if you like, but I think it's fine just the way it is.