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

Question

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

tkhunny · Answer

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.