Derivative of arcsec(x) and arccsc(x)

[NoOne0507]

I was trying to prove the derivatives of the inverse trig functions, but I ran into a problem when I tried doing it with arcsecant and arccosecant.

So the general process is this:

y = arcsec(x)
sec(y) = x
dy/dx * sec(y)tan(y) = 1
dy/dx = 1/[sex(y)tan(y)]

sec(y) = x
And for tan(y) we use the Pythagorean identities:

tan^2(y) = sec^2(y) - 1
tan(y) = [sec^(y) - 1] ^(1/2)

So dy/dx = 1/[x(x^2-1)^(1/2)]

However, my calculus book has one minor difference in it's derivative, an absolute value:
dy/dx = 1/[abs(x)(x^2-1)^(1/2)]

Where does this absolute value come from?

 

[AlephZero]

You missed the fact that $\tan^2 y = \sec^2 y - 1$ and therefore
$\tan y = \pm \sqrt{\sec^2 y - 1}$

But only one of the "$\pm$" solutions is the correct one, depending on which quadrant the angle is in.

The "abs(x)" gives the correct solution, if you always use the positive value of the square root.

 

[paulfr]

A similar situation involving absolute value is
Integral of 1/x dx = Ln |x| + C

The absolute value is added because one can not take the log of a negative number.

The absolute value seems to come out of nowhere but it is just a way of using Math symbols to say what the words would say.