# Derivative of arcsec(x) and arccsc(x)

1. Oct 28, 2011

### 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?

2. Oct 29, 2011

### 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.

3. Nov 1, 2011

### 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.