derivative of inverse trig function

• October 1st 2007, 07:01 PM
mstsoy
derivative of inverse trig function
y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

• October 1st 2007, 07:15 PM
Jhevon
Quote:

Originally Posted by mstsoy
y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

$\frac d{dx}\tan^{-1}u = \frac {u'}{1 + u^2}$ .........by the chain rule

$\frac d{dx} \csc^{-1} x = - \frac {1}{x \sqrt {x^2 - 1}}$
• October 1st 2007, 07:21 PM
mstsoy
no i have definitely tried working it out already.
i took derivative of both sides of the plus sign
the answer should come out to zero

but im not getting newhere near that.

im getting something like x/ [(x^2 -1)(x^2 -1)^2] - 1/[|x|(x^2-1)^2]

before simplifying, and after simplifying its not zero..urgh i cannot get it