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tsoya
Oct1-07, 10:04 PM
y= tan^(-1)[x^2-1]^(1/2) + csc^(-1)x

i cannot get to the answer, can someone help me?

well the answer should be zero. when i take the derivative of both parts (one of the tan inverse and one of the csc inverse) i dont get anywhere close to 0...

bob1182006
Oct1-07, 10:46 PM
could you show what you did?

and what type of answer are you getting? w/o doing it I think something near 1? maybe 1 minus something.

tsoya
Oct1-07, 10:57 PM
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

bob1182006
Oct1-07, 11:47 PM
Hm...

well the derivative of arctan x = 1/(1+x^2), but in your case you'll have to do the chain rule.
for csc^-1 I'm guessing that's inverse csc right? so then you'll want the derivative of 1/sin^-1 x which is arcsin to the power of negative 1 which will again need a use of the chain rule.

atqamar
Oct7-07, 03:01 PM
The answer I am getting is not zero or one.
You will have to use the chain rule multiple times. And keep in mind:
\frac{d}{dx} arccsc(x) = \frac{-1}{x\sqrt{x^2 - 1}}
\frac{d}{dx} arctan(x) = \frac{1}{x^2 + 1}.

Hurkyl
Oct7-07, 03:27 PM
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
Well, it almost works out to zero, doesn't it? Maybe you made a slight error in your calculation; have you rechecked them?

atqamar
Oct7-07, 04:44 PM
If the derivitive is graphed, it can be seen that after an approximate value of 2, the derivitive gets very close to 0.
This is the solution i got for the problem:

\frac{d}{dx}arccsc(x) + \sqrt{arctan(x^2 - 1)} = \frac{-1}{x\sqrt{x^2 - 1}} + \frac{x}{((x^2 - 1)^2 + 1)\sqrt{arctan(x^2 - 1)}}.