Derivative of inverse trif function

[tsoya]

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 don't get anywhere close to 0...

 

[bob1182006]

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]

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]

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]

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]

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]

If the derivative is graphed, it can be seen that after an approximate value of $$2$$, the derivative gets very close to $$0$$.
This is the solution i got for the problem:
$$\frac{d}{dx}arccsc(x) + \sqrt{arctan(x^2 - 1)} = \frac{-1}{x\sqrt{x^2 - 1}} + \frac{x}{((x^2 - 1)^2 + 1)\sqrt{arctan(x^2 - 1)}}$$.