View Full Version : derivative of inverse trif function
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.
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.
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}.
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?
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)}}.
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