Question

Find the derivative of y=arctan[radical(x^2 -1)] + arccsc(x) using implicit differentation. x>1

Answer 1

hi :)

lets first simplify the 1st term = arctan[radical(x^2 -1)]
put x = tan u

what u get ?

Answer 2

so am i just finding the derivative of the first part? I got \[\frac{ 2x }{ (x ^{2}-1)^2 }\]
Not sure if thats right; i used \[\frac{ d }{ dx }\tan ^{-1}u=\frac{ 1 }{ 1+u ^{2} }\times\frac{ d }{ dx }u\]

Answer 3

you missed the radical part ?

anyways, we can first simplify the 1st term before taking the derivative
if you don't simplify, it'll be quite complicated

Answer 4

oh ok, and my bad; i forgot about the radical haha
does it come out to \[\frac{ 2x }{ x^{2} } = \frac{ 2 }{ x }\] ?

Answer 5

i don't know how to find the derivative of arc csc. I only know how if its something like
y=arc csc(x)
csc(y)=x

Answer 6

I think it is easy if you use quotient rule. Even without simplifying. Do you know how to apply it @claborte ?

Answer 7

do i use quotient rule for the whole problem or just the arc csc part?

Answer 8

Ow wait! I didn't see the true problem.

Answer 9

yeah, its a doozy.

Answer 10

The derivative of csc(x)=csc(x)cot(x)