Question

How would you go about solving: (arctan(1/x))' ?

Answer 1

Without using that (arctan(u))'=1/(1+u^2).

Answer 2

can u rewrite in equation form? what is " ' " ?

Answer 3

Derivative.

Answer 4

so your real question is how we find
(arctan(u))'=1/(1+u^2) ?
cause when we know it we can use it

Answer 5

Basically. :D

Answer 6

y = arctan(x)
we want to find y'

now:
tan(y) = x
so take the derivative with respect to x
sec^2(y) * y' = 1
y' = cos^2(y)
y' = cos^2(arctan(x))

Answer 7

I never worked with the function sec. :/

Answer 8

sec = 1/cos

Answer 9

y' = cos^2(arctan(x))

according to my drawing :
arctan(x) = t
so
y' = cos^2(t)

now according to the drawing:
cos(t) = 1/sqrt(1+x^2)
so
cos^2(t) = 1/(1+x^2)

so
y' = 1/(1+x^2)

Answer 10

"I never worked with the function sec. :/"
so
you know
(tan(x))' = 1/(cos^2(x))
i just use
sec(x) = 1/cos(x)
so
(tan(x))' = sec^2(x)

Answer 11

That's a real cool alternative approach without using limits, what is the domain of the relation sec(x)=1/cos(x)?

Answer 12

this is just a definition..
instead of writing 1/cos(x)
we write sec(x)

Answer 13

Yes I know, but what is the domain?

Answer 14

Is it applicable theoretically even though it's derived from a geometric relation?

Answer 15

look at the function :
http://www.wolframalpha.com/input/?i=secx&dataset=

Answer 16

there is also
csc which is 1/sin

Answer 17

"Is it applicable theoretically even though it's derived from a geometric relation?"
you ask if the derivative of arctan ?
yes .. applicable

Answer 18

So both those relations above have the domain C?

Answer 19

what do you mean ?

Answer 20

I mean if the relation is applicable for all the possible values of t and x (even though your figure above might not even be a triangle).
The relation being sec(x)=1/cos(x)

Answer 21

Nevermind, I was just being curious about the relation sec(x)=1/(cos(x)); it was very beautiful.

Answer 22

the relation sec(x)=1/cos(x) is independent of the triangle.

Answer 23

Okay. :) Do you know a proof of this?

Answer 24

Nevermind, I was just being curious about the relation sec(x)=1/(cos(x)); it was very beautiful.

Answer 25

Okay. :) Do you know a proof of this?

Answer 26

the triangle was there to help us find what is
cos^2(arctan(x))
in terms of x.

Answer 27

Oooh.

Answer 28

we dont need any proof for
sec(x) = 1/(cos(x))
this is a definition ..

i can make a new one :
coolsector(x) = x^2 + 6

Answer 29

sec(x) is just a name for the function 1/cos(x)

Answer 30

I see! I got it mixed up. Thanks :)

Answer 31

yw :)