Question

how do you differenciate 1/invtan(x)?

Answer 1

differentiate \(\displaystyle \frac{1}{tan^{-1}x}\)?

Answer 2

yes

Answer 3

Do you mean \(\large{ \frac{1}{\arctan(x)} }\) or \(\large{\frac{1}{\cot (x)}}\)?

Answer 4

you can use chain rule, do you know it ?

Answer 5

nah :/

Answer 6

what is the derivative of 1/x first ?

Answer 7

-x^(-2)

Answer 8

right, so using chain rule,
y= 1/arctan x
let u= arctan x
then y=1/u
and dy/dx =dy/du*du/dx = (-u^{-2}) du/dx
what about du/dx =... ? (derivative of arctan x is ...?)

Answer 9

lol this is gonna sound really stupid but we actually havent studied arc yet, is there anther method with relying on chain rule?

Answer 10

without*

Answer 11

\(\arctan = tan^{-1}\)
also, we'll need chain rule for this.

Answer 12

Chain rule :\(dy/dx =(dy/du)*(du/dx)\)

Answer 13

arc tan is the same as inverse tangent u mean?

Answer 14

yes, exactly.

Answer 15

could you get the derivative now ?

Answer 16

yeah i understand the chain rule but i cant really follow the working out

Answer 17

clear till here ?
y= 1/arctan x
let u= arctan x
then y=1/u

Answer 18

yeah

Answer 19

then chain rule [ \(dy/dx =(dy/du)*(du/dx)\) ] tells you to just find 2 terms :
dy/du and
du/dx.
when y = 1/u, dy/du=...?
when u = arctan x , then du/dx =...?
then just multiply them.

Answer 20

so (-u)^-2 * 1/1+x^2 ?

Answer 21

and sub arctan x instead of u?

Answer 22

yes, you have to now resubstitute back u=arctan x

Answer 23

thats correct

Answer 24

thanks :)

Answer 25

welcome ^_^