how do you differenciate 1/invtan(x)?

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

yes

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

you can use chain rule, do you know it ?

nah :/

what is the derivative of 1/x first ?

-x^(-2)

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 ...?)

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

without*

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

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

arc tan is the same as inverse tangent u mean?

yes, exactly.

could you get the derivative now ?

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

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

yeah

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.

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

and sub arctan x instead of u?

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

thats correct

thanks :)

welcome ^_^

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