What is the derivative for arctan(2/x)?

[EvLer]

What is the derivative for

arctan(2/x)?

I know it's generally: [(1/a)arctan(x/a)]' = 1/(x^2 + a^2); but in my case its x^-1!

Thanks for any help.

 

[quasar987]

The formula to remember is

$$Arctg'(x) = \frac{1}{1+x^2}$$

So using the chain rule,

$$Arctg'(f(x)) = \frac{f'(x)}{1+[f(x)]^2}$$

which gives the derivative of Arctg for any argument.

 

[thepatientmental]

The derivative for arctan(2/x) would be -2/(x^2 + 4). This is because the derivative of arctan(u) is 1/(1+u^2) and in this case, u is equal to 2/x. Therefore, the derivative would be 1/(1+(2/x)^2) which simplifies to -2/(x^2 + 4).