# Math Help - help plz! tan^-1(s) = 1/1+s^2 how????

1. ## help plz! tan^-1(s) = 1/1+s^2 how????

I have a formula that I am trying to differentiate, but having lots of problems with it:
r(s)= tan^-1 s, root of 2 /2 log(1+s^2), s - tan^-1 s)
its derivates:
r'(s) = ( ( 1+s^2)^-1, root of 2 s(1+s^2)^-1, 1- (1+s^2)-1

= (1+s^2)^-1 (1, root of 2 s, s^2)

I dont know how the derivative of tan^-1(s) = 1/1+s^2, can anyone explain?

And cannot follow how all the derivatives work out, can anyone talk me through it? I have exam shortly and ally need to get to know these off by heart.

2. It's easy, first define $y=\arctan x\implies\tan y=x.$

Now, differentiate implicity, and yields $y'\sec^2y=1\implies y'=\frac1{1+\tan^2y}.$

And we're done.