• Oct 28th 2007, 07:52 AM
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.
• Oct 28th 2007, 08:05 AM
galactus
We mostly do not worry about deriving these each time. That's why they're memorized or looked up if needed. You can use the triangle to derive the derivative of arctan.

$\displaystyle y=tan^{-1}(x), \;\ x=tan(y)$

Differentiate the right equation implicitly:

$\displaystyle \frac{d}{dx}[x]=\frac{d}{dx}[tan(y)]$

$\displaystyle 1=sec^{2}(y)\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2 }(tan^{-1}(x))}$.........[1]

It follows from the triangle that:

$\displaystyle sec^{2}(tan^{-1}(x))=1+x^{2}$

And so, from [1], we have $\displaystyle \frac{dy}{dx}=\frac{1}{1+x^{2}}$
• Oct 28th 2007, 08:28 AM
Thanks galactus, much appreciated
!
• Oct 28th 2007, 09:43 AM
Krizalid
I recently posted a solution for your problem.

Double postin' is against the Rules, did you read them?
• Oct 28th 2007, 09:49 AM
You wanted to find $\displaystyle (\arctan x)',$ I posted the solution here.