1. ## differentiation skills help please!

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. 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 $\displaystyle \frac{dy}{dx}=\frac{1}{1+x^{2}}$

3. Thanks galactus, much appreciated
!

4. I recently posted a solution for your problem.

Double postin' is against the Rules, did you read them?

5. where exactly did you post it? My problem was for the whole equation, not just the first part. Do you have a link to your post? Or show me how I could have searched for it. If i know hwo to search properly for it, then I will next time.

6. You wanted to find $\displaystyle (\arctan x)',$ I posted the solution here.

7. I se now, yeah sorry for the double post. Can you help with my other question on the board about algebraic simplification?