how to find derivative of arctan(x+1) ? step by step explanation?

Question

accessdenied · Answer

Have you considered using implicit differentiation?

anonymous · Answer

No I thought I could use the chain rule… the original problem was f(x) = 7arctan(x+1) – ln(x^2 + 2x + 2) but i'm not sure on how to find the derivative of arctan(x+1)

jdoe0001 · Answer

$\bf \cfrac{d}{dx}[tan^{-1}(x)]=\cfrac{1}{1+x^2}$

accessdenied · Answer

Oh.  I was assuming this was more of a proof of the derivative we were looking towards.
I could still demonstrate that, but I believe it is usually provided as an identity somewhere.
  $ \dfrac{d}{dx} \arctan x = \dfrac{1}{x^2 + 1} $
You would need chain rule for $ \arctan (x+1)$, although conveniently th derivative of the inside is 1.

anonymous · Answer

I was looking at this derivative calculator for this problem what confused me was one of the steps it looks like this:
1/(x+1)^2 +1 times d/dx(x+1)

accessdenied · Answer

Chain rule would give us a step like this:
  $ \dfrac{d(\arctan (x+1)) }{dx} = \dfrac{d( \arctan(x + 1))}{d(x+1)} \times \dfrac{d(x+1)}{dx} $
The first factor uses the identity above, except for x+1 substituted for x.  The latter factor is the provided d/dx(x+1).

accessdenied · Answer

* If that way of writing it looks confusing, consider u = x + 1 and:
  $ \dfrac{d ( \arctan u)}{du} \times \dfrac{du}{dx} $

anonymous · Answer

alright thank you!

accessdenied · Answer

you're welcome! :)