What is the derivative of #arctan(x-1)#?

Question

5020 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-11T09:19:49

The answer is #(arctan(x+1))' = 1/((x+1)^2+1)#

Let's start with an uncommon method,

We have #theta = arctan(x+1)#

#tan(theta) = x+1#

Derivate both side

#theta'(tan^2(theta)+1) = 1#

#theta' = 1/(tan^2(theta)+1)#

But #tan(theta) = x+1#

#theta' = 1/((x+1)^2+1)#

Answer 2 · 2015-06-15T19:43:10

#d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))#

#=> 1/(1+(x-1)^2)*1#

#= 1/(1+(x-1)^2)#

#= 1/(x^2 - 2x + 2)#