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

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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

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

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

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

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

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