How do you differentiate $arctan (x - 1)$ $arctan(x-1)$ ?

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Answer 1 · 2017-02-09T00:32:13

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

From $y = arctan(x-1)$ you can say that $tan y = x-1$

And so by implicit differentiation $sec^2 y \ y' = 1$

Which means that:

$y' = 1/( sec^2 y) = 1/( tan^2 y + 1) = 1/( (x-1)^2 + 1 )= 1/( x^2 - 2x + 2)$

How do you differentiate arctan(x-1)arctan(x−1)arctan(x-1)?