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

1 Answer
May 19, 2016

#d/dx arctan(x-1) = 1/(x^2-2x+2)#

Explanation:

We can differentiate this using implicit differentiation:

Let #y = arctan(x-1)#

#=> tan(y) = x-1#

#=> d/dxtan(y) = d/dx(x-1)#

#=> sec^2(y)dy/dx = 1#

#=> dy/dx = 1/sec^2(y)#

Drawing a right triangle with an angle #y# such that #tan(y) = x-1#, we can find that #sec(y) = sqrt(x^2-2x+2)#. Plugging that in, we get our result:

#d/dx arctan(x-1) = 1/(x^2-2x+2)#


Note that the same process shows that the formula for #d/dxarctan(x) = 1/(x^2+1)#. If we already have that formula, then we can simply use the chain rule:

#d/dxarctan(x-1) = 1/((x-1)^2+1)(d/dx(x-1))#

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