How do you differentiate #arctan(1/x)#?

1 Answer
Nov 9, 2016

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

Explanation:

We can write #y=arctan(1/x) <=>tany=1/x#

# :. 1/tany=x #
# :. coty=x # ..... [1]

We can then differentiate implicitly:

# -csc^2y dy/dx= 1 #
# dy/dx= -1/csc^2y #

Using the identity # 1+cot^A-=csc^2A # we have

# dy/dx= -1/(1+cot^2y) #
# dy/dx= -1/(1+x^2) # (from [1])