What is the derivative of # y=cot (arctan x) #?

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Answer 1 · 2017-09-13T10:15:31

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

We have:

# y=cot (arctan x) #

Suppose we let:

# tanu = x <=> u = arctan x#

Then:

# cot u = 1/(tan u) = 1/x #

But, we established that #u = arctan x#, therefore:

# cot (arctan x) = 1/x #

Hence, we have:

# y = 1/x => dy/dx = -1/x^2 #