Question

How do you Find the derivative of #y=arctan((1-x)/(1+x) )#?

Answer 1

Let us find the derivative of `{1-x}/{1+x}`.

By Quotient Rule,
`({1-x}/{1+x})'={-1cdot(1+x)-(1-x)cdot1}/{(1+x)^2}={-2}/(1+x)^2`

By Chain Rule,
`y'=1/{1+({1-x}/{1+x})^2}cdot{-2}/(1+x)^2`

by multiplying the quotients together,
`={-2}/{(1+x)^2+(1-x)^2}`

by simplifying the denominator
`={-2}/{2(1+x^2)}`

by cancelling out 2's,
`=-{1}/{1+x^2}`