Question

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

Answer 1

`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])