Question

What's the derivative of `arctan(x/a)`?

Answer 1

`d/(dx)arctan(x/a)=a/(x^2+a^2)`

Explanation:

`y=arctan(x/a)` is equivalent to `tany=x/a`

Now taking derivative of both sides

`sec^2yxx(dy)/(dx)=1/a` or

`(dy)/(dx)=1/axx1/sec^2y`

= `1/axx1/(1+tan^2y)`

= `1/axx1/(1+x^2/a^2)`

= `1/axxa^2/(x^2+a^2)`

= `a/(x^2+a^2)`