Question

What is the derivative of `arctan (8/x^2)`?

Answer 1

`(-16x)/(x^4+64)`

Explanation:

By the chain rule:

`y=arctanu, u = 8/x^2`

`dy/(du) = 1/(1+u^2), (du)/dx = -16/x^3`

`dy/dx=dy/(du)*(du)/dx`

`= 1/(1+(8/x^2)^2)*-16/x^3`

`= -16/(x^3+64/x) = (-16x)/(x^4+64)`