Question

How do you differentiate `f(x)=arctan(1/(2x-1))` using the chain rule?

Answer 1

`f'(x)=-1/(2x^2-2x+1)`

Explanation:

According the Chain Rule, `d/dx[arctan(u)]=(u')/(1+u^2)`

If you want explanation of how to find this, please ask.

Therefore, `f'(x)=overbrace(d/dx[1/(2x-1)])^(color(blue)("write as " (2x-1)^(-1)))/(1+(1/(2x-1))^2`.

Simplify and get: `f'(x)=-1/(2x^2-2x+1)`