How to use implicit differentiate on arctan? y = $tan^{-1}$(2x-1).
Is $tan^{-1}$ arctan?
Is it $\frac{1}{x^2+1}$?
Does this mean the answer is  $\frac{1}{4x^-2+1}$, since the chain rule is 1?

$$\dfrac{d\{\arctan(2x-1)\}}{dx}=\dfrac{d\{\arctan(2x-1)\}}{d(2x-1)}\cdot\dfrac{d(2x-1)}{dx}=\dfrac1{1+(2x-1)^2}\cdot\dfrac{d(2x-1)}{dx}=?$$
