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?
 A: $$y=\tan^{-1}(2x-1)$$
$$\tan(y)=\tan(\tan^{-1}(2x-1))=2x-1$$
Differentiating with respect to x (using chain rule):
$$\frac{d\tan y}{dy}\frac{dy}{dx}=2$$
$$\frac{dy}{dx}=\frac{2}{\frac{d\tan y}{dy}}=\frac{2}{\sec^2y}=\frac{2}{\sec^2(\tan^{-1}(2x-1))}=2\cos^{2}(\tan^{-1}(2x-1))$$
But $\tan^{-1}(2x-1)$ is the angle that has $2x-1$ as tangent, so, by geometric logic, we find the cossine of such angle by the geometric definition of tangent and the simple formula 
$$\cos a=\frac{\mathrm{side of the triangle adjacent to the angle}}{\mathrm{hypotenuse}}$$
Now the geometrical interpretation of tangent gives:
$$(2x-1)^2+1^2=\mathrm{hypotenuse}^2$$
And substituting the hypotenuse value in the previous formula we find cos.
$$\cos(\tan^{-1}(2x-1))=\frac{1}{\sqrt{(2x-1)^2+1}}$$
$$\frac{dy}{dx}=\frac{2}{(2x-1)^2+1}=\frac{2}{4x^2+2-4x}=\frac{1}{2x^2-2x+1}$$
Here is a drawing to help with the geometrical part
A: $$\dfrac{d\{\arctan(2x-1)\}}{dx}=\dfrac{d\{\arctan(2x-1)\}}{d(2x-1)}\cdot\dfrac{d(2x-1)}{dx}=\dfrac1{1+(2x-1)^2}\cdot\dfrac{d(2x-1)}{dx}=?$$
