Find derivative of complicated inverse trig function

Derive:

\(y=\operatorname{arccot}x+\arctan \left(\frac{2+x}{1-2x}\right)\)

Before we start, recall:

\(\frac{d}{dx}\operatorname{arccot}x=\frac{-1}{1+x^2}\)
\(\frac{d}{dx}\arctan x=\frac{1}{1+x^2}\)

Now, let's return to the problem, the left side is easy, it becomes: \(\frac{dy}{dx}\), but the right side, specifically the tangent part requires the chain rule:

\(\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{\left(2+x\right)\left(-2\right)-\left(1-2x\right)}{\left(1-2x\right)^2}\right)\)

The derivative is done. Now we clean it:

\(\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{-4-2x-1+2x}{\left(1-2x\right)^2}\right)\)

Clean more:

\(\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{1}{1+\left(\frac{2+x}{1-2x}\right)^2}\left(\frac{-5}{\left(1-2x\right)^2}\right)\)

Clean more:

\(\frac{dy}{dx}=\frac{-1}{1+x^2}+\frac{-5}{\left[1+\left(\frac{2+x}{1-2x}\right)^2\right]\left(1-2x\right)^2}\)

If you want, you can combine more, but this is fine.