The derivative of arctan x is 1/(1+x²) — no trigonometry left in it at all. That is not a
coincidence, and the derivation explains it in three lines.
The inverse function rule
If y = arctan x then tan y = x. Differentiate both sides: sec²y · y' = 1, so y' = 1/sec²y. Now use
sec²y = 1 + tan²y = 1 + x², and the trigonometry disappears.
The same trick everywhere
arcsin gives 1/√(1−x²) by the same route through cos y = √(1−sin²y). The inverse hyperbolic
functions give 1/√(x²+1) and 1/(1−x²), differing from the circular versions only by signs.
Why it matters
Because these derivatives are algebraic, they turn up as the answers to integrals that contain no
trigonometry whatsoever. Recognising 1/(1+x²) on sight is worth more than remembering the derivative
in the forward direction.