Derivative of arctan(x)
Arctangent's derivative is a small algebraic miracle: an inverse trig function goes in, and a plain rational function with no trig at all comes out.
Step-by-step solution
Let y = arctan(x), so tan(y) = x.
sec²(y) · y′ = 1, hence y′ = 1/sec²(y).
The identity sec²(y) = 1 + tan²(y) = 1 + x² gives y′ = 1/(1 + x²).
Why it works
The disappearance of all trigonometry from the answer is what makes this derivative important rather than merely cute: it means the area under the innocent-looking curve 1/(1+x²) is measured by arctangent. Integrate it from −∞ to ∞ and you get exactly π — the reason this function (the "Cauchy distribution" shape) keeps appearing in probability and physics.
The formula also captures arctangent's shape precisely. The derivative peaks at 1 when x = 0 (the curve crosses the origin at 45°) and decays like 1/x² in both directions, which is why arctan flattens out toward its horizontal asymptotes at ±π/2 — always rising, never arriving.
Common mistakes
- Writing the derivative as −1/(1+x²) — that belongs to arccotangent. Arctan's derivative is positive everywhere, as an increasing function's must be.
- Confusing arctan(x) = tan⁻¹(x) with 1/tan(x) = cot(x). The superscript −1 means inverse function, not reciprocal.
- Forgetting the chain rule on arctan(inner): d/dx[arctan(3x)] = 3/(1 + 9x²).
Practice problems
Differentiate arctan(2x)
Answer: 2/(1 + 4x²).
Find the slope of arctan(x) at x = 1
Answer: 1/(1+1) = 1/2.
Differentiate x arctan(x)
Answer: arctan(x) + x/(1 + x²) — product rule.