Calculus I › Differentiation rules › full formula sheet
The derivative of arctan x
The inverse tangent’s derivative is 1/(1+x²) — implicit differentiation on tan y = x, collapsing through sec².
Notation: arctan x is the angle in (−π/2, π/2) whose tangent is x. Defined (and differentiable) for all real x.
Before this lesson: Implicit differentiation · Derivative of tan x
Where it comes from
Write y = arctan x as tan y = x and differentiate implicitly:
Before reading on: y = arctan x means tan y = x, so sec²y·dy/dx = 1. You need dy/dx in terms of x alone — which trig identity collapses sec²y back to x?
The graph confirms the shape: arctan rises (positive derivative), crosses 0 at 45° (slope 1 = 1/(1+0)), and flattens toward ±π/2 as x → ±∞ (1/(1+x²) → 0). The wrong guess 1/(1−x²) goes negative for |x| > 1 — but arctan never falls. Dead on arrival.
Derivation
The implicit proof, stated cleanly. The crux is collapsing sec²y back to x:
Why no sign issue? Unlike arcsin/arccos, there is no square root to sign — sec²y is always positive, and 1/(1+x²) is manifestly positive, matching arctan’s rise.
How to use it
The procedure:
- Plain arctan x → 1/(1+x²).
- Something inside? Chain: d/dx [arctan(u)] = u′/(1+u²). Example: d/dx [arctan(3x)] = 3/(1+9x²).
- Always positive: 1/(1+u²) > 0 everywhere — arctan always rises. A negative answer means an error.
- Defined everywhere: unlike arcsin/arccos, no domain restriction — the denominator 1+x² never vanishes.
Judgment calls
Do not confuse with arcsin: 1/(1+x²) vs 1/√(1−x²) — no root, plus sign. arctan(1/x)? Chain gives −1/(1+x²) — or note arctan(1/x) = π/2 − arctan x (x > 0), whose derivative is −1/(1+x²) directly.
Worked examples
Four problems, easiest first. Plus sign, no root.
Example 1 — d/dx [arctan(3x)]
- Layers: outer arctan u, inner u = 3x.
- Outside: 1/(1+(3x)²) = 1/(1+9x²).
- Inside’s derivative: 3. Multiply: = 3/(1+9x²).
Your turn: d/dx [arctan(2x)]
Answer: 2/(1+4x²)
Layers: outer arctan u, inner u = 2x. Outside: 1/(1+(2x)²) = 1/(1+4x²). Inside’s derivative: 2. Multiply: 2/(1+4x²).
Example 2 — d/dx [x·arctan x] (product)
- Structure: multiplied — product rule: f = x, g = arctan x.
- Derivatives: f′ = 1, g′ = 1/(1+x²).
- Assemble: = arctan x + x/(1+x²).
Your turn: d/dx [x²·arctan x] (product)
Answer: 2x arctan x + x²/(1+x²)
Product rule: f = x², g = arctan x. f′ = 2x, g′ = 1/(1+x²). Assemble: 2x arctan x + x²/(1+x²).
Example 3 — d/dx [arctan(1/x)] (the flip)
- Chain: = d/dx [1/x] / (1+(1/x)²) = (−1/x²) / ((x²+1)/x²).
- Simplify: = −1/(1+x²). (Why negative? arctan(1/x) falls as x grows — check the graph.)
- Cross-check: for x > 0, arctan(1/x) = π/2 − arctan x, whose derivative is −1/(1+x²). Matches.
Your turn: d/dx [arctan(x²)]
Answer: 2x/(1+x⁴)
Chain: d/dx [x²]/(1+(x²)²) = 2x/(1+x⁴). Always positive for x > 0 — arctan always rises.
Example 4 — the 1/2 slope: d/dx [arctan x] at x = 1
- Formula: 1/(1+1²) = 1/2.
- Geometry: at (1, π/4) the curve has visibly moderated from its 45° start — slope 1/2 is plausible.
- Limits: as x → ∞, 1/(1+x²) → 0 — the flattening toward π/2. The formula predicts the asymptote.
Your turn: The decaying slope: d/dx [arctan x] at x = √3
Answer: 1/4
Formula: 1/(1+3) = 1/4. At 0 the slope is 1, at 1 it’s 1/2, at √3 it’s 1/4 — decaying toward the π/2 asymptote, exactly as the graph shows.
Memorization tips
- Plus sign, no root: 1/(1+x²). The cleanest inverse-trig derivative — no sign ambiguity, no domain restriction.
- Flip and chain: tan y = x leads to sec²y times y′ = 1, so y′ = 1/(1+x²). Re-derive live in 30 seconds.
- Always positive: arctan always rises, so 1/(1+x²) is greater than 0 everywhere. A negative answer means an error, guaranteed.
- Steepest at 0: slope 1 at the origin, decaying to 0 at infinity. The formula is the flattening picture.
- arcsin vs arctan: 1/√(1−x²) vs 1/(1+x²). Root+minus vs no-root+plus — the contrasting pair.
- Defined everywhere: 1+x² never vanishes. No |x| < 1 fine print, unlike arcsin/arccos.
Final challenge
Five mixed questions — the sec² collapse, chains, and the 1/2 slope. Score 5/5 and it’s yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the derivative of arctan x?
d/dx[arctan x] = 1/(1+x²), for all real x. Proof: y = arctan x means tan y = x; implicit differentiation gives sec²y·dy/dx = 1, so dy/dx = 1/sec²y = 1/(1+tan²y) = 1/(1+x²).
Why is there no square root or sign issue?
The collapse goes through sec²y = 1+tan²y — no square root appears, so no sign choice. And 1/(1+x²) is manifestly positive, matching arctan’s always-rising graph.
What is d/dx[arctan(3x)]?
3/(1+9x²). Chain rule: 1/(1+(3x)²) times the inside’s derivative 3.
How is it different from arcsin’s derivative?
arcsin′ = 1/√(1−x²) (root, minus inside, domain |x|<1) vs. arctan′ = 1/(1+x²) (no root, plus, all real x). Different inverses, different shapes.
What is d/dx[arctan(1/x)]?
−1/(1+x²). Chain: (−1/x²)/(1+(1/x)²) simplifies to −1/(1+x²). Equivalently, arctan(1/x) = π/2 − arctan x (for x > 0), whose derivative is −1/(1+x²).
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