IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~10 min read

Differentiating Inverse Trigonometric Functions

arcsin, arccos, and arctan look like trig but their derivatives are surprisingly algebraic — square roots and rational expressions, no trig functions in sight. The reason: each derivation uses sin²y + cos²y = 1 to convert a “1/cos y” or “1/sec² y” into an expression in x. All three results are in the formula booklet.

📘 What you need to know

The three formula-booklet derivatives: d/dx[arcsin x] = 1/√(1−x²); d/dx[arccos x] = −1/√(1−x²); d/dx[arctan x] = 1/(1+x²).
arcsin and arccos derivatives are negatives of each other — they differ only by sign. This is the identity arcsin x + arccos x = π/2 differentiated.
Domains matter: arcsin and arccos require −1 ≤ x ≤ 1, and their derivatives are undefined at x = ±1 (vertical tangents). arctan is defined for ALL real x.
Why no trig in the answer: the inverse-function proof uses sin²y + cos²y = 1 to swap cos y for √(1 − x²), or sec² y for (1 + x²).
Chain rule for compound arguments: d/dx[arctan(u)] = (du/dx) / (1 + u²). The inner derivative pulls out as a factor in the numerator.
arctan derivative is bounded: 1/(1+x²) ≤ 1, with max 1 at x = 0. The graph of arctan never gets steeper than slope 1.
Use exact unit-circle values for specific-point questions: arcsin(1/2) = π/6, arccos(√3/2) = π/6, arctan(1) = π/4, etc.
Don’t memorise — quote. All three results are in the formula booklet. The risk is confusing the sign or putting the wrong expression under the square root.

The three derivatives — and the arcsin / arccos identity

Formula booklet — inverse trig derivatives ddx[arcsin x] = 1√(1 − x²) | ddx[arccos x] = −1√(1 − x²) | ddx[arctan x] = 11 + x²

The arcsin and arccos derivatives sum to zero — a curious-looking algebraic fact with a beautiful geometric reason.

The blue curve arcsin x and the orange curve arccos x together cover the strip from −π/2 to π. At every x ∈ [−1, 1], their y-values are symmetric about the dashed purple line y = π/4 — because arcsin x + arccos x = π/2. Differentiating that identity gives arcsin′ + arccos′ = 0, i.e., the derivatives are negatives of each other.

Why no trig in the answer: starting from sin y = x, the chain rule gives cos y · dy/dx = 1, so dy/dx = 1/cos y. The identity cos²y = 1 − sin²y = 1 − x² converts that into 1/√(1−x²) — fully algebraic.

🧭 Recipe — differentiate any inverse trig expression

Identify the base inverse trig function (arcsin, arccos, or arctan) and the argument (just x, linear, or nonlinear).
Quote the formula-booklet derivative: arcsin → 1/√(1−u²); arccos → −1/√(1−u²); arctan → 1/(1+u²), where u is the argument.
Apply chain rule if the argument u isn’t just x: multiply by du/dx. The inner derivative goes in the NUMERATOR.
Use product or quotient rule on top if needed — e.g., for x · arctan x the product rule applies, with d/dx[arctan x] = 1/(1+x²) plugged in.
For a specific point, substitute the x-value AFTER differentiating. Use exact values from the unit circle: arcsin(1/2) = π/6, arccos(√3/2) = π/6, arctan(1) = π/4, etc.

Chain rule for compound arguments

Linear argument (ax + b)

ddx[arctan(ax+b)] = a1 + (ax+b)²

inner derivative a pulls out into the numerator

General argument u(x)

ddx[arcsin u] = du/dx√(1 − u²)

du/dx goes in the numerator — the rest follows from the formula booklet

Worked examples

WE 1

Prove d/dx[arccos x] = −1/√(1 − x²)

By considering y = arccos x on its principal domain, show that ddx[arccos x] = −1√(1 − x²).

Step 1 — rewrite using the inverse relation y = arccos x ⟹ x = cos y, where 0 ≤ y ≤ π Step 2 — differentiate implicitly with respect to x d/dx[x] = d/dx[cos y] 1 = -sin y · (dy/dx) Step 3 — solve for dy/dx dy/dx = -1/sin y Step 4 — convert sin y to an expression in x using sin²y + cos²y = 1 sin²y = 1 – cos²y = 1 – x² on 0 ≤ y ≤ π, sin y ≥ 0, so sin y = +√(1 – x²) Step 5 — substitute dy/dx = -1/√(1 – x²) ✓ ∴ d/dx[arccos x] = -1/√(1 – x²) the “+” on the square root is forced by the principal-range argument: arccos’s range [0, π] is exactly where sin is non-negative, so we never have to worry about the negative root.

WE 2

Chain rule on arctan with a linear argument

Find dydx for y = arctan(3x − 2).

Step 1 — identify the inner function u and outer derivative u = 3x – 2 ⟹ du/dx = 3 d/du[arctan u] = 1/(1 + u²) Step 2 — apply chain rule dy/dx = (d/du[arctan u]) · (du/dx) = 1/(1 + (3x – 2)²) · 3 = 3/(1 + (3x – 2)²) dy/dx = 3/(1 + (3x − 2)²) the inner derivative 3 sits in the NUMERATOR of the answer, not the denominator. Pulling the 3 incorrectly into (1 + 3(3x−2)²) is wrong.

WE 3

Chain rule on arcsin with a NONLINEAR argument

Find dydx for y = arcsin(x²).

Step 1 — identify u and du/dx u = x² ⟹ du/dx = 2x d/du[arcsin u] = 1/√(1 – u²) Step 2 — apply chain rule dy/dx = 1/√(1 – (x²)²) · 2x = 2x / √(1 – x⁴) dy/dx = 2x / √(1 − x⁴) when squaring u = x² inside the square root, (x²)² = x⁴, not x². Watch carefully — composing a power with a power MULTIPLIES the exponents.

WE 4

Product rule with arctan

Find dydx for y = x · arctan x.

Step 1 — identify the two factors u = x, u’ = 1 v = arctan x, v’ = 1/(1 + x²) Step 2 — apply product rule: (uv)’ = u’v + uv’ dy/dx = (1)(arctan x) + (x)(1/(1 + x²)) = arctan x + x/(1 + x²) dy/dx = arctan x + x1 + x² the answer mixes a transcendental term (arctan x) and an algebraic one (x/(1+x²)) — that’s the natural form. Don’t try to “simplify” it further.

WE 5

Gradient at a specific x-value — exact form

Find the exact value of dydx for y = arccos x at the point where x = √3/2.

Step 1 — derivative from formula booklet dy/dx = -1/√(1 – x²) Step 2 — substitute x = √3/2 x² = (√3/2)² = 3/4 1 – x² = 1 – 3/4 = 1/4 √(1 – x²) = √(1/4) = 1/2 Step 3 — final value dy/dx = -1 / (1/2) = -2 dy/dx = −2 the point of tangency is (√3/2, π/6), since arccos(√3/2) = π/6. The negative gradient confirms arccos is decreasing — its derivative is negative everywhere on (−1, 1).

WE 6

Equation of tangent to y = arctan(2x) at x = 1/2

Find the equation of the tangent to the curve y = arctan(2x) at the point where x = 1/2.

Step 1 — find the y-coordinate y at x = 1/2: arctan(2 · 1/2) = arctan(1) = π/4 point of tangency: (1/2, π/4) Step 2 — find dy/dx using chain rule u = 2x, du/dx = 2 dy/dx = 2/(1 + (2x)²) = 2/(1 + 4x²) Step 3 — gradient at x = 1/2 dy/dx |_{x = 1/2} = 2/(1 + 1) = 2/2 = 1 Step 4 — tangent equation y – y₀ = m(x – x₀) y – π/4 = 1 · (x – 1/2) y = x – 1/2 + π/4 tangent: y = x − 1/2 + π/4 (or y = x + (π − 2)/4) keep π/4 as exact — don’t approximate to 0.785. The intercept has both a rational and a transcendental part, which is normal for inverse-trig tangent equations.

💡 Top tips

Quote the formula booklet for the three core derivatives — don’t rely on memory. Check the sign on arccos (it’s negative).
Inner derivative goes in the NUMERATOR for chain rule applied to inverse trig: d/dx[arctan(u)] = u′/(1 + u²), with u′ on top.
For arcsin and arccos, watch the domain restriction: the argument must be in [−1, 1]. arcsin(2x) only makes sense for −1/2 ≤ x ≤ 1/2.
Use exact unit-circle values at common points: arcsin(1/2) = π/6, arccos(1/2) = π/3, arctan(1) = π/4, arctan(√3) = π/3, etc.
Keep answers in exact form on Paper-1 questions: π, √3, etc., not their decimal approximations.

⚠ Common mistakes

Wrong sign on arccos: d/dx[arccos x] = −1/√(1−x²), NOT +1/√(1−x²). The minus sign is essential.
Putting the inner derivative in the wrong place: d/dx[arctan(3x)] = 3/(1+9x²), NOT 1/(3·(1+9x²)). The 3 goes ON TOP.
Squaring inside the square root wrong: in d/dx[arcsin(x²)] = 2x/√(1−x⁴), the (x²)² = x⁴, NOT x² or 2x².
Confusing arcsin′ with sin′: d/dx[arcsin x] = 1/√(1−x²); d/dx[sin x] = cos x. Completely different.
Trying to evaluate at x = ±1 for arcsin or arccos — the derivative is undefined there (vertical tangent). The curve EXISTS at x = ±1, but its slope doesn’t.

Up next: Differentiating Exponential & Logarithmic Functions. The base-e cases d/dx[e^x] = e^x and d/dx[ln x] = 1/x you’ve seen before — now we generalise to ANY base. The derivatives pick up a ln a factor: d/dx[a^x] = a^x ln a, and d/dx[log_a x] = 1/(x ln a). Both proved via implicit differentiation.

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