§ 01What Are Inverse Trig Functions?
The inverse trig functions undo what sine, cosine, and tangent do. Their derivatives turn out to involve algebraic (not trig) expressions, which is both surprising and useful.
You already know that sin(π/6) = 1/2. The inverse sine (arcsin) reverses this: arcsin(1/2) = π/6. Informally, arcsin(x) answers the question "which angle has sine equal to x?" The other five inverse trig functions work similarly for cosine, tangent, cosecant, secant, and cotangent.
Their derivatives look algebraic rather than trigonometric. The most important trio are:
d/dx [arctan x] = 1 / (1 + x²)
d/dx [arcsec x] = 1 / (|x| √(x² − 1))
Why do these look so algebraic? Because the derivation involves implicit differentiation plus the Pythagorean identity, which trades trig expressions for square roots. The triangle method below makes this completely transparent.
§ 02The Triangle Method — One System for All Six
Every inverse trig derivative can be derived with the same three-step process. Master this once and you can reproduce any of the six results from scratch in an exam.
Write y = arcsin(x). Rewrite as sin(y) = x, removing the "arc" by taking the trig function of both sides.
Differentiate both sides with respect to x. The left side gives cos(y) · dy/dx; the right side gives 1.
The expression contains cos(y), which involves an unknown angle y. Draw a right triangle labelled with sin(y) = x, then read off cos(y) using Pythagoras.
Replace the trig expression with the Pythagorean result. Solve for dy/dx to get the final algebraic formula.
This method works because every inverse trig function has a corresponding right-triangle picture that converts a trig ratio of y into an algebraic expression in x.
§ 03Deriving d/dx [arcsin x]
We derive the first and most important inverse trig derivative in full detail, showing exactly where the √(1 − x²) comes from.
Set y = arcsin(x). This means sin(y) = x, where y ∈ [−π/2, π/2].
We restrict the range to [−π/2, π/2] so that arcsin is a proper function (one-to-one).Differentiate both sides of sin(y) = x with respect to x:
cos(y) · dy/dx = 1 Left side uses the chain rule: d/dx[sin y] = cos(y) · dy/dx.Solve for dy/dx:
dy/dx = 1 / cos(y)We need cos(y) in terms of x. Draw a right triangle where sin(y) = x/1: so the opposite side is x and the hypotenuse is 1. By Pythagoras, the adjacent side is √(1 − x²).
cos(y) = adjacent / hypotenuse = √(1 − x²) / 1 = √(1 − x²) Since y ∈ [−π/2, π/2], cos(y) ≥ 0, so we take the positive root.Substitute back:
dy/dx = 1 / √(1 − x²)§ 04All Six Inverse Trig Derivatives
The same triangle-method process produces all six results. Note that each derivative comes in a paired "complementary" form: arcsin and arccos, arctan and arccot, arcsec and arccsc: each pair shares an absolute value and differs only in sign.
| Function | Derivative d/dx [f(x)] | Domain | Key triangle side |
|---|---|---|---|
| arcsin x | 1 / √(1 − x²) | −1 < x < 1 | adj = √(1−x²), hyp = 1 |
| arccos x | −1 / √(1 − x²) | −1 < x < 1 | Same triangle; opp sign |
| arctan x | 1 / (1 + x²) | All real x | adj = 1, opp = x, hyp = √(1+x²) |
| arccot x | −1 / (1 + x²) | All real x | Same triangle; opp sign |
| arcsec x | 1 / (|x| √(x² − 1)) | |x| > 1 | hyp = x, adj = 1, opp = √(x²−1) |
| arccsc x | −1 / (|x| √(x² − 1)) | |x| > 1 | Same triangle; opp sign |
Why arctan x is the most-used formula
Among the six, arctan appears most often in practice, it has no domain restrictions and its derivative 1/(1+x²) is a smooth rational function. Integrals of the form ∫ dx/(1+x²) and ∫ dx/(a²+x²) both give arctan results, making it central to integration techniques.
§ 05With the Chain Rule — Composite Forms
In practice, the argument of an inverse trig function is rarely just x. When it is a composite expression like 3x, x², or (x+1)/2, the chain rule must be applied.
If f(x) = arcsin(g(x)), then by the chain rule:
The "a-version" of arctan is worth memorising: it appears constantly when integrating rational functions by partial fractions, where irreducible quadratics (a² + x²) end up in denominators.
Example: d/dx [arctan(3x²)]
Here g(x) = 3x², so g′(x) = 6x. Applying the chain rule:
Students sometimes write d/dx[arcsin(3x)] = 1/√(1 − 9x²) and stop. The chain rule requires multiplying by the inner derivative 3, giving the correct answer 3/√(1 − 9x²).
§ 06Interactive Graph — Function and Derivative
Hover over the graph to read off exact values at any x. The gold curve is the inverse trig function itself; the teal curve is its derivative.
§ 07Fifteen Worked Exercises
Work through each example yourself before revealing the solution. Use the hint button if you need a nudge in the right direction.
- Set sin(y) = x and differentiate both sides w.r.t. x: cos(y) · dy/dx = 1
- Solve for dy/dx: dy/dx = 1 / cos(y)
- From the triangle with sin(y) = x, hyp = 1: cos(y) = √(1 − x²): dy/dx = 1 / √(1 − x²)
- Set tan(y) = x, differentiate: sec²(y) · dy/dx = 1 ⟹ dy/dx = 1 / sec²(y)
- Use the identity sec²(y) = 1 + tan²(y) = 1 + x²: dy/dx = 1 / (1 + x²)
- Set cos(y) = x, differentiate: −sin(y) · dy/dx = 1 ⟹ dy/dx = −1 / sin(y)
- Triangle for arccos: cos(y) = x, hyp = 1, so opp = √(1 − x²), thus sin(y) = √(1 − x²): dy/dx = −1 / √(1 − x²)
- Outer function: arctan(u) → derivative 1/(1+u²). Inner function: u = 5x → du/dx = 5.
- Chain rule: d/dx [arctan(5x)] = 5 / (1 + (5x)²)
- Simplify: = 5 / (1 + 25x²)
- Outer: arcsin(u) → 1/√(1−u²). Inner: u = 2x → du/dx = 2.
- Chain rule: d/dx [arcsin(2x)] = 2 / √(1 − 4x²)
- Domain restriction: 1 − 4x² > 0 ⟹ |x| < 1/2. = 2 / √(1 − 4x²)
- Outer: arctan(u), inner: u = x², so u′ = 2x.
- Chain rule: dy/dx = 2x / (1 + (x²)²) = 2x / (1 + x⁴)
- dy/dx = 2x / (1 + x⁴)
- Product rule: dy/dx = u′v + uv′ with u = x, v = arcsin(x).
- u′ = 1, v′ = 1/√(1 − x²).
- dy/dx = 1 · arcsin(x) + x · 1/√(1 − x²)
- dy/dx = arcsin(x) + x / √(1 − x²)
- d/dx [arctan(x)] = 1/(1+x²).
- d/dx [arctan(1/x)]: let u = 1/x, u′ = −1/x². = (−1/x²) / (1 + 1/x²) = (−1/x²) / ((x²+1)/x²) = −1/(1+x²)
- Add: dy/dx = 1/(1+x²) + (−1/(1+x²)) = 0
- The derivative is 0, confirming that arctan(x) + arctan(1/x) = π/2 for x > 0 (a constant). dy/dx = 0
- u = √x, u′ = 1/(2√x).
- Chain rule: d/dx [arcsin(√x)] = (1/(2√x)) / √(1 − x)
- Simplify: = 1 / (2√x · √(1 − x)) = 1 / (2√(x(1−x)))
- Product rule: u = eˣ (u′ = eˣ), v = arctan(x) (v′ = 1/(1+x²)).
- dy/dx = eˣ · arctan(x) + eˣ · 1/(1+x²)
- Factor: = eˣ [arctan(x) + 1/(1+x²)]
- Let u = (1+x)/(1−x). By the quotient rule: u′ = [(1)(1−x) − (1+x)(−1)] / (1−x)² = [1−x+1+x] / (1−x)² = 2/(1−x)²
- Chain rule with arctan: dy/dx = u′ / (1 + u²) = [2/(1−x)²] / [1 + (1+x)²/(1−x)²]
- Simplify denominator: 1 + u² = [(1−x)² + (1+x)²]/(1−x)² = [1−2x+x² + 1+2x+x²]/(1−x)² = (2+2x²)/(1−x)² = [2/(1−x)²] · [(1−x)²/(2+2x²)] = 2 / (2+2x²) = 1/(1+x²)
- Remarkably, this equals the derivative of arctan(x) itself: confirming the identity arctan((1+x)/(1−x)) = arctan(x) + π/4. dy/dx = 1 / (1 + x²)
- Quotient rule: u = arcsin(x), v = arccos(x).
- u′ = 1/√(1−x²), v′ = −1/√(1−x²).
- dy/dx = [u′v − uv′] / v² = [(1/√(1−x²))·arccos(x) − arcsin(x)·(−1/√(1−x²))] / [arccos(x)]²
- Factor 1/√(1−x²): = [arccos(x) + arcsin(x)] / [√(1−x²) · (arccos(x))²]
- Using the identity arcsin(x) + arccos(x) = π/2: dy/dx = π / [2√(1−x²) · (arccos x)²]
- Outer: arctan(u) → 1/(1+u²). Inner: u = sin x → u′ = cos x.
- dy/dx = cos(x) / (1 + sin²(x))
- This cannot be simplified further without trigonometric substitution. dy/dx = cos x / (1 + sin²x)
- Point on curve: x = 1, y = arctan(1) = π/4. Point is (1, π/4).
- Slope: dy/dx = 1/(1+x²). At x = 1: m = 1/(1+1) = 1/2.
- Tangent line y − y₀ = m(x − x₀): y − π/4 = (1/2)(x − 1)
- Rearrange: y = x/2 − 1/2 + π/4
- Product rule: u = arcsin(x), v = arccos(x). u′ = 1/√(1−x²), v′ = −1/√(1−x²).
- dy/dx = arccos(x)/√(1−x²) − arcsin(x)/√(1−x²)
- Factor: = [arccos(x) − arcsin(x)] / √(1−x²)
- At x = 0: arccos(0) = π/2, arcsin(0) = 0, √(1−0) = 1. dy/dx |_(x=0) = (π/2 − 0) / 1 = π/2
§ 08Common Mistakes
These are the errors that appear most often in homework and exams. Being aware of them before they happen is half the battle.
arcsin(x) is the inverse function of sin, it returns an angle. It is not 1/sin(x), which is csc(x). These are completely different functions. The notation sin⁻¹(x) is the source of this confusion, which is why arcsin is preferred.
d/dx [arctan(x³)] = 1/(1+x⁶) is wrong. The correct answer is 3x²/(1+x⁶). Every inner function must contribute its derivative as a multiplicative factor.
arccos and arccot have the same denominator as arcsin and arctan respectively, but a negative sign out front. Forgetting the minus sign loses a mark instantly.
arcsin(x) and arccos(x) are only defined for |x| ≤ 1, so their derivatives only exist for |x| < 1. Always state the domain when differentiating these functions, especially when the argument is itself a function of x.
When extracting cos(y) from the triangle for arcsin, cos(y) is positive because y ∈ [−π/2, π/2]. If you mistakenly take the negative root, the sign of the final derivative is wrong. Always check the range of the inverse function to determine the sign of the triangle side.
§ 0910-Question Quiz
Test your understanding of inverse trig derivatives. Select your answer for each question, then press Submit to see your score with full explanations.