Inverse Trigonometric Derivatives

Inverse trigonometric functions — arcsin, arccos, arctan, arcsec, arccsc, and arccot — are the "reverse" operations of the standard trig functions. Where sin(π/6) = 1/2, the inverse gives arcsin(1/2) = π/6. Because these functions appear constantly in physics, engineering, and geometry (describing angles from ratios), knowing how to differentiate them is an essential calculus skill. Deriving the derivatives of inverse trig functions uses a clever chain-rule argument: start with an implicit relationship like y = arcsin(x), rewrite it as sin(y) = x, differentiate both sides with respect to x using implicit differentiation, and then use the Pythagorean identity to express the result purely in terms of x. The answers always involve radical expressions such as √(1 − x²), reflecting the geometric constraint that angles must stay in specific ranges. The six standard formulas you will master in this topic are compact and elegant. Once memorized and understood, they combine with the chain rule to allow differentiation of complicated composite expressions like arctan(3x²) or arcsin(e^x), opening up techniques for integration (inverse-trig antiderivatives) later in the course.

Key takeaways

Implicit Differentiation of arcsin(x)
Deriving arccos(x)
Deriving arctan(x)
Summary of All Six Formulas
Applying the Chain Rule

Implicit Differentiation of arcsin(x)

Let y = arcsin(x), which means sin(y) = x with y ∈ [−π/2, π/2]. Differentiating both sides with respect to x: cos(y) · (dy/dx) = 1, so dy/dx = 1/cos(y). Since sin²(y) + cos²(y) = 1 and cos(y) ≥ 0 on the principal branch, cos(y) = √(1 − sin²(y)) = √(1 − x²). Therefore d/dx[arcsin(x)] = 1/√(1 − x²).

Deriving arccos(x)

A near-identical argument applies to y = arccos(x), meaning cos(y) = x with y ∈ [0, π]. Differentiating: −sin(y) · (dy/dx) = 1, so dy/dx = −1/sin(y). On this branch sin(y) ≥ 0, so sin(y) = √(1 − cos²(y)) = √(1 − x²). Thus d/dx[arccos(x)] = −1/√(1 − x²). Note it is the negative of the arcsin derivative — this makes sense because arcsin(x) + arccos(x) = π/2 (a constant), so their derivatives must sum to zero.

Deriving arctan(x)

Let y = arctan(x), so tan(y) = x with y ∈ (−π/2, π/2). Differentiating implicitly: sec²(y) · (dy/dx) = 1. Using the identity sec²(y) = 1 + tan²(y) = 1 + x², we get dy/dx = 1/(1 + x²). This is one of the most frequently used results in calculus. It has no square root and is defined for all real x.

Summary of All Six Formulas

The complete list of inverse trig derivatives is: d/dx[arcsin(x)] = 1/√(1−x²), d/dx[arccos(x)] = −1/√(1−x²), d/dx[arctan(x)] = 1/(1+x²), d/dx[arccot(x)] = −1/(1+x²), d/dx[arcsec(x)] = 1/(|x|√(x²−1)), d/dx[arccsc(x)] = −1/(|x|√(x²−1)). The arcsec and arccsc formulas use |x| because the principal branch requires x > 1 or x < −1, and the sign of x affects the sign of the derivative without the absolute value.

Applying the Chain Rule

In practice, the argument of an inverse trig function is often a function u(x) rather than just x. The chain rule gives d/dx[arctan(u)] = (1/(1+u²)) · (du/dx). For example, d/dx[arctan(3x²)] = 1/(1+(3x²)²) · 6x = 6x/(1+9x⁴). Always identify the inner function u, differentiate the outer inverse-trig function using the standard formula evaluated at u, then multiply by du/dx.

The Key Equation

\frac{d}{dx}[\arctan(u)] = \frac{1}{1+u^{2}}\cdot\frac{du}{dx}

u — Inner function (dimensionless): Any differentiable function of x that serves as the argument of arctan.
\frac{du}{dx} — Derivative of inner function (dimensionless): The rate of change of u with respect to x, found by standard differentiation rules.
1+u^{2} — Denominator term (dimensionless): Comes from the Pythagorean identity sec²(y) = 1 + tan²(y) used in the implicit differentiation derivation.

Worked Example

Problem. Find dy/dx if y = arcsin(4x³) − 5 arctan(√x).

\frac{dy}{dx} = \frac{12x^{2}}{\sqrt{1-16x^{6}}} - \frac{5}{2\sqrt{x}\,(1+x)}

Answer. dy/dx = 12x²/√(1−16x⁶) − 5/(2√x · (1+x))

Figure

Diagram

Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered

Gottfried Wilhelm Leibniz

1646–1716 · German

Leibniz independently co-invented calculus alongside Newton, introducing the dy/dx notation used universally today. He systematically developed rules for differentiating transcendental functions, laying the algebraic framework that makes formulas like d/dx[arctan(x)] = 1/(1+x²) expressible and manipulable. His notation made the chain rule and implicit differentiation natural operations, which are the exact tools used to derive every inverse-trig derivative formula. Leibniz was also a diplomat, philosopher, and inventor of a mechanical calculator — he reportedly designed the machine so it could multiply and divide, not just add, making it more powerful than Pascal's earlier device. He conducted all this work while holding full-time positions at royal courts.

Implicit Differentiation — Every inverse trig derivative is derived by implicit differentiation — it is the core technique that transforms sin(y)=x into dy/dx = 1/√(1−x²).
The Chain Rule — When inverse trig functions have composite arguments like arctan(3x²), the chain rule is mandatory to multiply by the inner derivative.
Pythagorean Trigonometric Identities — The identity sin²θ + cos²θ = 1 and its relatives are used algebraically in every single derivation to eliminate y and express results in terms of x.
Inverse Trig Antiderivatives — Reading the derivative formulas in reverse gives integration formulas: ∫1/√(1−x²)dx = arcsin(x)+C and ∫1/(1+x²)dx = arctan(x)+C, used extensively in integral calculus.
Logarithmic Differentiation — Like inverse trig differentiation, logarithmic differentiation is another advanced technique that uses implicit methods to differentiate otherwise unwieldy functions.