Derivatives of Inverse Trigonometric Functions

Core Idea

The derivatives of the inverse trig functions produce algebraic expressions: d/dx[arcsin(x)] = 1/sqrt(1 - x²), d/dx[arccos(x)] = -1/sqrt(1 - x²), d/dx[arctan(x)] = 1/(1 + x²). These are derived using implicit differentiation (e.g., if y = arcsin(x), then sin(y) = x, differentiate implicitly). These derivatives appear frequently as results of integration, making them important both for differentiation and for recognizing integral forms.

How It's Best Learned

Derive each using implicit differentiation and Pythagorean identities. Practice with chain rule applications: d/dx[arctan(3x)], d/dx[arcsin(x²)]. Recognize the integral forms: integral of 1/(1 + x²) dx = arctan(x) + C.

Common Misconceptions

• Confusing the derivative of arcsin with the derivative of arccos (they differ by a sign).
• Forgetting the chain rule when the argument is not simply x.
• Not recognizing 1/(1 + x²) as the derivative of arctan in integration problems.

Explainer

To find the derivative of arcsin(x), you don't need a new rule — you need implicit differentiation, which you already know. Let y = arcsin(x). By definition of the inverse, this means sin(y) = x, where y ∈ [-π/2, π/2]. Differentiate both sides with respect to x using the chain rule: cos(y) · (dy/dx) = 1. Solve for dy/dx: dy/dx = 1/cos(y). The result should be in terms of x, not y, so substitute back. From the Pythagorean identity, cos²(y) = 1 - sin²(y) = 1 - x², and since y ∈ [-π/2, π/2] where cosine is non-negative, cos(y) = √(1 - x²). Therefore d/dx[arcsin(x)] = 1/√(1 - x²). The entire derivation is just implicit differentiation plus a Pythagorean identity substitution.

The same strategy yields the other derivatives. For arctan: let y = arctan(x), so tan(y) = x. Differentiating implicitly: sec²(y) · dy/dx = 1, so dy/dx = 1/sec²(y). Using the identity sec²(y) = 1 + tan²(y) = 1 + x², we get d/dx[arctan(x)] = 1/(1 + x²). For arccos, the derivation mirrors arcsin, but the derivative of cosine introduces a minus sign: d/dx[arccos(x)] = -1/√(1 - x²). Notice that arcsin(x) + arccos(x) = π/2 (a constant), so their derivatives must sum to zero — the two formulas are negatives of each other, which serves as a self-consistency check.

These algebraic-looking results are surprising at first — why does differentiating a trigonometric function produce something involving square roots? The answer is that the inverse functions "undo" the original trig functions, and the algebraic expressions encode the geometry of those functions' slopes at each point. But the key practical payoff is in integration: encountering 1/√(1 - x²) in an integral should immediately trigger recognition of arcsin. The antiderivatives ∫1/√(1 - x²)dx = arcsin(x) + C and ∫1/(1 + x²)dx = arctan(x) + C are among the most commonly occurring results in calculus, and they appear frequently as the answers to trigonometric substitution problems.

When the argument is a composite function, the chain rule applies as always. For d/dx[arctan(3x)]: the outer derivative formula gives 1/(1 + (3x)²) and the inner derivative of 3x is 3, so the answer is 3/(1 + 9x²). For d/dx[arcsin(x²)]: outer derivative is 1/√(1 - (x²)²) = 1/√(1 - x⁴), inner derivative is 2x, so the answer is 2x/√(1 - x⁴). The common error is applying the formula correctly but forgetting to multiply by the inner derivative — always check whether the argument is just x or something more complex.