Differentiating Inverse Trigonometric Functions

Start with theta = arcsin(x). That means sin(theta) = x, and you want d(theta)/dx. Differentiate both sides with respect to x: cos(theta) times d(theta)/dx = 1, so d(theta)/dx = 1/cos(theta). The problem is that answer is in terms of theta, not x. You need cos(theta) rewritten in terms of x, and that is where the right triangle comes in: build a triangle where theta is an angle, the opposite side is x, and the hypotenuse is 1. The Pythagorean theorem forces the adjacent side to be √(1-x²), so cos(theta) = √(1-x²)/1. Substitute that back in and you get d/dx[arcsin(x)] = 1/√(1-x²) — pure algebra, no theta, no trig symbols left.

The same triangle method builds every inverse trig derivative: d/dx[arctan(x)] = 1/(1+x²), d/dx[arcsin(x)] = 1/√(1-x²), and so on. Once you have memorized these base formulas, every problem in this unit is a chain rule problem: treat the inside function as u, apply the formula, then multiply by du/dx. For arctan(3x), u = 3x, so the derivative is 1/(1+(3x)²) times 3, which simplifies to 3/(1+9x²).

This topic sits right after implicit differentiation because the triangle trick is really implicit differentiation in disguise — you differentiate an equation that defines theta implicitly, then solve for d(theta)/dx. It also previews later units: inverse trig derivatives show up constantly in integration, since 1/√(1-x²) and 1/(1+x²) are antiderivative patterns you will need to recognize instantly.