Differentiating Inverse Trigonometric Functions

Inverse trig functions answer backwards questions: arcsin(x) asks what angle has sine x. To find the slope of that answer, you don't invert sine's derivative directly, you draw a right triangle. Say theta = arctan(x). Then tan(theta) = x, so picture a right triangle with opposite side x and adjacent side 1, making the hypotenuse √(1+x²). Since theta is defined by this triangle, you can find cos(theta) = 1/√(1+x²) and use implicit differentiation on tan(theta) = x: sec²(theta) times dtheta/dx = 1, so dtheta/dx = cos²(theta) = 1/(1+x²). The same triangle trick on arcsin and arccos produces square roots, because those ratios involve the hypotenuse directly, while tangent's ratio of two legs cancels the square root away.

Once you have the three base formulas, derivative of arcsin(x) is 1/√(1-x²), arccos(x) is -1/√(1-x²), and arctan(x) is 1/(1+x²), everything else is chain rule bookkeeping. For y = arctan(3x) + arcsin(x²), treat 3x and x² as the inner function u, differentiate the outer inverse trig formula with x replaced by u, then multiply by du/dx. That gives 3/(1+9x²) for the first term and 2x/√(1-x⁴) for the second.

This topic sits next to implicit differentiation because that's exactly how these formulas get derived, and it previews why domain restrictions matter so much for inverse functions in general. The square root in arcsin's derivative also reconnects to vertical asymptotes and undefined slopes from early limits work: wherever the expression under a square root hits zero, the tangent line goes vertical instead of just steep.