Worked example · Derivatives
Derivatives of the inverse trigonometric functions
arcsin x has slope 1/√(1 − x²), arctan x has slope 1/(1 + x²). Drag the secant on either graph, then see where the formulas come from.
The derivatives of the three main inverse trigonometric functions are algebraic, with no trigonometry left in them. For |x| < 1, arcsin x has slope 1/√(1 − x²) and arccos x has slope −1/√(1 − x²); for every x, arctan x has slope 1/(1 + x²).
| function | derivative | valid for |
|---|---|---|
| arcsin x | 1/√(1 − x²) | −1 < x < 1 |
| arccos x | −1/√(1 − x²) | −1 < x < 1 |
| arctan x | 1/(1 + x²) | all x |
| arccot x | −1/(1 + x²) | all x |
| arcsec x | 1/(|x|√(x² − 1)) | |x| > 1 |
| arccsc x | −1/(|x|√(x² − 1)) | |x| > 1 |
The first rows are derived in OpenStax §3.7, Derivatives of Inverse Functions[1]. The arcsec and arccsc rows depend on which branch a textbook chooses for the inverse; the version above uses the common choice of angles in [0, π]. The rest of this page derives the first three rows two ways, reads them off the graphs, and gives the chain-rule versions. The same idea, differentiating an inverse function, gives the Derivative of ln x.
arcsin x, by implicit differentiation
Let y = arcsin x, the angle between −π/2 and π/2 whose sine is x. Then sin y = x. Differentiate both sides with respect to x; the left side needs the chain rule:
The last step replaces cos y by √(1 − sin² y) = √(1 − x²). The positive root is the right one because y lies between −π/2 and π/2, where the cosine is not negative. That choice of range is not a technicality: it is what makes arcsin a function at all, and it is why the derivative is positive. The slope of the inverse is the reciprocal of the slope of the original, read at the matching point, the same reflection argument that gives ln x from ex[2].
The same result from a triangle
Draw a right triangle with hypotenuse 1 and one side x, opposite the angle y. Then sin y = x, and Pythagoras gives the adjacent side √(1 − x²), which is cos y. For arctan, use legs 1 and x: tan y = x, the hypotenuse is √(1 + x²), and cos y = 1/√(1 + x²). Implicit differentiation of tan y = x gives sec² y · y′ = 1, so
The triangle is worth remembering because it converts any expression like cos(arcsin x) or sec(arctan x) into algebra, which is the step every one of these derivations needs. The derivative of tan used here, sec² y, is worked out on the Derivative of tan x page.
What the graphs show
The arctan graph rises everywhere and flattens toward its horizontal asymptotes at ±π/2, so its slope is positive, largest in the middle, and fades to 0 far out; 1/(1 + x²) is exactly that bump, equal to 1 at 0 and 1/2 at ±1. The arcsin graph is the sine curve between −π/2 and π/2, reflected in y = x. The sine is flat at the ends of that stretch, so its reflection is vertical there: the slope 1/√(1 − x²) grows without bound as x → ±1. At x = 0.5 it is 1/√0.75 ≈ 1.1547; at 0.99 it is already about 7.09.
With the chain rule
Inside a composite, multiply by the derivative of the inner function[1]. arcsin(2x) has slope 2/√(1 − 4x²), valid for |x| < 1/2. arctan(x²) has slope 2x/(1 + x⁴). arctan(1/x) has slope −1/(1 + x²) for x ≠ 0, the negative of arctan's own slope; that is because arctan x + arctan(1/x) is constant on each side of 0 (π/2 for x > 0, −π/2 for x < 0). The pattern of a sum whose derivative vanishes, as in the prediction above, is a useful check on any of these results.
Why these matter: integrals
Read backwards, the table is a table of antiderivatives, and that is where most students meet these functions again. ∫ dx/(1 + x²) = arctan x + C, so the area under 1/(1 + x²) from 0 to 1 is arctan 1 = π/4 ≈ 0.785398. That integral is a classic test for numerical rules, and the Trapezoidal rule calculator page uses it. Spivak builds the trigonometric functions themselves from integrals like these, defining π through the area of a circle and the inverse functions first[3].