Derivative Of Sin Inverse X: The Proof Your Teacher Skipped

Last Updated: Written by Dr. Carolina Mello Dias
derivative of sin inverse x the proof your teacher skipped
derivative of sin inverse x the proof your teacher skipped
Table of Contents

Derivative of sin inverse x Made Clear (No Implicit Confusion)

The derivative of arcsin(x) with respect to x is $$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$$ for all x in the domain $$-1 < x < 1$$. At the endpoints $$x = -1$$ and $$x = 1$$, the derivative does not exist due to vertical tangents, while for |x| > 1, arcsin(x) is not real, so the derivative is not defined in the real-number sense. This crisp result is independent of any interpretation and holds under standard calculus rules.

To illuminate the result, consider the geometric intuition: arcsin(x) is the inverse of sin(y) restricted to y in $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. Small changes in x near a given x produce changes in y that scale by the reciprocal of the cosine of y, which, by the Pythagorean identity, yields the $$\frac{1}{\sqrt{1 - x^2}}$$ factor. This aligns with the chain rule applied to the identity $$\sin(y) = x$$ and $$y = \arcsin(x)$$.

Key derivation steps

Assume x = $$\sin(y)$$ with $$y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ so that $$y = \arcsin(x)$$. Differentiating both sides with respect to x gives:

  1. dx/dy = $$\cos(y)$$.
  2. Thus dy/dx = 1 / $$\cos(y)$$ = 1 / $$\sqrt{1 - \sin^2(y)}$$ = 1 / $$\sqrt{1 - x^2}$$.
  3. Hence $$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$$ for -1 < x < 1.

This sequence makes explicit why the derivative becomes unbounded as x approaches ±1, matching the well-known behavior of arcsin near the endpoints of its real-domain interval.

derivative of sin inverse x the proof your teacher skipped
derivative of sin inverse x the proof your teacher skipped

Mathematical table of values

x arcsin(x) derivative
0 0 1
0.5 $$\frac{\pi}{6}$$ $$\frac{1}{\sqrt{0.75}}$$ ≈ 1.155
-0.5 -$$\frac{\pi}{6}$$ $$\frac{1}{\sqrt{0.75}}$$ ≈ 1.155
0.9 $$\arcsin(0.9)$$ ≈ 1.1198 1 / $$\sqrt{1 - 0.81}$$ = 1 / $$\sqrt{0.19}$$ ≈ 2.294

Common questions (FAQ)

Note: This article presents the derivative in a way aligned with rigorous calculus and practical education leadership, reflecting our Marist Education Authority commitment to clarity, reliability, and actionable insights for school leadership, teachers, and policymakers across Latin America.

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Dr. Carolina Mello Dias

Dr. Carolina Mello Dias holds a Ph.D. in Education Leadership from the University of São Paulo, with a concentration in Catholic and Marist pedagogy.

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