Derivative of arcsin(x): A Fully Explained Tutorial

A complete step-by-step proof of $d/dx(\arcsin x) = 1/\sqrt{1-x^2}$, with every subtle point explained.

Main Result (click to collapse)
$$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \quad -1 \lt x \lt 1$$

Background: What Is arcsin(x)?

Before diving into the proof, it is essential to understand what arcsin actually is — and what its inputs and outputs represent.

arcsin as the Inverse of Sine

The sine function takes an angle (in radians) and returns a ratio (a pure number between −1 and 1). For example:

$$\sin\!\left(\frac{\pi}{6}\right) = 0.5$$

The arcsin function does the exact reverse: it takes a ratio as input and returns an angle in radians as output. For example:

$$\arcsin(0.5) = \frac{\pi}{6}$$

Domain and Range

Property Value What It Means
Domain (valid inputs) $-1 \leq x \leq 1$ x is a sine ratio — never outside this range
Range (outputs) $y \in \left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$ y is an angle in radians

What Units Does x Have?

x does not contain radian values. This is a common source of confusion.

  • x (the input) is a dimensionless ratio, like 0.5 or −0.3. It represents something like the ratio of opposite to hypotenuse in a right triangle.
  • y (the output) is an angle measured in radians.

The Graph of y = arcsin(x)

When you plot arcsin(x):

Restricting sin(x) to Make arcsin Possible

The sine function (gray, dashed) is periodic and fails the horizontal line test — many different angles produce the same output. The highlighted segment [−π/2, π/2] (blue, solid) is the principal range: the one portion where sin is strictly increasing and therefore invertible. This is the segment arcsin is built from.

Why Does arcsin Need a Restricted Range?

The sine function is not one-to-one across its full domain — it produces the same output at infinitely many different angles. For example:

$$\sin\!\left(\frac{\pi}{6}\right) = \sin\!\left(\frac{5\pi}{6}\right) = 0.5$$

For arcsin to be a valid function, it must return a single unique angle for each input. The principal range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ is the standard restriction used because it is centered around 0 and covers all possible sine output values exactly once.

This restriction is not just a convention — it is what makes the proof work, as you will see in Step 6.

The Proof, Step by Step

Step 1 — Introduce a Variable

$$\text{Let } y = \arcsin x$$

What is happening: We give the output of arcsin a name, $y$, so we can work with it algebraically. At this point, $y$ is an angle in radians satisfying $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, and $x$ is the ratio input satisfying $-1 \leq x \leq 1$.

Step 2 — Rewrite Using the Definition of Inverse

$$\sin y = x$$

What is happening: By definition, if $y = \arcsin x$, then $\sin y = x$. This just restates the inverse relationship: arcsin "undoes" sine.

Why this is useful: We have now converted a statement about arcsin into a statement about the ordinary sine function, which is much easier to differentiate.

Step 3 — Differentiate Both Sides Implicitly

$$\frac{d}{dx}(\sin y) = \frac{d}{dx}(x) \quad\Longrightarrow\quad \cos(y)\,\frac{dy}{dx} = 1$$

What is happening: We differentiate both sides of $\sin y = x$ with respect to $x$.

  • The right side is trivial: $\dfrac{d}{dx}(x) = 1$.
  • The left side requires the Chain Rule, because $y$ is itself a function of $x$.

The Chain Rule in detail:

$$\frac{d}{dx}(\sin y) = \cos(y) \cdot \frac{dy}{dx}$$

Think of it as: "derivative of the outside function (cos y), times the derivative of the inside function (dy/dx)."

This is called implicit differentiation — we differentiate an equation that defines $y$ implicitly in terms of $x$, rather than having an explicit formula like $y = \ldots$

Step 4 — Solve for dy/dx

$$\cos(y)\,\frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{\cos(y)}$$

What is happening: Simple algebra — divide both sides by $\cos(y)$.

The subtle condition: This division is only valid when $\cos(y) \neq 0$. Cosine equals zero at $y = \pm\dfrac{\pi}{2}$, which correspond to the endpoints $x = \pm 1$ of the domain. This is why the proof states $-1 \lt x \lt 1$ (strict inequality) — the derivative does not exist at the endpoints.

We now have $\dfrac{dy}{dx}$ expressed in terms of $y$. The goal of the remaining steps is to convert this into an expression in terms of $x$.

Step 5 — Apply the Pythagorean Identity

$$\cos^2 y + \sin^2 y = 1 \quad\Longrightarrow\quad \cos y = \pm\sqrt{1 - \sin^2 y}$$

What is happening: The fundamental Pythagorean identity from trigonometry is rearranged to express $\cos y$ in terms of $\sin y$.

The ± problem: Taking the square root of both sides introduces a sign ambiguity. We get two possible answers:

$$\cos y = +\sqrt{1 - \sin^2 y} \qquad \text{or} \qquad \cos y = -\sqrt{1 - \sin^2 y}$$

We must determine which sign is correct. This is handled in Step 6 — the most subtle step in the proof.

cos(y) is Non-Negative on the Principal Range

Over the principal range [−π/2, π/2], cos(y) is always ≥ 0 — it touches zero only at the endpoints and reaches its maximum of 1 at y = 0. This is why we choose the positive square root in the next step: the negative root never applies within this range.

Step 6 — Choose the Correct Sign (The Critical Step)

$$\text{Since } y = \arcsin x \text{ implies } y \in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right], \text{ we take } \cos y = \sqrt{1 - \sin^2 y}$$

This step is the most subtle in the entire proof. Here is why the positive root is the right choice.

The Unit Circle Argument

Recall that on the unit circle, $\cos y$ equals the x-coordinate of the point at angle $y$. Now examine the principal range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$:

  • Angles from $0$ to $\dfrac{\pi}{2}$ lie in the 1st quadrant → x-coordinates are positive → $\cos y \gt 0$
  • Angles from $-\dfrac{\pi}{2}$ to $0$ lie in the 4th quadrant → x-coordinates are positive → $\cos y \gt 0$
  • At the endpoints $y = \pm\dfrac{\pi}{2}$: $\cos\!\left(\pm\dfrac{\pi}{2}\right) = 0$

Conclusion: Throughout the entire principal range of arcsin, $\cos y \geq 0$. It is never negative. Therefore, we take the positive square root.

If arcsin used a different principal range — for example, $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$ — then $y$ would fall in the 2nd and 3rd quadrants, where cosine is negative. The negative root would be required, and the final derivative would have the wrong sign. The proof works cleanly only because the chosen principal range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ guarantees $\cos y \geq 0$ throughout.

Step 6 is essentially saying:

"I know which sign to pick because the definition of arcsin has already told me exactly where $y$ lives on the unit circle — and in that region, cosine is always non-negative."

Step 7 — Substitute sin y = x

$$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$

What is happening: We use the fact established in Step 2 that $\sin y = x$. Substituting this into the expression for $\cos y$ converts everything into terms of $x$ — the original variable.

Reminder: $x$ here is just a plain dimensionless number between −1 and 1. There are no radian units in $x$ itself.

Step 8 — Substitute Back and Conclude

$$\frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1-x^2}}$$
$$\boxed{\frac{d}{dx}\big(\arcsin x\big) = \frac{1}{\sqrt{1-x^2}}}$$

What is happening: We substitute the result of Step 7 into the expression for $\dfrac{dy}{dx}$ found in Step 4. The proof is complete.

Summary of Key Subtle Points

Step Subtle Point Why It Matters
Step 2 Rewriting arcsin as sin Converts the hard problem into an easier one
Step 3 Chain Rule on $\sin y$ $y$ is a function of $x$, so the chain rule is required
Step 4 Division by $\cos y$ Only valid at interior points; explains why $x = \pm 1$ are excluded
Step 5 ± ambiguity from square root Sets up the critical sign decision
Step 6 Principal range forces $\cos y \geq 0$ Justifies taking the positive root; the most subtle step
Step 7 $\sin y = x$ replacement Bridges $y$-expressions back to $x$-expressions

A Note on the Domain Restriction

The final formula $\dfrac{1}{\sqrt{1-x^2}}$ becomes undefined when $x = \pm 1$ because the square root equals zero and division by zero is undefined. This is consistent with the geometry: at the endpoints of its graph, arcsin has a vertical tangent, meaning the slope is infinite — not a finite derivative.