Derivative of arcsin(x)
The derivative of the inverse sine is $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}$, valid for $-1 < x < 1$. At the endpoints $x=\pm1$ the derivative does not exist: the graph has vertical tangents there.
See the derivative of arcsin(x) step by step in the calculator →
📈 Graph of arcsin(x) and its derivative
🔍 Why is the derivative of arcsin(x) equal to $\frac{1}{\sqrt{1-x^2}}$?
Let $y = \arcsin x$, so $\sin y = x$ and $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Differentiating implicitly gives $\cos y \cdot y' = 1$, hence
We may take the positive square root because $\cos y \ge 0$ for $y$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ – the range of arcsin.
🔗 Derivative of arcsin(x) with an inner function
If the argument is a function $u(x)$, the chain rule adds the inner derivative:
✏️ Worked examples
| Function | Derivative | Note | |
|---|---|---|---|
| $\arcsin(2x)$ | $\frac{2}{\sqrt{1-4x^2}}$ | Valid for $|x| < \frac12$. | steps → |
| $\arcsin\left(\frac{x}{3}\right)$ | $\frac{1}{\sqrt{9-x^2}}$ | In general $\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{a^2-x^2}}$ for $a>0$. | steps → |
| $\arcsin(\sqrt{x})$ | $\frac{1}{2\sqrt{x}\sqrt{1-x}}$ | Inner derivative $\frac{1}{2\sqrt x}$, and $1-(\sqrt x)^2 = 1-x$. | steps → |
| $x\arcsin x$ | $\arcsin x + \frac{x}{\sqrt{1-x^2}}$ | Product rule. | steps → |
⚠️ Common mistakes
Forgetting the square root: the derivative is $\frac{1}{\sqrt{1-x^2}}$, not $\frac{1}{1-x^2}$.
Confusing $\sin^{-1}x$ with $\csc x = \frac{1}{\sin x}$.
❓ Frequently asked questions
What is the derivative of sin^-1(x)?
$\sin^{-1}x$ means $\arcsin x$; its derivative is $\frac{1}{\sqrt{1-x^2}}$.
Where is arcsin(x) not differentiable?
At $x = 1$ and $x = -1$, the ends of its domain, where the tangent is vertical.
What is the second derivative of arcsin(x)?
$\frac{x}{(1-x^2)^{3/2}}$.