AP Calculus AB and BC

Derivative of x arcsin x: Answer, Proof, Mistakes

Q: What is the derivative of x arcsin x?

It is \arcsin x + \frac{x}{\sqrt{1 - x^2}}, from the product rule.

Q: Why does the derivative fail at x = 1 and x = -1?

Because \sqrt{1 - x^2} = 0 there, so the second term divides by zero. The function itself has values at those points; only the tangent line is vertical.

Q: What is the integral of arcsin x?

It is x\arcsin x + \sqrt{1 - x^2} + C, by integration by parts. Differentiating it returns \arcsin x, because the two terms with \sqrt{1 - x^2} cancel.

The derivative of x arcsin x with respect to x is arcsin x + x/sqrt(1 - x^2). The product rule differentiates x to 1 and arcsin x to 1/sqrt(1 - x^2), so both terms stay in the answer. The formula holds on the open interval from -1 to 1, since the square root in the denominator is zero at both endpoints.

\frac{d}{dx}\left[x\arcsin x\right] = \arcsin x+\frac{x}{\sqrt{1-x^2}}

The proof: product rule on x arcsin x

Set $f(x) = x$ and $g(x) = \arcsin x$ . Then $f'(x) = 1$ , and $g'(x) = \frac{1}{\sqrt{1 - x^2}}$ is the inverse sine rule from Topic 3.4.

\frac{d}{dx}\left(x\arcsin x\right) = 1\cdot\arcsin x + x\cdot\frac{1}{\sqrt{1 - x^2}} = \arcsin x + \frac{x}{\sqrt{1 - x^2}}

Nothing combines, so the finished derivative mixes an inverse trig term with an algebraic one. That mixture is normal for products of this shape and is not a sign of an error. It also means the problem sits in two units at once: the product rule is Topic 2.8, while the derivative of $\arcsin x$ is not taught until Topic 3.4.

Domain: the derivative stops short of the endpoints

$\arcsin x$ is defined on the closed interval $[-1, 1]$ , but the derivative divides by $\sqrt{1 - x^2}$ , which is zero at $x = 1$ and $x = -1$ . So the derivative exists only on the open interval $-1 < x < 1$ .

Near the ends the slope runs off: as $x \to 1^{-}$ the derivative goes to $+\infty$ , and as $x \to -1^{+}$ it goes to $-\infty$ . The graph reaches $x = \pm 1$ with vertical tangents.

Both factors are odd, so $x\arcsin x$ is even. Its only critical point is $x = 0$ , where the derivative is $\arcsin 0 + 0 = 0$ , and that point is the minimum, with value 0.

Common mistakes

Using $\frac{1}{1 + x^2}$ , the $\arctan$ derivative, instead of $\frac{1}{\sqrt{1 - x^2}}$ .
Dropping the $\arcsin x$ term and answering $\frac{x}{\sqrt{1 - x^2}}$ .
Splitting the root and writing $\frac{x}{1 - x}$ . The radical covers the whole expression $1 - x^2$ , so it cannot be pulled apart.
Evaluating at a point such as $x = 2$ . Outside $[-1, 1]$ neither $x\arcsin x$ nor its derivative exists.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of x arcsin x?

It is $\arcsin x + \frac{x}{\sqrt{1 - x^2}}$ , from the product rule.

Why does the derivative fail at x = 1 and x = -1?

Because $\sqrt{1 - x^2} = 0$ there, so the second term divides by zero. The function itself has values at those points; only the tangent line is vertical.

What is the integral of arcsin x?

It is $x\arcsin x + \sqrt{1 - x^2} + C$ , by integration by parts. Differentiating it returns $\arcsin x$ , because the two terms with $\sqrt{1 - x^2}$ cancel.