AP Calculus AB and BC

Derivative of sqrt(x^2+1): Answer, Proof, Mistakes

Q: What is the derivative of sqrt(x^2+1)?

It is \frac{x}{\sqrt{x^2+1}}. Write it as (x^2+1)^{1/2} and apply the chain rule: \frac{1}{2}(x^2+1)^{-1/2}\cdot 2x simplifies to \frac{x}{\sqrt{x^2+1}}.

Q: Where is the derivative zero?

Only at x = 0. The numerator x is zero there while the denominator \sqrt{x^2+1} = 1 is nonzero, so the curve has a minimum at x = 0.

Q: Is the derivative ever undefined?

No. Since x^2 + 1 \geq 1, the denominator \sqrt{x^2+1} is never zero, so \frac{x}{\sqrt{x^2+1}} is defined for every real x.

The derivative of sqrt(x^2+1) with respect to x is x/sqrt(x^2+1). Write the root as (x^2+1)^(1/2) and use the chain rule: the power rule gives (1/2)(x^2+1)^(-1/2), and multiplying by the inner derivative 2x leaves x/sqrt(x^2+1). The slope is 0 at x = 0 and approaches 1 for large positive x.

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

The proof: chain rule on (x^2+1)^(1/2)

Rewrite the square root as a power so the power rule applies: $\sqrt{x^2+1} = (x^2+1)^{1/2}$ . The inner function is $u = x^2 + 1$ with $u' = 2x$ .

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

The $\frac{1}{2}$ from the power rule and the $2$ from the inner derivative cancel, leaving a clean $\frac{x}{\sqrt{x^2+1}}$ . Notice $x^2 + 1 \geq 1 > 0$ for all $x$ , so the derivative is defined everywhere, with no domain gaps.

What the slope tells you

The sign of $\frac{x}{\sqrt{x^2+1}}$ is the sign of $x$ , since the denominator is always positive. So the curve falls for $x < 0$ , has a minimum at $x = 0$ , and rises for $x > 0$ .

For large $|x|$ , $\sqrt{x^2+1} \approx |x|$ , so the slope approaches $1$ as $x \to \infty$ and $-1$ as $x \to -\infty$ . The graph straightens toward the slant asymptotes $y = \pm x$ of the hyperbola.

Common mistakes

Answering $\frac{1}{2\sqrt{x^2+1}}$ and forgetting the inner derivative $2x$ . Without the chain rule factor the $x$ in the numerator never appears.
Differentiating the square root as if the inside were just $x$ , giving $\frac{1}{2\sqrt{x}}$ -style answers. The inside is $x^2 + 1$ , so its derivative $2x$ belongs in the numerator.
Botching the exponent: from $(x^2+1)^{1/2}$ the new exponent is $-\frac{1}{2}$ , not $\frac{1}{2}$ or $-\frac{3}{2}$ .
Splitting the root as $\sqrt{x^2} + \sqrt{1}$ . Square roots do not distribute over addition, so $\sqrt{x^2+1} \neq x + 1$ .

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 sqrt(x^2+1)?

It is $\frac{x}{\sqrt{x^2+1}}$ . Write it as $(x^2+1)^{1/2}$ and apply the chain rule: $\frac{1}{2}(x^2+1)^{-1/2}\cdot 2x$ simplifies to $\frac{x}{\sqrt{x^2+1}}$ .

Where is the derivative zero?

Only at $x = 0$ . The numerator $x$ is zero there while the denominator $\sqrt{x^2+1} = 1$ is nonzero, so the curve has a minimum at $x = 0$ .

Is the derivative ever undefined?

No. Since $x^2 + 1 \geq 1$ , the denominator $\sqrt{x^2+1}$ is never zero, so $\frac{x}{\sqrt{x^2+1}}$ is defined for every real $x$ .