AP Calculus AB and BC

Derivative of sqrt(x): Answer, Proof, and Mistakes

Q: What is the derivative of \sqrt{x}?

It is \frac{1}{2\sqrt{x}}, valid for x > 0. Rewrite \sqrt{x} as x^{1/2} and use the power rule: the exponent \frac{1}{2} comes down as a coefficient and drops to -\frac{1}{2}, giving \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.

Q: Why is the 2 in the denominator?

Because the power rule brings the exponent \frac{1}{2} down as a coefficient. That \frac{1}{2} multiplies x^{-1/2} = \frac{1}{\sqrt{x}}, so the product is \frac{1}{2}\cdot\frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}, with both the 2 and the root below the bar.

Q: Is \sqrt{x} differentiable at x = 0?

No. The function is continuous at 0 because \sqrt{x} \to 0 as x \to 0^+, matching its value \sqrt{0} = 0; but the graph has a vertical tangent there, so the slope is undefined. The derivative \frac{1}{2\sqrt{x}} exists only for x > 0.

Q: How do you differentiate \sqrt{g(x)}?

Use the chain rule: \frac{d}{dx}\left[\sqrt{g(x)}\right] = \frac{g'(x)}{2\sqrt{g(x)}}. For example, \frac{d}{dx}\left[\sqrt{3x}\right] = \frac{3}{2\sqrt{3x}} and \frac{d}{dx}\left[\sqrt{x^2 + 1}\right] = \frac{x}{\sqrt{x^2 + 1}}.

The derivative of the square root of x is 1 over 2 times the square root of x, written f'(x) = 1/(2 sqrt(x)). It comes from the power rule: rewrite sqrt(x) as x^(1/2), so the derivative is (1/2)x^(-1/2). The slope is defined for x > 0 and grows without bound as x approaches 0.

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

The proof: it is just the power rule

The square root is a power in disguise. Rewrite $\sqrt{x}$ as $x^{1/2}$ , then apply the power rule $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$ with $n = \frac{1}{2}$ .

\frac{d}{dx}\left[x^{1/2}\right] = \frac{1}{2}\,x^{1/2 - 1} = \frac{1}{2}\,x^{-1/2} = \frac{1}{2}\cdot\frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}

Rewrite before you differentiate

Radicals and variables in a denominator are power-rule problems in disguise. Turn every square root into a fractional exponent, $\sqrt{x} = x^{1/2}$ , and every $\frac{1}{x^k}$ into $x^{-k}$ , before you touch the derivative. Skipping that step is where most coefficient and sign errors begin.

Why the slope blows up at x = 0

The function $\sqrt{x}$ is defined at $x = 0$ , where $\sqrt{0} = 0$ , but its derivative is not. As $x$ shrinks toward $0$ , $\sqrt{x}$ shrinks with it, so $\frac{1}{2\sqrt{x}}$ grows without bound.

\lim_{x \to 0^+} \frac{1}{2\sqrt{x}} = +\infty

Geometrically, the graph leaves the origin with a vertical tangent line, and a vertical line has no finite slope. So $\sqrt{x}$ is continuous at $0$ but not differentiable there. The domain of $f'(x) = \frac{1}{2\sqrt{x}}$ is $x > 0$ , one endpoint narrower than the domain $x \ge 0$ of $\sqrt{x}$ itself.

Where √x shows up on the AP exam

You rarely differentiate a bare $\sqrt{x}$ on the exam. The plain case is a one-line power-rule check, but square roots earn their points inside larger expressions, where they force a chain rule or hide in a distance formula.

Context	What you do	CED location
A bare $\sqrt{x}$ term	Power rule with $n = \frac{1}{2}$	Unit 2, Topic 2.5 (Applying the Power Rule)
A composite $\sqrt{g(x)}$	Chain rule: outside times inside	Unit 3, Topic 3.1 (The Chain Rule)
Distance $\sqrt{x^2 + y^2}$ in related rates	Chain rule while differentiating with respect to time	Unit 4, Topics 4.4 to 4.5
Arc length integrand, $\sqrt{1 + [f'(x)]^2}$ (BC)	Set up the integral, then evaluate	Unit 8, Topic 8.13

The differentiation units are heavily weighted. Unit 2 (fundamental derivative rules) is worth 10 to 15 percent of the AB exam and 5 to 10 percent of BC; Unit 3 (chain rule and composites) is worth 5 to 10 percent on both. Knowing $\frac{d}{dx}\left[\sqrt{x}\right]$ cold frees time for the harder composites built on top of it.

Common mistakes

Misplacing the coefficient $\frac{1}{2}$ . The derivative is $\frac{1}{2\sqrt{x}}$ , not $2\sqrt{x}$ or $\frac{\sqrt{x}}{2}$ . The coefficient $\frac{1}{2}$ multiplies $x^{-1/2}$ , so both the $2$ and the root land in the denominator, never in the numerator.
Forgetting to subtract $1$ from the exponent. Differentiating $x^{1/2}$ gives $\frac{1}{2}x^{-1/2}$ , not $\frac{1}{2}x^{1/2}$ . The new exponent is $\frac{1}{2} - 1 = -\frac{1}{2}$ .
Dropping the chain rule. For $\sqrt{g(x)}$ the answer is $\frac{g'(x)}{2\sqrt{g(x)}}$ ; writing $\frac{1}{2\sqrt{g(x)}}$ alone leaves out the inside derivative $g'(x)$ .
Reporting a slope at $x = 0$ . There is none, since the tangent is vertical, so $f'(0)$ does not exist even though $\sqrt{0} = 0$ .
Confusing the derivative with the integral. $\frac{d}{dx}\left[\sqrt{x}\right] = \frac{1}{2\sqrt{x}}$ , while $\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$ . Adding $1$ to the exponent is integration; subtracting $1$ is differentiation.

Two chain-rule composites

Every composite $\sqrt{g(x)}$ reuses the same outside derivative, $\frac{1}{2\sqrt{g(x)}}$ , multiplied by the inside derivative $g'(x)$ .

Take $\frac{d}{dx}\left[\sqrt{x^2 + 1}\right]$ . Here $g(x) = x^2 + 1$ and $g'(x) = 2x$ .

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

Take $\frac{d}{dx}\left[\sqrt{\sin x}\right]$ . Here $g(x) = \sin x$ and $g'(x) = \cos x$ .

\frac{d}{dx}\left[\sqrt{\sin x}\right] = \frac{1}{2\sqrt{\sin x}}\cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}}

The pattern never changes: differentiate the root, then multiply by the derivative of whatever sits under it.

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 $\sqrt{x}$ ?

It is $\frac{1}{2\sqrt{x}}$ , valid for $x > 0$ . Rewrite $\sqrt{x}$ as $x^{1/2}$ and use the power rule: the exponent $\frac{1}{2}$ comes down as a coefficient and drops to $-\frac{1}{2}$ , giving $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ .

Why is the $2$ in the denominator?

Because the power rule brings the exponent $\frac{1}{2}$ down as a coefficient. That $\frac{1}{2}$ multiplies $x^{-1/2} = \frac{1}{\sqrt{x}}$ , so the product is $\frac{1}{2}\cdot\frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}$ , with both the $2$ and the root below the bar.

Is $\sqrt{x}$ differentiable at $x = 0$ ?

No. The function is continuous at $0$ because $\sqrt{x} \to 0$ as $x \to 0^+$ , matching its value $\sqrt{0} = 0$ ; but the graph has a vertical tangent there, so the slope is undefined. The derivative $\frac{1}{2\sqrt{x}}$ exists only for $x > 0$ .

How do you differentiate $\sqrt{g(x)}$ ?

Use the chain rule: $\frac{d}{dx}\left[\sqrt{g(x)}\right] = \frac{g'(x)}{2\sqrt{g(x)}}$ . For example, $\frac{d}{dx}\left[\sqrt{3x}\right] = \frac{3}{2\sqrt{3x}}$ and $\frac{d}{dx}\left[\sqrt{x^2 + 1}\right] = \frac{x}{\sqrt{x^2 + 1}}$ .