# Square Root Function √x: Graph, Domain, Derivative and Integral

> The square root function f(x) = √x: its domain and graph, the derivative 1/(2√x), the integral (2/3)x^(3/2) + C, and the vertical tangent at 0.

Source: https://practicecalculus.com/functions/square-root-function
Updated: 2026-11-03

$$ f(x) = \sqrt{x} = x^{1/2} $$

## At a glance

- **Domain:** $[0, \infty)$
- **Range:** $[0, \infty)$
- **Derivative:** $\frac{1}{2\sqrt{x}}$
- **Integral:** $\frac{2}{3}x^{3/2} + C$
- **Not differentiable at:** $x = 0$ (vertical tangent)
- **Inverse:** $x^2$ for $x \ge 0$

## Key takeaways

- The square root function $\sqrt{x} = x^{1/2}$ is defined only for $x \ge 0$ and returns the nonnegative root.
- Its derivative is $\frac{1}{2\sqrt{x}}$, which blows up at $x = 0$, so the graph has a vertical tangent there.
- Its integral is $\frac{2}{3}x^{3/2} + C$, from the reverse power rule.
- Rewrite roots as fractional powers before differentiating or integrating.

The **square root function** $f(x) = \sqrt{x}$ returns the nonnegative number whose square is $x$. It is the inverse of $x^2$ on $x \ge 0$, the power function with exponent $\frac{1}{2}$, and the standard example of a function that is continuous at a point but not differentiable there.

## Formula

$$f(x) = \sqrt{x} = x^{1/2}, \qquad x \ge 0$$

By convention $\sqrt{x}$ means the **principal** (nonnegative) root, so $\sqrt{9} = 3$, not $\pm 3$. As a consequence, $\sqrt{x^2} = |x|$ for every real $x$.

## Graph of the square root function

```graph
title: 'Graph of y = √x and its derivative y = 1/(2√x)'
caption: 'The square root rises quickly near $0$ and then flattens. Its derivative is huge near $0$ and shrinks toward $0$ as $x$ grows.'
x: [-0.5, 9]
y: [-0.5, 4]
functions:
  - expr: sqrt(x)
    label: y = √x
  - expr: 1/(2sqrt(x))
    label: y = 1/(2√x) (derivative)
    dashed: true
points:
  - [0, 0, '(0, 0)']
  - [4, 2, '(4, 2)']
```

## Key properties

- **Domain and range:** both $[0, \infty)$.
- **Increasing** on its whole domain, since its derivative is positive for $x > 0$.
- **Concave down:** $f''(x) = -\frac{1}{4}x^{-3/2} < 0$ for $x > 0$, so it grows ever more slowly.
- **Continuous at $0$** from the right, with $\lim_{x \to 0^+}\sqrt{x} = 0$.
- **Not differentiable at $0$**, where the tangent is vertical.
- **Unbounded but slow:** $\sqrt{x} \to \infty$ as $x \to \infty$, slower than any positive power $x^p$ with $p > \frac{1}{2}$.
- **Reflection of a parabola:** the graph is the right half of $y = x^2$ reflected over the line $y = x$.

## Derivative of √x

Using the [power rule](/rules/power-rule) on $x^{1/2}$:

$$\frac{d}{dx}\sqrt{x} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}, \qquad x > 0$$

With the [chain rule](/rules/chain-rule) this becomes the general pattern

$$\frac{d}{dx}\sqrt{g(x)} = \frac{g'(x)}{2\sqrt{g(x)}}$$

## Integral of √x

$$\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$$

Add one to the exponent ($\frac{1}{2} + 1 = \frac{3}{2}$) and divide by $\frac{3}{2}$, which is the same as multiplying by $\frac{2}{3}$.

## Worked examples

**Example 1: tangent line.** At $x = 4$, $f(4) = 2$ and $f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4}$. The tangent line is $y = 2 + \frac{1}{4}(x - 4) = \frac{1}{4}x + 1$.

**Example 2: linear approximation.** Using that tangent line, $\sqrt{4.2} \approx \frac{1}{4}(4.2) + 1 = 2.05$. The true value is about $2.0494$. See [linear approximation](/blog/linear-approximation).

**Example 3: chain rule.** $\frac{d}{dx}\sqrt{3x + 1} = \frac{3}{2\sqrt{3x + 1}}$.

**Example 4: definite integral.** $\int_0^4 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^4 = \frac{2}{3}(8) = \frac{16}{3}$.

**Example 5: domain.** $\sqrt{4 - x^2}$ requires $4 - x^2 \ge 0$, so the domain is $[-2, 2]$. Its graph is the upper half of the circle $x^2 + y^2 = 4$.

## Common mistakes

1. **Splitting a root over a sum.** $\sqrt{x^2 + 9} \ne x + 3$. Roots split over products, $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a, b \ge 0$, but never over sums.
2. **Writing the derivative as $\frac{1}{\sqrt{x}}$ or $2\sqrt{x}$.** The factor is $\frac{1}{2}$ and the root stays in the denominator.
3. **Integrating to $\frac{3}{2}x^{3/2}$.** Divide by $\frac{3}{2}$; do not multiply.
4. **Assuming $\sqrt{x^2} = x$.** It equals $|x|$, which matters when $x$ can be negative.

## How it connects

Square roots appear in arc length, distance formulas and [trig substitution](/blog/trig-substitution) integrals like $\int \sqrt{1 - x^2}\,dx$. They are one member of the [power functions](/functions/power-functions), and $\sqrt{x^2} = |x|$ links them to the [absolute value function](/functions/absolute-value-function).

## Frequently asked questions

### What is the derivative of the square root of x?

Write $\sqrt{x} = x^{1/2}$ and apply the power rule: $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. It is defined for $x > 0$.

### What is the integral of the square root of x?

$\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$.

### Is the square root function differentiable at 0?

No. The derivative $\frac{1}{2\sqrt{x}}$ grows without bound as $x \to 0^+$, so the graph has a vertical tangent at the origin and no finite slope there.

### What is the domain of the square root function?

The domain is $x \ge 0$, because a negative number has no real square root. For $\sqrt{g(x)}$, solve $g(x) \ge 0$.

### How do you differentiate the square root of a function?

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