# Derivative of √x: Formula, Proof and Examples

> The derivative of √x is 1/(2√x). See the power rule and limit proofs, the chain rule for √u, why x = 0 fails, and examples like √(x² + 1).

Source: https://practicecalculus.com/derivatives/sqrt-x
Updated: 2026-11-06

$$ \frac{d}{dx}\,\sqrt{x} = \frac{1}{2\sqrt{x}} $$

## At a glance

- **Derivative:** $\dfrac{1}{2\sqrt{x}}$ for $x > 0$
- **As a power:** $\dfrac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{-1/2}$
- **Chain rule form:** $\dfrac{d}{dx}\sqrt{u} = \dfrac{u'}{2\sqrt{u}}$
- **At x = 0:** Not differentiable, the tangent is vertical
- **Integral version:** $\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$

## Key takeaways

- The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, valid for $x > 0$.
- Rewrite $\sqrt{x}$ as $x^{1/2}$ and use the power rule: $\frac{1}{2}x^{-1/2}$.
- For a square root of anything, $\frac{d}{dx}\sqrt{u} = \frac{u'}{2\sqrt{u}}$.
- At $x = 0$ the graph has a vertical tangent, so the derivative does not exist there.

The **derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$**, for every $x > 0$. Roots are just fractional powers, so the power rule handles them.

$$\frac{d}{dx}\,\sqrt{x} = \frac{1}{2\sqrt{x}}$$

## Graph of √x and its derivative

```graph
title: 'Graph of y = √x and its derivative y = 1/(2√x)'
caption: 'The square root rises steeply near $0$ and flattens out. Its derivative is large near $0$ and shrinks toward $0$.'
x: [-0.5, 6]
y: [-0.5, 3]
functions:
  - expr: sqrt(x)
    label: y = √x
    domain: [0, 6]
  - expr: 1/(2sqrt(x))
    label: y = 1/(2√x) (derivative)
    dashed: true
    domain: [0.03, 6]
points:
  - [4, 2, '(4, 2)']
  - [4, 0.25, 'slope 1/4']
```

The derivative is always positive, so $\sqrt{x}$ always increases. But it increases more and more slowly, which is why the curve bends down.

## Proof with the power rule

Rewrite the root as a power, $\sqrt{x} = x^{1/2}$, and use the [power rule](/rules/power-rule):

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

## Proof from the limit definition

$$f'(x) = \lim_{h \to 0}\frac{\sqrt{x + h} - \sqrt{x}}{h}$$

Plugging in $h = 0$ gives $\frac{0}{0}$, so multiply top and bottom by the conjugate $\sqrt{x + h} + \sqrt{x}$:

$$\frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{1}{\sqrt{x + h} + \sqrt{x}}$$

Now let $h \to 0$ to get $\frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$.

## Why x = 0 is a problem

At $x = 0$ the formula would divide by zero. The difference quotient there is $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$, which grows without bound as $h \to 0^+$. The tangent line at the origin is vertical, so $\sqrt{x}$ is not [differentiable](/learn/calculus-1/the-derivative/differentiability) at $0$.

## The chain rule version: derivative of √u

$$\frac{d}{dx}\,\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot u' = \frac{u'}{2\sqrt{u}}$$

| Function | Inside $u$ | Derivative |
| --- | --- | --- |
| $\sqrt{3x}$ | $3x$ | $\frac{3}{2\sqrt{3x}}$ |
| $\sqrt{x^2 + 1}$ | $x^2 + 1$ | $\frac{x}{\sqrt{x^2 + 1}}$ |
| $\sqrt{1 - x^2}$ | $1 - x^2$ | $\frac{-x}{\sqrt{1 - x^2}}$ |
| $\sqrt{\sin x}$ | $\sin x$ | $\frac{\cos x}{2\sqrt{\sin x}}$ |

## Worked examples

**Example 1: combine powers first.** Differentiate $x\sqrt{x}$. Since $x\sqrt{x} = x^{3/2}$:

$$\frac{d}{dx}\,x^{3/2} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}$$

**Example 2: a quotient.** $\frac{1}{\sqrt{x}} = x^{-1/2}$, so its derivative is $-\frac{1}{2}x^{-3/2} = -\frac{1}{2x\sqrt{x}}$.

**Example 3: a tangent line.** At $x = 9$, $\sqrt{9} = 3$ and the slope is $\frac{1}{6}$. The tangent is $y = 3 + \frac{1}{6}(x - 9)$. This line is the basis for estimating $\sqrt{10} \approx 3 + \frac{1}{6} \approx 3.167$, a classic [linear approximation](/blog/linear-approximation). The true value is about $3.162$.

## Common mistakes

1. **Forgetting the 2.** The answer is $\frac{1}{2\sqrt{x}}$, not $\frac{1}{\sqrt{x}}$.
2. **Lowering the exponent the wrong way.** $\frac{1}{2} - 1 = -\frac{1}{2}$, not $-\frac{3}{2}$ or $\frac{3}{2}$.
3. **Dropping $u'$.** $\frac{d}{dx}\sqrt{x^2 + 1}$ needs the factor $2x$ on top.
4. **Plugging in $x = 0$.** The formula has no value at $0$. Say the derivative does not exist there instead of writing $\frac{1}{0}$.

The reverse direction is on the [integral of √x](/integrals/sqrt-x) page. The [square root function](/functions/square-root-function) page covers its graph and domain.

## Frequently asked questions

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

It is $\frac{1}{2\sqrt{x}}$. Writing $\sqrt{x} = x^{1/2}$, the power rule gives $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

### Is √x differentiable at 0?

No. The function is defined at $0$, but $\frac{1}{2\sqrt{x}}$ blows up as $x \to 0^+$. The graph has a vertical tangent there.

### What is the derivative of 2√x?

It is $2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$.

### What is the derivative of √(2x)?

By the chain rule, $\frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$.