Derivative of √x: 1/(2√x) with Proof and Examples

Quick answer:

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

The square root is one of the first functions where students have to rewrite something before differentiating, and it is also the first common example of a function that is defined at a point but has no derivative there. This guide covers two proofs, a picture that makes the formula obvious, seven worked examples, the usual mistakes and a set of practice problems.

Why it works: the intuition

Think of \( x \) as the area of a square. Then \( \sqrt{x} \) is the length of its side. Suppose the area grows by a tiny amount \( dx \). How much does the side grow?

If the side is \( s \) and it grows by \( ds \), the square gains two thin strips of size \( s \times ds \) along two edges (plus a tiny corner piece you can ignore). So the new area is about

$$s^2 + 2s\,ds$$

The added area \( 2s\,ds \) must equal \( dx \), so \( \frac{ds}{dx} = \frac{1}{2s} = \frac{1}{2\sqrt{x}} \). That is the derivative, found without any algebra rules at all.

The picture also explains the shape of the graph. A big square has long edges, so a little extra area spreads over a lot of edge and the side barely grows. That is why \( \sqrt{x} \) keeps rising but gets flatter and flatter: the slope is \( \frac12 \) at \( x = 1 \), \( \frac14 \) at \( x = 4 \) and only \( \frac{1}{20} \) at \( x = 100 \).

Method 1: the power rule

Rewrite the root as a power: \( \sqrt{x} = x^{1/2} \). The power rule says bring the exponent down and subtract one:

$$\frac{d}{dx}x^{1/2} = \tfrac12 x^{-1/2} = \frac{1}{2\sqrt{x}}$$

The negative exponent means “one over,” which is why the root ends up in the denominator. This rewrite-then-differentiate habit works for every root and every power of a root.

Method 2: the limit definition

Start from the definition of the derivative:

$$\lim_{h\to0}\frac{\sqrt{x+h} - \sqrt{x}}{h}$$

Plugging in \( h = 0 \) gives \( \frac00 \), so you need to rearrange first. Multiply top and bottom by the conjugate \( \sqrt{x+h} + \sqrt{x} \). The top becomes a difference of squares, and the \( x \) terms cancel:

$$\begin{aligned} &\lim_{h\to0}\frac{(x+h) - x}{h\left(\sqrt{x+h} + \sqrt{x}\right)} \\ &= \lim_{h\to0}\frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt x} \end{aligned}$$

The conjugate trick is worth learning in its own right; it shows up in many limit problems involving roots.

Chain rule: derivative of √(g(x))

When there is an expression under the root, the chain rule multiplies by its derivative:

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

In words: derivative of the inside, over two times the original root.

Example 1. \( \frac{d}{dx}\sqrt{x^2 + 1} = \frac{2x}{2\sqrt{x^2+1}} = \frac{x}{\sqrt{x^2+1}} \). The 2s cancel, which happens often when the inside has an \( x^2 \) term.

Example 2. \( \frac{d}{dx}\sqrt{3x - 2} = \frac{3}{2\sqrt{3x-2}} \). The inside derivative 3 goes on top.

Example 3. \( x\sqrt{x} = x^{3/2} \), so its derivative is \( \frac32 x^{1/2} = \frac{3\sqrt x}{2} \). Combining powers first is faster than the product rule.

Example 4. \( \frac{1}{\sqrt x} = x^{-1/2} \) has derivative \( -\frac12 x^{-3/2} = -\frac{1}{2x^{3/2}} \). Rewriting avoids the quotient rule entirely.

Example 5: a tangent line. Find the tangent line to \( y = \sqrt x \) at \( x = 9 \). The point is \( (9, 3) \) and the slope is \( \frac{1}{2\cdot 3} = \frac16 \), so

$$y = 3 + \frac{1}{6}(x - 9)$$

For more on this, see the equation of a tangent line guide.

Example 6: a quotient. Differentiate \( \frac{\sqrt x}{x + 1} \) with the quotient rule. After combining the terms in the numerator over a common denominator, the result simplifies nicely:

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

The derivative is zero at \( x = 1 \), so that is where the function reaches its maximum value of \( \frac12 \).

Example 7 (exam level): distance. A point moves along the x-axis, and \( D = \sqrt{x^2 + 9} \) is its distance from \( (0, 3) \). How fast does the distance change with \( x \) when \( x = 4 \)? The chain rule gives \( \frac{x}{\sqrt{x^2 + 9}} \), which at \( x = 4 \) is \( \frac{4}{5} \). Distance formulas always contain a square root, so this pattern is everywhere in optimization problems.

Other roots

The same idea works for any root. For the cube root, \( \sqrt[3]{x} = x^{1/3} \):

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

Unlike \( \sqrt{x} \), the cube root is defined for negative numbers too. Its derivative is undefined only at \( x = 0 \), where the graph has a vertical tangent.

What happens at x = 0?

The formula \( \frac{1}{2\sqrt x} \) blows up as \( x \to 0^{+} \). Geometrically, the graph of \( \sqrt x \) starts at the origin with a vertical tangent line, so \( \sqrt x \) is not differentiable at 0 even though it is defined there. In the square picture, a square of zero size has no edge to spread new area over, so its side has to grow infinitely fast at first.

Useful application: approximating square roots

The derivative powers the linear approximation trick. Near \( x = 4 \):

$$\sqrt{4.1} \approx \sqrt4 + \frac{1}{2\sqrt4}(0.1) = 2 + 0.025 = 2.025$$

The true value is about 2.02485, so the estimate is off by only 0.00015. The estimate is slightly too big because the graph bends downward, so the tangent line sits above the curve. You can run the same computation for other numbers with the linear approximation calculator.

Common mistakes

• Dropping the 2. The derivative is \( \frac{1}{2\sqrt x} \), not \( \frac{1}{\sqrt x} \). The \( \frac12 \) comes from the exponent.
• Subtracting wrongly in the exponent. \( \frac12 - 1 = -\frac12 \), not \( -\frac32 \) or \( \frac32 \). Write the subtraction out if in doubt.
• Splitting a root over a sum. \( \sqrt{x^2 + 1} \) is not \( x + 1 \), so you cannot differentiate it as \( 1 \). Use the chain rule on the whole root.
• Forgetting the inside derivative. \( \frac{d}{dx}\sqrt{5x + 1} \) is \( \frac{5}{2\sqrt{5x+1}} \); leaving out the 5 is the most common chain rule slip.
• Plugging in \( x = 0 \). The derivative does not exist there, so a slope of “infinity” or “zero” is not a valid answer.

Where it’s used

Square roots appear wherever distances, speeds and standard sizes appear, so their derivative is everywhere in applied problems:

• Distance and geometry. Every distance formula has a root, so minimizing a distance or finding how fast it changes always involves the chain rule version of this derivative.
• Physics. The period of a simple pendulum grows like the square root of its length, and the time for an object to fall from rest grows like the square root of the height. The derivative tells you how sensitive those times are to small changes in length or height.
• Arc length and surface area. The arc length formula integrates a square root, and simplifying it often starts by differentiating one.
• Estimation. As shown above, a tangent line to \( \sqrt x \) gives fast, accurate mental estimates of roots near perfect squares.

In all of these, the key feature is the same: the root grows quickly for small inputs and slowly for large ones, and \( \frac{1}{2\sqrt x} \) measures exactly how quickly.

Try it yourself

Type any root expression below to see each rule applied, or use the full derivative calculator.

Practice problems

Try these before checking the answers.

1. \( \frac{d}{dx}\sqrt{5x + 1} \)
2. \( \frac{d}{dx}\sqrt{\sin x} \)
3. \( \frac{d}{dx}\,4\sqrt[4]{x} \)
4. \( \frac{d}{dx}\sqrt{1 - x^2} \)
5. \( \frac{d}{dx}\,\sqrt{x}\,(x + 2) \)
6. \( \frac{d}{dx}\sqrt{2x} \)

Answers: (1) \( \frac{5}{2\sqrt{5x+1}} \); (2) \( \frac{\cos x}{2\sqrt{\sin x}} \); (3) \( x^{-3/4} \); (4) \( -\frac{x}{\sqrt{1-x^2}} \); (5) \( \frac{3x + 2}{2\sqrt x} \); (6) \( \frac{1}{\sqrt{2x}} \).

FAQ

Is the derivative of √x equal to 1/√x?

No, it is half of that: \( \frac{1}{2\sqrt x} \). The factor \( \frac12 \) comes from the exponent.

What is the integral of √x?

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

Is √x differentiable at 0?

No. The function is defined at 0, but the graph has a vertical tangent there, so the slope is unbounded and the derivative does not exist.

What is the second derivative of √x?

Differentiate \( \frac12 x^{-1/2} \) to get \( -\frac14 x^{-3/2} = -\frac{1}{4x^{3/2}} \). It is negative, so the graph is concave down for all \( x > 0 \).

What is the derivative of √(2x)?

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

Related: derivative of ln x.

Further reading

• Paul’s Online Notes: Differentiation Formulas — rewriting roots as fractional powers and applying the power rule, with many examples.
• Paul’s Online Notes: The Definition of the Derivative — limit-definition examples, including a square root worked with the conjugate trick.
• Paul’s Online Notes: Linear Approximations — using tangent lines to estimate roots and other values.