Derivative of √x

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

Differentiate again: \frac{d^2}{dx^2}\left[\sqrt{x}\right] = -\frac{1}{4 x^{\frac{3}{2}}}.

Plug into the derivative: f'(3.6) = 0.263523. That is the slope of the tangent line there; drag the graph to see it.

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

Write √x as x^(1/2), then apply the power rule. The slope becomes infinite as x → 0⁺, where the curve starts out vertical.

Power rule
d/dx[xⁿ] = n·xⁿ⁻¹.
$\frac{d}{dx}\left[\sqrt{x}\right] = \frac{1}{2 \sqrt{x}}$

Drag the graph · at x = 3.91 the slope is 0.25286, the height of the dashed curve.

Questions students ask

What is the derivative of √x?

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

What is the second derivative of √x?

Differentiate again: $\frac{d^2}{dx^2}\left[\sqrt{x}\right] = -\frac{1}{4 x^{\frac{3}{2}}}$ .

What is the slope of √x at x = 3.6?

Plug into the derivative: $f'(3.6) = 0.263523$ . That is the slope of the tangent line there; drag the graph to see it.

How do you find the derivative of √x?

Write √x as x^(1/2), then apply the power rule. The slope becomes infinite as x → 0⁺, where the curve starts out vertical.

What is the integral of √x?

See the integral of √x page for the antiderivative with steps.