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Why is the derivative of x² equal to 2x?

A short, rigorous derivation from the limit definition, with a geometric interpretation.

The derivative of $x^2$ is $2x$ because its difference quotient simplifies to $2x+h$ , which tends to $2x$ as $h\to0$ .

From the definition

Start with $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$ For $f(x)=x^2$ , $\frac{(x+h)^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h.$ Letting $h$ tend to zero gives $f'(x)=2x$ .

What it means

The parabola’s slope grows linearly with $x$ . At $x=3$ the tangent slope is $6$ ; at $x=-2$ it is $-4$ ; at the vertex $x=0$ it is zero.

Quick check

Between $x=3$ and $x=3.01$ , $x^2$ rises from $9$ to $9.0601$ . The ratio $0.0601/0.01=6.01$ is already close to $2\cdot3=6$ .

Frequently asked questions

Do I need the power rule?

No. The proof uses only the derivative definition and expansion of a squared binomial.

Is the derivative of x² always positive?

No. It is 2x: negative for x<0, zero at x=0, and positive for x>0.