Bring the exponent down as a factor, then reduce the exponent by one.
\[\frac{d}{dx}x^{n} = n x^{n-1}\]
What it says
For a whole number \(n\), the binomial theorem expands \((x + h)^{n}\) as \(x^{n} + n x^{n-1}h + \left(\text{terms with } h^{2} \text{ and higher}\right)\). Subtracting \(x^{n}\), dividing by \(h\) and letting \(h \to 0\) kills every term but \(n x^{n-1}\).
The rule is true for every real exponent, not only positive integers — negative, fractional and irrational exponents all obey it. That makes it the workhorse for roots and reciprocals, once they are rewritten as powers: \(\sqrt{x} = x^{1/2}\) and \(\frac{1}{x^{3}} = x^{-3}\).
The pattern is worth memorising in words rather than symbols: the old exponent becomes the coefficient, and the new exponent is one less.
When it applies
The base is the variable and the exponent is a constant.
Roots and reciprocals, after rewriting them with exponents.
Together with the chain rule when the base is a function: \(\left(u^{n}\right)' = n u^{n-1}u'\).
Five worked examples
Every line is the step the calculator would show, in the order it applies them. Each graph
is live: hover it to read both curves and see the tangent whose slope is the derivative,
drag to pan, scroll to zoom.
1
Example 1
\[\frac{d}{dx}\left[x^{5}\right]\]
The exponent \(5\) comes down in front and drops to \(4\).
\[\frac{d}{dx}\left[x^{5}\right] = 5x^{5-1}\]
Tidy the exponent.
\[= 5x^{4}\]
Answer
\[5 x^{4}\]
f is odd and very steep away from the origin. f′ is even and never dips below the axis, so f increases everywhere, touching zero only at \(x = 0\). That flat point is not a turning point — the curve carries on rising through it.
The plainest case: a positive whole-number exponent.
Both curves are cut off at the left, since neither exists for negative x. f′ plunges from a great height toward the axis, dropping below f near \(x = 0.5\).
Notice the derivative blows up as \(x \to 0^{+}\): the square-root graph is vertical at the origin.