A worked derivation of f′(x) = a·n·xⁿ⁻¹ directly from the limit definition of the derivative for an integer exponent, showing exactly where each term in the formula comes from.
Why Deriving the Power Rule Once Makes It More Than a Memorized Formula
This site's Derivative & Limit Evaluator applies the power rule f'(x) = a·n·xⁿ⁻¹ directly, as most practical calculus work does — but working through where that formula actually comes from, starting from the limit definition of the derivative, turns it from an arbitrary rule to memorize into a formula whose every term has a clear, traceable origin. This derivation is shown here for a positive integer exponent n, the case that generalizes most transparently using basic algebra.
Starting From the Limit Definition
Begin with the formal limit definition applied to f(x) = a·xⁿ: f'(x) = lim(h→0) [a·(x+h)ⁿ − a·xⁿ] / h. The coefficient a factors out of both terms in the numerator immediately, leaving f'(x) = a · lim(h→0) [(x+h)ⁿ − xⁿ] / h — so the derivation reduces to working out what happens to [(x+h)ⁿ − xⁿ] / h as h approaches zero, with the coefficient a simply carried along unchanged throughout.
Expanding (x+h)ⁿ Using the Binomial Theorem
The binomial theorem expands (x+h)ⁿ as a sum: xⁿ + n·xⁿ⁻¹·h + [n(n−1)/2]·xⁿ⁻²·h² + ... + hⁿ, where every term after the first two carries a factor of h raised to the second power or higher. This expansion is the key algebraic tool that makes the rest of the derivation possible — it breaks (x+h)ⁿ into a piece that exactly matches xⁿ (which will cancel), a piece proportional to h (which will become the answer), and a collection of higher-order terms in h (which will vanish in the limit).
Why the xⁿ Term Cancels Immediately
Substituting this binomial expansion into the numerator [(x+h)ⁿ − xⁿ] gives [xⁿ + n·xⁿ⁻¹·h + (higher-order h terms) − xⁿ], and the leading xⁿ term cancels exactly against the subtracted xⁿ — this cancellation is precisely why the difference quotient does not simply blow up or reduce to zero; it is engineered by the subtraction in the original derivative definition to leave behind exactly the terms that actually depend on h.
Dividing by h and Isolating the Terms That Survive the Limit
After the xⁿ terms cancel, the numerator is n·xⁿ⁻¹·h plus terms containing h² and higher powers. Dividing the entire expression by h gives n·xⁿ⁻¹ plus terms that still contain at least one remaining positive power of h (since dividing hᵏ by h for k ≥ 2 still leaves at least h¹ remaining). This is the critical structural point of the whole derivation: every remaining term after this division still has an h factor attached, while the n·xⁿ⁻¹ term does not.
Why Every Remaining Term Vanishes as h Approaches Zero
As h approaches zero, any term that still contains a factor of h (to any positive power) approaches zero as well, since it is being multiplied by a quantity shrinking toward zero. The n·xⁿ⁻¹ term, having no remaining h factor at all after the division, is unaffected by this limiting process and survives unchanged. Taking the limit as h → 0 therefore leaves exactly n·xⁿ⁻¹ as the entire result of the bracketed limit.
Reassembling the Full Power Rule
Recombining this result with the coefficient a that was factored out at the very start gives f'(x) = a · n·xⁿ⁻¹ — precisely the power rule formula this site's calculator applies directly. Every piece of the familiar formula now has a clear origin: the a carries through unchanged from the original coefficient, the n comes from the binomial expansion's second term, and the exponent reduction from n to n−1 comes from exactly one factor of xⁿ⁻¹ remaining after the algebraic cancellation and division worked out above.