Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Continuity and derivatives of positive-base real powers

Statement

For a>0a>0, the function xaxx\mapsto a^x is continuous on R\mathbb R and (ax)=axloga.(a^x)'=a^x\log a. For αR\alpha\in\mathbb R, the function xxαx\mapsto x^\alpha is continuous and differentiable on (0,)(0,\infty), with (xα)=αxα1.(x^\alpha)'=\alpha x^{\alpha-1}.

Facts & Assumptions

Given: A positive base aa, a real exponent α\alpha, and x>0x>0.

[L1]

ax=exp(xloga)a^x=\exp(x\log a) and xα=exp(αlogx)x^\alpha=\exp(\alpha\log x) (Real powers for positive bases, with the zero-base positive-exponent convention).

[L2]

log(x)=1/x\log'(x)=1/x on (0,)(0,\infty) and exp(u)=exp(u)\exp'(u)=\exp(u) (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The exponential function is smooth and (exp)=exp(\exp)'=\exp).

Proof

technique · direct
1.1

The chain rule applied to ax=exp(xloga)a^x=\exp(x\log a) gives (ax)=exp(xloga)loga=axloga(a^x)'=\exp(x\log a)\log a=a^x\log a.

L1L2L3
1.2

The chain rule applied to xα=exp(αlogx)x^\alpha=\exp(\alpha\log x) gives (xα)=exp(αlogx)α/x=αxα1(x^\alpha)'=\exp(\alpha\log x)\alpha/x=\alpha x^{\alpha-1}.

L1L2L3
2.1

Both functions are continuous on their stated domains because the displayed derivatives exist there.

step 1.1step 1.2L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources