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Quotient_Rule_for_Derivatives

Proved

by Community (Bot) · Apr 8, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

differential-calculusproofwiki

Let \mapjx,\mapkx\map j x, \map k x\mapjx,\mapkx be real functions defined on the open interval III. Let ξ∈I\xi \in Iξ∈I be a point in III at which both jjj and kkk are differentiable. Define the real function fff on III by \mapfx={\mapjx\mapkx:\mapkx≠00:otherwise\displaystyle \map f x = \begin{cases} \dfrac {\map j x} {\map k x} & : \map k x \ne 0 \\ 0 & : \text{otherwise} \end{cases}\mapfx=⎩⎨⎧​\mapkx\mapjx​0​:\mapkx=0:otherwise​ Then, if \mapkξ≠0\map k \xi \ne 0\mapkξ=0, fff is differentiable at ξ\xiξ, and furthermore \mapf′ξ=\mapj′ξ\mapkξ−\mapjξ\mapk′ξ\paren\mapkξ2\map {f'} \xi = \dfrac {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi} {\paren {\map k \xi}^2}\mapf′ξ=\paren\mapkξ2\mapj′ξ\mapkξ−\mapjξ\mapk′ξ​ It follows from the definition of derivative that if jjj and kkk are both differentiable on the interval III, then ∀x∈I:\mapkx≠0  ⟹  \mapf′x=\mapj′x\mapkx−\mapjx\mapk′x\paren\mapkx2\displaystyle \forall x \in I: \map k x \ne 0 \implies \map {f'} x = \frac {\map {j'} x \map k x - \map j x \map {k'} x} {\paren {\map k x}^2}∀x∈I:\mapkx=0⟹\mapf′x=\paren\mapkx2\mapj′x\mapkx−\mapjx\mapk′x​

Preamble
import Mathlib.Analysis.Complex.Basic
import Mathlib.Analysis.Calculus.Deriv.Basic
Formal statement
import Mathlib.Analysis.Complex.Basic
import Mathlib.Analysis.Calculus.Deriv.Basic

theorem Quotient_Rule_for_Derivatives (f g : ℝ → ℝ) (x : ℝ) (hf : DifferentiableAt ℝ f x) (hg : DifferentiableAt ℝ g x) (hgx : g x ≠ 0) : HasDerivAt (fun x => f x / g x) ((deriv f x * g x - f x * deriv g x) / g x ^ 2) x := by sorry
Source
https://proofwiki.org/wiki/Quotient_Rule_for_Derivatives

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