Definition D1
f is differentiable at x if the limit lim (h→0) [f(x+h) − f(x)]/h exists, and its value is written f′(x).
Proof.
(1)a − b means a + (−b), and F/h means F · h⁻¹. The difference quotient is defined only for h ≠ 0.convention
(2)Setting t = x + h, the same limit may also be written lim (t→x) [f(t) − f(x)]/(t − x).convention
(3)If a limit exists, its value is unique. Hence the notation f′(x) is justified.A2
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Definition D2
φ is continuous at u₀ if lim (u→u₀) φ(u) = φ(u₀).
Proof.
In ε-δ terms: for every ε > 0 there is η > 0 such that |u − u₀| < η implies |φ(u) − φ(u₀)| < ε. The case u = u₀ is included — there the left side is 0.A3, convention
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Definition D3
Set u₀ = f(x), and define φ(u) = [g(u) − g(u₀)]/(u − u₀) for u ≠ u₀, φ(u₀) = g′(u₀).
Proof.
(1)There may be h for which Δu = f(x+h) − f(x) equals 0, and then [g(u) − g(u₀)]/(u − u₀) is undefined.convention
(2)φ is the function obtained by assigning the value g′(u₀) at that one point, extending the domain to include u₀.(1), convention
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Lemma L1
If g is differentiable at u₀, then φ is continuous at u₀.
Proof.
(1)For u ≠ u₀, φ(u) = [g(u) − g(u₀)]/(u − u₀).D3
(2)Since g is differentiable at u₀, lim (u→u₀) [g(u) − g(u₀)]/(u − u₀) = g′(u₀)assumption, D1
(3)The limit considers only u with 0 < |u − u₀|. Hence the value in (2) does not depend on how φ(u₀) is defined.A3
(4)lim (u→u₀) φ(u) = g′(u₀) = φ(u₀)(1), (2), (3), D3
(5)Therefore φ is continuous at u₀.(4), D2
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Lemma L2
If lim (h→0) w(h) = u₀ and φ is continuous at u₀, then lim (h→0) φ(w(h)) = φ(u₀).
Proof.
(1)Let ε > 0.assumption
(2)Since φ is continuous at u₀, there is η > 0 such that |u − u₀| < η implies |φ(u) − φ(u₀)| < ε.(1), assumption, D2
(3)Since lim w = u₀, there is δ > 0 such that 0 < |h| < δ implies |w(h) − u₀| < η.(2), assumption, A3
(4)Therefore 0 < |h| < δ implies |φ(w(h)) − φ(u₀)| < ε.(2), (3)
(5)Line (2) holds also for h with w(h) = u₀. With a mere limit instead of continuity, such h could not be handled.(2), D2
(6)Therefore lim (h→0) φ(w(h)) = φ(u₀).(4), (5), A3
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Lemma L3
For every h, g(f(x+h)) − g(f(x)) = φ(f(x+h)) · [f(x+h) − f(x)].
Proof.
(1)Case f(x+h) ≠ u₀: φ(f(x+h)) = [g(f(x+h)) − g(u₀)] · [f(x+h) − u₀]⁻¹D3
(2)Multiplying both sides by f(x+h) − u₀ gives φ(f(x+h)) · [f(x+h) − f(x)] = g(f(x+h)) − g(f(x))(1), E, A5, A6, A7, A8
(3)Case f(x+h) = u₀: the left side is g(u₀) − g(u₀) = 0.A10
(4)The right side is φ(u₀) · 0 = 0, so the two sides agree.(3), A9, A10
(5)In both cases the equality holds.(2), (4)
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Theorem T1
If f is differentiable at x and g is differentiable at u₀ = f(x), then g∘f is differentiable at x and (g∘f)′(x) = g′(f(x)) · f′(x).
Proof.
(1)Suppose f and g are differentiable at x and u₀ respectively.assumption, D1
(2)If h ≠ 0, then [g(f(x+h)) − g(f(x))]/h = φ(f(x+h)) · [f(x+h) − f(x)]/hL3, E, A5
(3)Since f is differentiable at x, lim (h→0) f(x+h) = f(x) = u₀(1), A4
(4)Applying L1 and L2 to w(h) = f(x+h) gives lim (h→0) φ(f(x+h)) = φ(u₀) = g′(u₀)(3), L1, L2, D3
(5)lim (h→0) [f(x+h) − f(x)]/h = f′(x)(1), D1
(6)Taking limits on both sides of (2) gives lim (h→0) [g(f(x+h)) − g(f(x))]/h = g′(u₀) · f′(x)(2), (4), (5), E, A1
(7)Therefore (g∘f)′(x) = g′(f(x)) · f′(x).(6), D1
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