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), F/h means F · h⁻¹, and g² means g · g.convention
(2)If a limit exists, its value is unique. Hence the notation f′(x) is justified.A5
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Lemma L1
If g is differentiable at x and g(x) ≠ 0, then there exists δ > 0 such that g(x+h) ≠ 0 for all h with 0 < |h| < δ.
Proof.
(1)Since g is differentiable at x, lim (h→0) g(x+h) = g(x)assumption, A7
(2)Since g(x) ≠ 0, we have |g(x)| > 0.assumption, A9
(3)Take ε = |g(x)|. Then there is δ > 0 such that 0 < |h| < δ implies |g(x+h) − g(x)| < |g(x)|.(1), (2), A6
(4)If g(x+h) = 0, then |g(x+h) − g(x)| = |−g(x)| = |g(x)|, contradicting (3).(3), A9, A15
(5)Therefore 0 < |h| < δ implies g(x+h) ≠ 0.(4)
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Lemma L2
If g is differentiable at x and g(x) ≠ 0, then 1/g is differentiable at x and (1/g)′(x) = −g′(x)/g(x)².
Proof.
(1)Suppose g is differentiable at x and g(x) ≠ 0.assumption, D1
(2)If 0 < |h| < δ, then g(x+h) ≠ 0, and hence g(x+h)g(x) ≠ 0.L1, A17
(3)For such h, 1/g(x+h) − 1/g(x) = [g(x) − g(x+h)] · [g(x+h)g(x)]⁻¹ = −[g(x+h) − g(x)] · [g(x+h)g(x)]⁻¹(2), A18, A13, A11, A12, A14, A10, A16
(4)Multiplying both sides by h⁻¹ gives [1/g(x+h) − 1/g(x)]/h = −{[g(x+h) − g(x)]/h} · [g(x+h)g(x)]⁻¹(3), E, A11, A12, A16
(5)lim (h→0) [g(x+h) − g(x)]/h = g′(x)(1), D1
(6)lim (h→0) g(x+h) = g(x)(1), A7
(7)lim (h→0) [g(x+h)g(x)]⁻¹ = 1/[g(x)g(x)] = 1/g(x)² (the limit of the denominator is not 0)(6), A1, A4, A2
(8)Taking limits on both sides of (4) gives lim (h→0) [1/g(x+h) − 1/g(x)]/h = −g′(x) · 1/g(x)²(4), (5), (7), E, A1, A3
(9)Therefore (1/g)′(x) = −g′(x)/g(x)².(8), D1
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Theorem T1
If f and g are differentiable at x and g(x) ≠ 0, then f/g is differentiable at x and (f/g)′(x) = [f′(x)g(x) − f(x)g′(x)] / g(x)².
Proof.
(1)f/g means f · (1/g).convention
(2)By L2, 1/g is differentiable at x and (1/g)′ = −g′/g².L2
(3)Applying the product rule to f and 1/g gives (f · 1/g)′ = f′ · (1/g) + f · (1/g)′(1), (2), A8
(4) = f′ · g⁻¹ + f · (−g′ · (g²)⁻¹)(3), (2)
(5)f′ · g⁻¹ = f′ · g · g⁻¹ · g⁻¹ = f′g · (g²)⁻¹A13, A11, A12, A14, A18
(6)Therefore (f/g)′ = [f′g − fg′] · (g²)⁻¹ = (f′g − fg′)/g².(4), (5), A10, A11, A12, A16
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