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)If a limit exists, its value is unique. Hence the notation f′(x) is justified.A4
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Lemma L1
The identity function e(x) = x is differentiable at every x and e′(x) = 1.
Proof.
(1)If h ≠ 0, then [(x + h) − x] · h⁻¹ = h · h⁻¹ = 1A11, A12, A13, A14, A9
(2)lim (h→0) 1 = 1A3
(3)Therefore e′(x) = 1.(1), (2), D1
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Theorem T1
If f and g are differentiable at x, then fg is differentiable at x and (fg)′(x) = f′(x)g(x) + f(x)g′(x).
Proof.
(1)Suppose f and g are differentiable at x.assumption, D1
(2)Since f(x+h)g(x) + (−f(x+h)g(x)) = 0, adding these two terms does not change the value.A13, A14
(3)f(x+h)g(x+h) − f(x)g(x) = [f(x+h)g(x+h) − f(x+h)g(x)] + [f(x+h)g(x) − f(x)g(x)](2), A11, A12
(4) = f(x+h)[g(x+h) − g(x)] + [f(x+h) − f(x)]g(x)(3), A6, A8
(5)If h ≠ 0, multiplying both sides by h⁻¹ gives [f(x+h)g(x+h) − f(x)g(x)]/h = f(x+h) · [g(x+h) − g(x)]/h + [f(x+h) − f(x)]/h · g(x)(4), E, A6, A7, A8
(6)lim (h→0) [g(x+h) − g(x)]/h = g′(x), lim (h→0) [f(x+h) − f(x)]/h = f′(x)(1), D1
(7)lim (h→0) g(x) = g(x)A3
(8)Since f is differentiable at x, lim (h→0) f(x+h) = f(x)(1), A5
(9)Taking limits on both sides of (5) gives lim (h→0) [f(x+h)g(x+h) − f(x)g(x)]/h = f(x)g′(x) + f′(x)g(x)(5), (6), (7), (8), E, A1, A2
(10)Therefore (fg)′(x) = f′(x)g(x) + f(x)g′(x).(9), A12, D1
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Corollary C
(fg)′ = f′g′ does not hold in general.
Proof.
(1)Let f(x) = g(x) = x. Then f′(x) = g′(x) = 1.L1
(2)By T1, (fg)′(x) = 1 · x + x · 1 = x + x = 2x.T1, A6, A8, A10, convention
(3)f′(x)g′(x) = 1 · 1 = 1.(1), A10
(4)At x = 1, (2) gives 2 and (3) gives 1, and 1 ≠ 2.(2), (3), A10, A15
(5)Therefore the two values are not equal.(4)
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