Definition D1
f is differentiable at x means that 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 the limit exists, its value is unique. Therefore the notation f′(x) is justified.A2
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Definition D2
The h → 0− limit of the difference quotient is called the left derivative f′−(x), and the h → 0+ limit the right derivative f′+(x).
Proof.
Each is the limit obtained by sending h to 0 from one side only. It means two things together: that the one-sided limit exists, and what its value is.convention
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Theorem T1
f is differentiable at x if and only if f′−(x) and f′+(x) both exist and are equal, in which case f′(x) is their common value.
Proof.
(1)For h ≠ 0, let Q(h) = [f(x+h) − f(x)]/h.convention, D1
(2)f being differentiable at x means that lim (h→0) Q(h) exists.(1), D1
(3)That limit exists if and only if the left and right limits both exist and are equal.(2), A1
(4)The left limit of Q is f′−(x) and the right limit is f′+(x).(1), D2
(5)Therefore f being differentiable at x is equivalent to f′−(x) = f′+(x).(2), (3), (4)
(6)The limit value is unique, so that single common value is f′(x).(5), A2, D1
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Corollary C1
g(x) = |x| is not differentiable at 0, because the left and right derivatives differ.
Proof.
(1)|t| means t if t ≥ 0 and −t if t < 0. Therefore |0| = 0 and 0 + h = h.convention, A20, A22
(2)If h > 0, then [g(0+h) − g(0)]/h = h · h⁻¹ = 1.(1), A20, A21, A22, A14
(3)If h < 0, then [g(0+h) − g(0)]/h = (−h) · h⁻¹ = (−1)(h · h⁻¹) = −1.(1), A19, A16, A14, A18
(4)The difference quotient is the constant 1 for h > 0 and the constant −1 for h < 0, so f′+(0) = 1, f′−(0) = −1.(2), (3), A3, D2
(5)1 ≠ −1, so the two values differ, and g is not differentiable at 0.(4), A23, T1
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Corollary C2
The difference quotient of g(x) = x^(1/3) at 0 is positive on both sides, yet g is not differentiable at 0.
Proof.
(1)x^(1/3) means the real number s with s³ = x, and there is exactly one such s. Therefore g(0) = 0.convention, A5
(2)If h ≠ 0 and u = h^(1/3), then h = u³ and u ≠ 0.(1)
(3)Q(h) = [g(0+h) − g(0)]/h = u · (u³)⁻¹ = u · u⁻¹(u²)⁻¹ = (u²)⁻¹(2), A20, A21, A22, A15, A16, A14, A18
(4)u ≠ 0, so 0 < u², and therefore Q(h) = (u²)⁻¹ > 0, regardless of the sign of h.(3), A9, A6
(5)The sign of the difference quotient is the same on both sides. What splits here is not the sign.(4)
(6)Suppose lim (h→0) Q(h) exists. Then there are δ > 0 and M > 0 with Q(h) ≤ M for every h with 0 < |h| < δ.assumption, A4, (4)
(7)0 < M⁻¹, so s = min(δ, M⁻¹, 1) is positive. Let u = s/2; then 0 < u < δ, u < M⁻¹, u < 1.(6), A6, A10, A11, A12
(8)0 < u < 1, so u² = u · u < u · 1 = u, and for the same reason u³ < u².(7), A7, A17, A18
(9)Let h = u³; then 0 < h < u < δ, so h lies in the range of (6).(7), (8), A12
(10)u² < u < M⁻¹, and multiplying both sides by the positive number M(u²)⁻¹ gives M < (u²)⁻¹ = Q(h).(3), (7), (8), A12, A7, A8, A6, A14, A16, A17, A18, E
(11)(6) says Q(h) ≤ M. This cannot hold together with M < Q(h).(6), (9), (10), A13
(12)Therefore lim (h→0) Q(h) does not exist, and g is not differentiable at 0.(11), D1
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