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PROOF 135

Prove.

The derivative of xⁿ

On differentiable functions.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

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.A7

∎

Definition D2

For a natural number n, set x¹ = x and x^(k+1) = xᵏ · x.

Proof.

Therefore, if k ≥ 2, then hᵏ = h² · h^(k−2).convention, A12

∎

Lemma L1

If n is a natural number and h ≠ 0, there is a polynomial R such that [(x+h)ⁿ − xⁿ]/h = nxⁿ⁻¹ + h · R(x,h).

Proof.

(1)By the binomial theorem, (x+h)ⁿ = Σ (k=0…n) nCk xⁿ⁻ᵏ hᵏA1, D2
(2)The k = 0 term is nC0 xⁿ = xⁿ, and the k = 1 term is nC1 xⁿ⁻¹h = n xⁿ⁻¹ h.(1), A2, A15
(3)For k ≥ 2, hᵏ = h² h^(k−2), so every such term has h² as a factor. Write the sum of those terms as h² R(x,h).(1), D2, A11, A12, A13
(4)Therefore (x+h)ⁿ = xⁿ + n xⁿ⁻¹ h + h² R(x,h)(1), (2), (3), A17, A18
(5)Subtracting xⁿ from both sides gives (x+h)ⁿ − xⁿ = n xⁿ⁻¹ h + h² R(x,h)(4), E, A17, A18, A19, A20
(6)h ≠ 0, so multiplying both sides by h⁻¹ gives [(x+h)ⁿ − xⁿ]/h = n xⁿ⁻¹ + h R(x,h)(5), E, A11, A12, A13, A14, A15

∎

Theorem T1

If n is a natural number, f(x) = xⁿ is differentiable at every x and f′(x) = n xⁿ⁻¹.

Proof.

(1)For h ≠ 0, the difference quotient of f is n xⁿ⁻¹ + h R(x,h).L1, D1
(2)lim (h→0) h = 0A5
(3)R(x,h) is a polynomial in h, so each term is a constant times a power of h, and its limit exists.A3, A4, A5, A6, D2
(4)Therefore lim (h→0) h R(x,h) = 0 · lim R(x,h) = 0(2), (3), A4, A16
(5)lim (h→0) n xⁿ⁻¹ = n xⁿ⁻¹A6
(6)Adding the two limits gives lim (h→0) [n xⁿ⁻¹ + h R(x,h)] = n xⁿ⁻¹(4), (5), A3, A20
(7)Therefore f is differentiable at x and f′(x) = n xⁿ⁻¹.(1), (6), A7, D1

∎

Corollary C

If m is a natural number and x ≠ 0, then (x⁻ᵐ)′ = −m x⁻ᵐ⁻¹.

Proof.

(1)x⁻ᵐ means (xᵐ)⁻¹.A9
(2)Let g(x) = xᵐ; then g′(x) = m xᵐ⁻¹.T1
(3)By the quotient rule, (x⁻ᵐ)′ = (1/g)′ = −g′/g² = −m xᵐ⁻¹ · ((xᵐ)²)⁻¹(1), (2), A8
(4)(xᵐ)² = xᵐ xᵐ = x²ᵐ, and xᵐ⁻¹ · (x²ᵐ)⁻¹ = xᵐ⁻¹ x⁻²ᵐ = x⁻ᵐ⁻¹.A10, A9, D2
(5)Therefore (x⁻ᵐ)′ = −m x⁻ᵐ⁻¹.(3), (4), A12, A13
(6)The meaning of x⁻ᵐ is a convention chosen so that the law of exponents is preserved. Under that convention, the form of the derivative also matches that of natural-number exponents.(5), A9, A10

∎

Remark

Can A1 be removed from this proof?

The exponent n comes down as a coefficient in exactly one place, line (2) of L1. Check where that n comes from.