We prove the theorem by using mathematical induction on 'n'.
Suppose that 'p' and 'q' are functions defined and differentiable on I. Then we know that 'pq' is differentiable on I and that
(pq)'(x) = p'(x).q(x) + p(x) q'(x) ∀ x ∈ I.
This proves the theorem for n = 1.
Assume, as induction hypothesis, that the theorem is true for n – 1 ≥ 1. If p and q are functions defined on I having (n – 1)th order derivatives on I, then pq has (n – 1)th order derivative on I and
(pq)(n – 1)(x) = p(n – 1)(x) q(x) + (n – 1)C1 p(n – 2)(x) q'(x) + ... + (n – 1)Cr p(n – r – 1)(x)qr(x) + ... + p(x) q(n – 1)(x)
Now suppose that 'f' and 'g' are defined on I and have nth order derivatives on I. Then 'f' and 'g' have (n – 1)th order derivatives on I. Therefore, by induction hypothesis 'fg' has (n – 1)th order derivative on I and
(fg)(n – 1)(x) = f(n – 1)(x) g(x) + (n – 1)C1f(n – 2)(x) g'(x) + ... + (n – 1)Crf(n – r – 1)(x)gr(x) + ... + f(x)g(n – 1)(x) -------- (i)
Now the right side of (i) is a finite sum of terms of the form
(n – 1)Cr f(n – r – 1)(x) gr(x), where 0 ≤ r ≤ n – 1
Since both 'f' and 'g' have nth order derivatives on I and since 0 ≤ r ≤ n – 1, we get that f(n – 1 – r) and gr are differentiable on I and so is their product.
Since a finite sum of differentiable functions is differentiable, we get that right side of (i) is a differentiable function.
Hence the left side of (i) is also differentiable.
Now differentiating (i) w.r.t. 'x', we get
(fg)(n)(x) | = | (fg)(n – 1)(x) |
= | [f(n)(x) g(x) + f(n – 1)(x) g'(x)] + (n – 1)C1 [f(n – 1)(x) g'(x) + f(n – 2)(x) g''(x)] + ... + (n – 1)Cr [f(n – r)(x) g(r)(x) + f(n – r – 1)(x) g(r + 1)(x)] + ... + [f'(x) g(n – 1)(x) + f(x) g(n)(x)] | |
= | f(n)(x) g(x) + [1 + (n – 1)C1] f(n – 1)(x) g'(x) + [(n – 1)C1 + (n – 1)C2]f(n – 2) (x) g''(x) + ... + [(n – 1)Cr – 1 + (n – 1)Cr] f(n – r)(x) g(r)(x) + ... + f(x) g(n)(x), (by collecting similar terms) | |
= | f(n)(x) g(x) + nC1f(n – 1)(x) g'(x) + ... + nCr f(n – r)(x) g(r)(x) + ... + f(x) g(n)(x) | |
(since (n – 1)Cr – 1 + (n – 1)Cr = nCr ∀ r = 1, 2, ...., n). |
This proves the theorem for 'n'. Hence by mathematical induction the theorem is true for all positive integers 'n' satisfying the hypothesis of the theorem.