 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/2/solution

n = 0

is the identity

uv = uv

. Suppose the formula holds for

n

and differentiate every term using the product rule:

(uv)^{(n + 1)} = \sum_{r = 0}^{n} (r n) [u^{(r + 1)} v^{(n - r)} + u^{(r)} v^{(n - r + 1)}] .

For an interior index

r

u^{(r)} v^{(n + 1 - r)}

becomes

(r - 1 n) + (r n) = (r n + 1)

by Pascal's identity. The two endpoint coefficients are one. This proves

(uv)^{(n)} = r = 0 \sum n (r n) u^{(r)} v^{(n - r)}, (r n) = \frac{n !}{r ! ( n - r )!} .

Only finitely many derivatives are involved; it suffices that both functions are

n

times differentiable.

For

u = x (x + 1)

and

v = e^{2 x}

u

beyond order two vanish. The three surviving terms of the generalized Leibniz rule give

\frac{d ^{n}}{d x ^{n}} [x (x + 1) e^{2 x}] = e^{2 x} [2^{n} (x^{2} + x) + n 2^{n - 1} (2 x + 1) + n (n - 1) 2^{n - 2}] = 2^{n} e^{2 x} [x^{2} + (n + 1) x + \frac{n ( n + 1 )}{4}] .

The simplified expression works for every integer

n \geq 0

, including the two cases where one or more of the three terms are absent.

For

u = x^{2}

and

v = sin x

, use

d^{m} (sin x) / d x^{m} = sin (x + mπ /2)

. Again only three terms survive. With

T = x + nπ /2

, they are

\frac{d ^{n}}{d x ^{n}} (x^{2} sin x) = x^{2} sin T + 2 n x sin (T - π /2) + n (n - 1) sin (T - π) .

Since

sin (T - π /2) = - cos T

and

sin (T - π) = - sin T

, this becomes

\frac{d ^{n}}{d x ^{n}} (x^{2} sin x) = [x^{2} - n (n - 1)] sin (x + \frac{nπ}{2}) - 2 n x cos (x + \frac{nπ}{2}) .

The factors

n

and

n (n - 1)

remove the nonexistent terms when

n = 0

1

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