Leibnitz's Theorem

1. Statement

If u and v are functions of x, each possessing derivatives up to the nth order, then the nth derivative of their product is

(uv)_n = u_n v + nC_1 u_{n-1} v_1 + nC_2 u_{n-2} v_2 + ... + nC_r u_{n-r} v_r + ... + u v_n

where $nC_r$ is the binomial coefficient n!/(r! (n − r)!) and $u_k = d^k u / dx^k$ (with $u_0 = u$ , $v_0 = v$ ).

This is exactly the binomial expansion pattern applied to derivatives: the "coefficients" are binomial coefficients, the powers of a are replaced by successive derivatives of u, and the powers of b by successive derivatives of v.

For small n the formula reads:

n = 1: (uv)₁ = u₁v + uv₁ — the ordinary product rule.
n = 2: (uv)₂ = u₂v + 2u₁v₁ + uv₂.
n = 3: (uv)₃ = u₃v + 3u₂v₁ + 3u₁v₂ + uv₃.

2. Proof by Induction (standard short version)

The result is true for n = 1. Assume it true for n and differentiate once more:

(uv)_{n+1} = u_{n+1}v + (nC_1 + nC_0) u_n v_1 + (nC_2 + nC_1) u_{n-1} v_2 + ... + u v_{n+1}

Using Pascal's rule $nC_r + nC_{r-1} = (n+1)C_r$ each bracket collapses to the next binomial coefficient, giving the statement for n + 1. Hence true for all n.

3. When to Use It — the Practical Rule

Use Leibnitz when y is a product of two functions of different kinds: polynomial × exponential, polynomial × trig, polynomial × log, exponential × trig.

The trick that makes such questions quick: choose the polynomial as u, because after a few differentiations its derivatives become zero, and the infinite-looking series truncates. For u = x², u₃ = u₄ = … = 0, so only three terms survive and the answer is a closed formula in n.

y	u (chosen)	v	Why
x² e^(2x)	x²	e^(2x)	u₃ = 0 → 3 terms
x³ sin 2x	x³	sin 2x	u₄ = 0 → 4 terms
x² log x	x²	log x	3 terms; v_r of log x is a standard result
eˣ sin x	either	either	no truncation — both have infinite derivatives, use the general exponential-trig result

   Find yₙ for  y = x² e^(2x)

   Take  u = x²  ⇒  u₁ = 2x , u₂ = 2 , u₃ = 0
         v = e^(2x) ⇒  v_r = 2ʳ e^(2x)

   Since u_{n-r} = 0 for n - r > 2, only r = n, n-1, n-2 contribute:

        yₙ = u_{n} v + nC_1 u_{n-1} v_1 + nC_2 u_{n-2} v_2
           = x²·2ⁿ e^(2x) + n·(2x)·2^(n-1) e^(2x) + [n(n-1)/2]·2·2^(n-2) e^(2x)

        yₙ = 2^(n-2) e^(2x) [ 4x² + 4nx + n(n-1) ]

   Verification — for n = 2 :  2⁰ e^(2x)[4x² + 8x + 2] = 2e^(2x)(2x² + 4x + 1)
   Direct:  y′  = 2x e^(2x) + 2x² e^(2x) = 2e^(2x)(x + x²)
            y″  = 2[2e^(2x)(x + x²) + e^(2x)(1 + 2x)] = 2e^(2x)(2x² + 4x + 1)  ✓

7. Comparison of the Two Standard Tools

Situation	Tool
Single standard function (e^(ax), sin(ax+b), 1/(ax+b), …)	Standard nth-derivative table
Product of a polynomial with e^(ax), sin, cos or log	Leibnitz's theorem (series truncates)
Product of e^(ax) with sin/cos	General result (a² + b²)^(n/2) e^(ax) sin(bx + c + nθ)
Product of two trig functions of the same type	Product-to-sum identities, then the table
Rational function	Partial fractions, then the table

8. Exam Points

State Leibnitz's theorem and prove it by induction — 3-5 marks, almost certain.
Find the nth derivative of a product such as x²eˣ, x²log x, x³sin x, x eˣ cos x using the theorem — 7 marks, the most frequent question of this topic.
Verify the theorem for n = 2 or n = 3 on a given product — 3-5 marks.
Choose u and v appropriately and explain why a polynomial is taken as u — 3 marks.
Find yₙ for e^(ax) sin(bx + c) using the general result — 5 marks.
Use the theorem to prove an identity involving nth derivatives — 5 marks.

Leibnitz's Theorem