The Leibnitz Rule is an extended version of the product rule used in differentiation. It helps us find the nth derivative of the product of two functions. In simple terms, when two functions are multiplied, and we want to take their derivative multiple times, Leibnitz Rule gives us a clear formula to do that.

This rule was introduced by the famous German mathematician and philosopher Gottfried Wilhelm Leibnitz.

We prove this rule using two basic tools in calculus:

1. The product rule – used for the first derivative of a product. If
   y = u · v,
   then
   dy/dx = u · dv/dx + v · du/dx
2. Mathematical induction – a method used to prove formulas that work for all natural numbers.

So, the Leibnitz Rule is a powerful method for finding higher-order derivatives when two functions are multiplied together.

In this article, we will learn about the Leibnitz Theorem, its statement and prove it with the help of mathematical induction. We will also see some solved examples in the end.

Leibnitz Theorem

Leibnitz’s Rule is a more advanced version of the product rule in calculus. It helps us find the nth derivative (like the second, third, and so on) of two functions that are multiplied together. If both functions, say u(x) and v(x), can be differentiated n times, then their product u(x) × v(x) can also be differentiated n times using this rule.

This rule works for many types of functions, like polynomials, trigonometric functions (like sine and cosine), exponential functions, and logarithmic functions.

Leibnitz’s Rule is important because it helps in real-life situations where quantities change over time. For example, speed is the rate of change of distance, and acceleration is the rate of change of speed. Using derivatives, we can find these changing values exactly at a specific moment, which is very useful in science and engineering.

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Consider the two functions u(x) and v(x) that can be differentiated n times. The formula of the nth derivative of the product of the two functions will be given by:

\(\frac{d^n}{dx^n}(uv)\)=\((u(x).v(x))^n = {\sum_{r=0}^n}({^nC_r})u^{(n – r)}(x).v^r(x)\) =

Where,

\(^nC_r = \frac{n!}{r!.(n – r)!}\)

\(n! = 1 × 2 v 3 × 4 × ….(n – 1) × n\)

On substituting n=1 in this formula we get product rule

\((uv)′=u′v+uv′\)

Leibnitz Theorem Proof

Now let’s see the working and proof of the Leibnitz Theorem.

Derivation of Leibnitz Theorem

Leibnitz theorem is derived from the generalization of the product rule of derivatives. Let u′, u′′, u′′′,… and v′, v′′, v′′′, be the higher order derivatives of the functions u(x) and v(x) respectively. Let us multiply these two functions to get u(x).v(x). For simplicity let′s write uv. Let′s differentiate it now.

First Derivative:

\((uv)′ = u′v+uv′\)

Now if we differentiate the above expression again, we get the second derivative.

Second Derivative:

\((uv)′′ = [(uv)′]′ =(u′v+uv′)′\)

\(= (u′v)′+(uv′)′\)

\(= u′′v + u′v′ + u′ v′ + uv′′\)

\(= u′′v + 2u′v′ + uv′′\)

Third Derivative:

\((uv)′′′ = [(uv)′′]′\)

\(= (u′′v + 2u′v′ + uv′′)′\)

\(= (u′′v)′ + (2u′v′)′ + (uv′′)′\)

\(= u′′′v + u′′v′ + 2u′′v′ + 2u′v′′ + u′v′′ + uv′′′\)

\(= u′′′v + 3u′′v′ + 3u′v′′ + uv′′′\)

Leibnitz Theorem gives us a formula that allows us to generalize the product rule to directly find any given \(n^{th}\) order derivative of the product of two functions. \(\frac{d^n}{dx^n}(uv)\). Lets look at the proof for the Leibnitz theorem.

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Proof of Leibnitz Theorem

The formula of Leibniz theorem can be proved by the Principle of Mathematical Induction. Let us consider two functions \(u(x)\) & \(v(x)\), and they have the derivatives up to the \((n)^{th}\) order, then we have to prove that

\(\frac{d^n}{dx^n}(uv)=\sum _{i=0}^n \binom{n}{ i} u^{(i)}v^{(n-i)}\)

where \(u^{(k)}\) is k-th derivative with respect to \(x\) and \(\binom{n}{ i}\) is just a different representation number of possible combinations equal to \(\frac{n!}{i!(n – i)!}\)

Now, first we check if the theorem holds for n=1,

So, for n=1 and i = 0,1

LHS = \(\frac{d}{dx}(uv) = u’v + vu’\) (by product rule)

R.H.S = \(\sum _{i=0}^1\binom{1}{i}u^{(i)}v^{(1-i)}\) = uv’ + vu’

LHS=RHS

Thus the theorem holds for n=1.

Now let’s assume the theorem holds for some \(1<m < n\). Thus,

\(\frac{d^m}{dx^m}(uv)=\sum_{i=0}^{m}\binom {m}{i}u^{(m)}v^{(m-i)} =s\).

For the theorem to be true we need to show it holds for \(m+1<n\).

\(\frac{d^{m+1}}{dx^{m+1}}(uv)=\frac{d}{dx}(s)\).

\(\frac{d^{m+1}}{dx^{m+1}}(uv)=\frac{d^{m}}{dx^{m}}\left(\frac{d}{dx}(uv)\right)=\frac{d^{m}}{dx^{m}}(u’v)+\frac{d^{m}}{dx^{m}}(uv’)\)

\(\frac{d^{m+1}}{dx^{m+1}}(uv)=\sum_{i=0}^{m}\binom{m}{i}u^{(i+1)}v^{(m-i)}+\sum_{i=0}^{m} \binom {m}{i}u^{(i)}v^{(m+1-i)}\)

\(=u^{(m+1)}v+\sum_{i=1}^{m}\binom{m}{i-1}u^{(i)}v^{(m+1-i)}+\sum_{i=1}^{m}\binom {m}{i}u^{(i)}v^{(m+1-i)}+uv^{(m+1)}\)

\(=u^{(m+1)}v+\sum_{i=1}^{m}\left(\binom{m}{i-1}+\binom{m}{i}\right)u^{(i)}v^{(m+1-i)}+uv^{(m+1)}\)

\(=\sum_{i=0}^{m+1}\binom {m+1}{i}u^{(i)}v^{(m+1-i)}\)

Hence Proved, for n=\((m+1)\). Hence the theorem is proved by induction

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Properties of Leibnitz Rule

The Leibnitz Rule (also called Leibnitz Theorem) helps to find the nth derivative of the product of two functions. Below are some key properties:

1. Product of Functions
   The rule is used when two differentiable functions are multiplied together, like:
   y = f(x) ⋅ g(x)
2. Combination of Derivatives
   The nth derivative of the product is a sum of terms involving derivatives of both functions up to order n:
   dⁿ/dxⁿ [f(x)g(x)] = Σ (from k = 0 to n) [C(n, k) × f⁽ⁿ⁻ᵏ⁾(x) × g⁽ᵏ⁾(x)]
3. Use of Binomial Coefficients
   Each term uses a binomial coefficient, written as:
   C(n, k) = n! / (k!(n−k)!)
4. Symmetry in Derivatives
   The derivatives in each term are taken in a balanced way — one goes from n to 0, the other from 0 to n.
5. Linear Operation
   The rule follows the property of linearity. That means derivatives can be taken one term at a time.
6. Special Case (n = 1)
   When n = 1, it becomes the normal product rule:
   d/dx [f(x)g(x)] = f′(x)g(x) + f(x)g′(x)
7. Applicable to Differentiable Functions
   The rule only works if both f(x) and g(x) are differentiable up to the nth order.

Applications of Leibnitz Rule

The Leibnitz Rule is widely used in calculus and helps to find the nth derivative of a product of two functions. Some important applications include:

1. Higher Order Derivatives
   It is mainly used to compute the nth derivative of the product of two differentiable functions without having to repeatedly apply the product rule.
2. Solving Problems in Differential Calculus
   Many problems involving patterns in derivatives or where functions are multiplied benefit from using Leibnitz Rule directly.
3. Mathematical Proofs and Theorems
   The rule plays a key role in proving theorems and identities involving derivatives, especially in advanced mathematics.
4. Series Expansion
   The Leibnitz Rule is useful while expanding functions into Taylor or Maclaurin series when those functions are products.
5. Physics and Engineering
   In fields like physics and engineering, it helps solve problems involving motion, wave equations, and signal processing where functions depend on time and position.
6. Computer Algorithms
   Some symbolic mathematics software and algorithms use the rule to automate differentiation in symbolic computation.
7. Economics and Statistics
   It is also applied in economics for finding marginal rates and in statistics for deriving formulas involving expectations and variances of functions.

Solved Examples on Leibnitz Theorem

Now let’s see some solved derivatives on Leibnitz Theorem

Solved Example 1: Find the nth derivative of \(y=e^x(2x+3)^3\)

Solution:

Let \(e^x = u\) and \((2x+3)^2 = v\)

According to the Leibnitz Theorem Formula,

\((u(x).v(x))^n = {\sum_{r=0}^n}({^nC_r})u^{(n – r)}(x).v^r(x)\)

\({d^n\over{dx^n}}(uv) = ^nC_0u_nv_0 + ^nC_1u_{n-1}v_1 + ^nC_2u_{n-2}v_2 + ^nC_3u_{n-3}v_3…\)

\({d^n\over{dx^n}}[e^x(2x+3)^3]= ^nC_0[e^x(2x+3)^3] + ^nC_1e^x(2x+3)^2(2)+^C_2.12e^x(2x+3)+^nC_3e^x(48)\)

\({d^n\over{dx^n}}[e^x(2x+3)^3]=e^x[(2x+3)^3+6n(2x+3)^2+12n(n-1)(2x+3)+8n(n-1)(n-2)]\)

Solved Example 2: Find the nth derivative of \((ax+b)sinx\)

Solution:

Let \(u = sinx\) and \(v = (ax+b)\)

According to the Leibnitz Theorem Formula,

\((u(x).v(x))^n = {\sum_{r=0}^n}({^nC_r})u^{(n – r)}(x).v^r(x)\)

\({d^n\over{dx^n}}(uv) = ^nC_0u_nv_0 + ^nC_1u_{n-1}v_1 + ^nC_2u_{n-2}v_2 + ^nC_3u_{n-3}v_3…\)

\({d^n\over{dx^n}}[sinx(ax+b)]=(ax+b)sin(x+n{\pi\over{2}})+^nC_1sin(x+(n-1){\pi\over{2}}).a+^nC_2sin(x+(n-2){\pi\over{2}}).0\)

\({d^n\over{dx^n}}[sinx(ax+b)]=(ax+b)sin(x+n{\pi\over{2}})+{n!\over{1!(n-1)!}}sin(x+(n-1){\pi\over{2}}).a\)

\({d^n\over{dx^n}}[sinx(ax+b)]=(ax+b)sin(x+n{\pi\over{2}})+{a.n.(n-1)!\over{(n-1)!}}\)

Solved Example 3: Find the 4th-order derivative of the function y = xsinhx

Solution:

Let \(u=sinhx\) and \(v=x\)

According to the Leibnitz Theorem Formula,

\((u(x).v(x))^n = {\sum_{r=0}^n}({^nC_r})u^{(n – r)}(x).v^r(x)\)

\({d^n\over{dx^n}}(uv) = ^nC_0u_nv_0 + ^nC_1u_{n-1}v_1 + ^nC_2u_{n-2}v_2 + ^nC_3u_{n-3}v_3…\)

\((sinhx)′ = coshx\)

\((sinhx)′′ = sinhx\)

\((sinhx)′′′ = coshx\)

\((sinhx)′′′′ = sinhx\)

\({d^4\over{dx^4}}(xsinhx)=^4C_0sinhx.(x) + ^4C_1coshx.(1)\)

\({d^4\over{dx^4}}(xsinhx)=xsinhx+4coshx\)

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