Leibniz’s Theorem is a fundamental concept in calculus that generalizes the product rule of differentiation and helps us find the n^th derivative of the product of two functions. It is a powerful tool in mathematical analysis, particularly when dealing with functions that change smoothly.

This theorem plays a crucial role in modeling instantaneous rates of change in various mathematical and real-world scenarios. In this article, you will learn the formula of the Leibnitz Theorem, the proof, and the derivation of the Leibnitz Theorem.

What is Leibnitz’s Theorem?

Leibniz’s rule is like an expanded version of the product rule. It says that if two functions, u(x) and v(x), can change smoothly a bunch of times (we call it being differentiable), then their multiplication, u(x) times v(x), can also change smoothly the same number of times.

Leibniz’s theorem helps us find the n^th derivative of the product of two functions, u(x) and v(x), which can be differentiated many times. The formula for the nth derivative of the product is shown like this:

The n^th derivative of (u(x) × v(x)) is equal to the sum of (n choose r) times u’s (n-r)^th derivative times v’s r^th derivative, where r goes from 0 to n.

Here, (n choose r) is calculated as n! / (r! × (n-r)!), and n! means multiplying all the numbers from 1 to n.

When we set n to 1 in this formula, we get the product rule, which says that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Who is Gottfried Wilhelm Leibniz?

Gottfried Wilhelm Leibniz, a German mathematician and philosopher who lived in the 17th and 18th centuries, contributed significantly to various fields, including mathematics. Leibniz is well-known for developing calculus independently of Sir Isaac Newton. The formula known as Leibniz’s Theorem, which plays a role in calculus, is named after him.

Leibniz’s Theorem provides a systematic way to expand a function into an infinite series of terms, involving its derivatives. It helps mathematicians analyze and understand functions in a more detailed manner. Leibniz’s contributions to mathematics, including this theorem, have had a lasting impact and remain fundamental in the study of calculus.

Leibnitz Theorem Formula

The formula for Leibniz’s theorem, which helps find the nth derivative of the product of two functions u(x) and v(x), is expressed as:

[Tex]\frac{d^n}{dx^n} (u(x) \cdot v(x)) = \sum_{r=0}^{n} \binom{n}{r} \cdot u^{(n-r)}(x) \cdot v^{(r)}(x) [/Tex]

In this formula:

• dⁿ/dxⁿ represents the nth derivative with respect to x.
• [Tex]\binom{n}{r} [/Tex] is the binomial coefficient, calculated as [Tex]\frac{n!}{r! \cdot (n-r)!} [/Tex]
• u^kx denotes the kth derivative of the function u(x).
• v^kx denotes the kth derivative of the function v(x).
• The summation is over all values of r from 0 to n.

N^th Derivative in Leibniz’s Formula

For a function f(x) that is (n + 1)-times differentiable on an interval containing (a) and (x), Leibniz’s Theorem states:

[Tex]f(x) = f(a) + f'(a)(x-a) + \frac{f”(a)}{2!}(x-a)^2 + \frac{f”'(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) [/Tex]

In this formula:

• f(x) is the original function.
• f(a) is the function’s value at \( x = a \).
• f'(a) is the first derivative of \( f(x) \) evaluated at \( x = a \).
• f”(a) is the second derivative of \( f(x) \) evaluated at \( x = a \), and so on.

R_n(x) is the remainder term, and it can be expressed as:

[Tex]R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} [/Tex]

Here, (c) is some value between (a) and (x).

This formula helps to expand a function around a specific point (a) by considering its derivatives up to the nth order.

Leibnitz Rule Proof

Leibniz’s rule can be proven using mathematical induction. If we have two functions, f(x) and g(x), that can be smoothly changed many times (differentiable), we start by showing that the product rule holds true for n = 1:

Derivation of Leibnitz Theorem

Step 1: Base Case (n = 1)

Start with the product of two differentiable functions, f(x) and g(x), and apply the product rule:

(f(x)⋅g(x))′ = f′(x)⋅g(x) + f(x)⋅g′(x)

This is the basic product rule and serves as the base case.

Step 2: Inductive Hypothesis

Assume that Leibniz’s rule holds for some positive integer (n):

[Tex](f(x) \cdot g(x))^n = \sum_{r=0}^{n} \binom{n}{r} f^{(n-r)}(x) \cdot g^{(r)}(x) [/Tex]

This is inductive hypothesis.

Step 3: Inductive Step (n + 1)

Now, we want to show that the rule holds for n+1:

[Tex](f(x) \cdot g(x))^{n+1} [/Tex]

Using the binomial theorem, expand this expression:

[Tex](f(x) \cdot g(x))^n \cdot (f(x) \cdot g(x)) [/Tex]

Apply the inductive hypothesis to the first part and the product rule to the second part:

[Tex]\sum_{r=0}^{n} \binom{n}{r} f^{(n-r)}(x) \cdot g^{(r)}(x) \cdot (f(x) \cdot g(x))’ [/Tex]

Now, expand the product rule in the second term:

[Tex]\sum_{r=0}^{n} \binom{n}{r} f^{(n-r)}(x) \cdot g^{(r+1)}(x) + \sum_{r=0}^{n} \binom{n}{r} f^{(n-r+1)}(x) \cdot g^{(r)}(x) [/Tex]

Combine like terms:

[Tex]\sum_{r=0}^{n+1} \left(\binom{n}{r} f^{(n-r)}(x) \cdot g^{(r+1)}(x) + \binom{n}{r-1} f^{(n-r+1)}(x) \cdot g^{(r)}(x)\right) [/Tex]

Factor out terms and simplify:

[Tex]\sum_{r=0}^{n+1} \binom{n+1}{r} f^{(n+1-r)}(x) \cdot g^{(r)}(x) [/Tex]

By the principle of mathematical induction, the expression holds true for all positive integral values of \(n\). Therefore, Leibniz’s rule is proven.

Also, Check

• Euler’s Theorem
• Rolle’s Theorem and Lagrange’s Mean Value Theorem
• Derivative Rules

Solved Examples on Leibnitz Rule

Example 1: Let u(x)=3x²+2x and v(x)=e^x. Using Leibniz’s Rule, find the second derivative of the product u(x)⋅v(x).

Solution:

Let u(x) = 3x² + 2x and v(x) = e^x

u'(x) = 6x + 2

u”(x) = 6

v'(x) = e^x

v”(x) = e^x

Applying Leibniz’s Rule

(uv)” = 6e^x + 2(6x + 2)e^x + (3x² + 2x)e^x

On simplifying the expression we get,

(uv)” = 12e^x + (3x² + 14x + 2)e^x

Example 2: Consider the functions f(x)=sin(x) and g(x)=x². Determine the third derivative of the product f(x)⋅g(x) using Leibniz’s Rule.

Solution:

Consider f(x) = sin(x) and g(x) = x²

f'(x) = cos(x)

f”(x) = -sin(x)

f”'(x) = -cos(x)

g'(x) = 2x

g”(x) = 2

g”'(x) = 0

Applying Leibniz’s Rule

(fg)”’ = -cos(x) · x² + 3(-sin(x) · 2x) + 3(cos(x) · 2)

On simplifying the expression we get,

(fg)”’ = -x²cos(x) – 6x sin(x) + 6cos(x)

Practice Problems on Leibnitz Theorem

1. Find the n^th derivative of f(x)=x³ sin(x).

2. For the function g(x)=e^x cos(x), find the coefficients of the nth derivative at x=0.

3. Apply Leibniz’s Theorem to find the x⁴ term in the expansion of (1+x)⁵ .

4. Approximate the value of √1.1 using Leibniz’s Theorem with a third-degree Taylor polynomial centered at x=1. Estimate the error in your approximation.

Summary

Leibnitz’s Theorem, also known as the Leibniz rule for differentiation under the integral sign, is a powerful tool in calculus that provides a method for differentiating an integral whose limits and integrand depend on the variable of differentiation.

This theorem elegantly combines differentiation and integration, allowing for the calculation of derivatives of integrals with variable limits. It finds applications in various fields, including physics, engineering, and probability theory, simplifying complex problems by breaking them down into more manageable parts. Leibnitz’s Theorem exemplifies the depth and utility of mathematical principles in providing solutions to real-world problems.

Leibnitz Theorem – FAQs

What is Newton Leibnitz’s Theorem?

Also known as the Fundamental Theorem of Calculus, it states that if F(x) is the antiderivative of f(x), then ∫_a^bf(x) dx = F(b) – F(a). It is different from Leibnitz Theorem.

Who made Leibniz rule?

The Leibnitz rule, formulated by Gottfried Wilhelm Leibniz.

What is the Leibnitz Method of Successive Differentiation?

The Leibnitz method of successive differentiation involves repeatedly applying the product rule to differentiate a product of functions.

How do you prove Leibnitz Theorem?

We can prove Leibnitz theorem using the mathematical induction, and it is discussed in the article above.

What is the General Formula of the Leibnitz Rule?

The general formula for Leibniz’s Rule, applied to the nth derivative of the product of two functions u(x) and v(x), is given by:

[Tex](uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)}(x) \cdot v^{(k)}(x) [/Tex]

What is the Conclusion of the Leibniz Theorem?

The conclusion of Leibniz’s Theorem is that the n^th derivative of the product of any two differentiable functions is the sum of all possible combinations of the derivatives of each function in descending order.