Leibnitz Theorem

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.

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.

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:

\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)

In this formula:

• dⁿ/dxⁿ represents the nth derivative with respect to x.
• \binom{n}{r} is the binomial coefficient, calculated as \frac{n!}{r! \cdot (n-r)!}
• 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:

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)

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:

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

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):

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

This is inductive hypothesis.

Step 3: Inductive Step (n + 1)

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

(f(x) \cdot g(x))^{n+1}

Using the binomial theorem, expand this expression:

(f(x) \cdot g(x))^n \cdot (f(x) \cdot g(x))

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

\sum_{r=0}^{n} \binom{n}{r} f^{(n-r)}(x) \cdot g^{(r)}(x) \cdot (f(x) \cdot g(x))'

Now, expand the product rule in the second term:

\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)

Combine like terms:

\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)

Factor out terms and simplify:

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

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

Related Article:

• 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

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

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

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

Question 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.