Question

Leibniz theorem explanation and formula

Answer 1

The Leibniz formula expresses the derivative on nth order of the product of two functions. Suppose that the functions u(x) and v(x) have the derivatives up to nth order. Consider the derivative of the product of these functions.

The first derivative is described by the well known formula:

(uv)′=u′v+uv′.

Differentiating this expression again yields the second derivative:

(uv)′′=[(uv)′]′=(u′v+uv′)′=(u′v)′+(uv′)′=u′′v+u′v′+u′v′+uv′′=u′′v+2u′v′+uv′′.

Likewise, we can find the third derivative of the product uv:

(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′′′.

It is easy to see that these formulas are similar to the binomial expansion raised to the appropriate exponent. Assuming that the terms with zero exponent u0 and v0 correspond to the functions u and v themselves, we can write the general formula for the derivative of nth order of the product of functions uv as follows:

(uv)(n)=n∑i=0(ni)u(n−i)v(i),

where (ni) denotes the number of i-combinations of n elements.

This formula is called the Leibniz formula and can be proved by induction.

Proof.
Suppose that the functions u and v have the derivatives of (n+1)th order. Using the recurrence relation, we write the expression for the derivative of (n+1)th order in the following form:

y(n+1)=[y(n)]′=[(uv)(n)]′=[n∑i=0(ni)u(n−i)v(i)]′.

After differentiation we obtain:

y(n+1)=n∑i=0(ni)u(n−i+1)v(i)+n∑i=0(ni)u(n−i)v(i+1).

Both sums in the right-hand side can be combined into a single sum. Indeed, take an intermediate index 1≤m≤n. The first term when i=m is written as

(nm)u(n−m+1)v(m),

and the second term when i=m−1 is as follows:

(nm−1)u(n−(m−1))v((m−1)+1)=(nm−1)u(n−m+1)v(m).

The sum of these two terms is given by

(nm)u(n−m+1)v(m)+(nm−1)u(n−m+1)v(m)=[(nm)+(nm−1)]⋅u(n−m+1)v(m).

It is known from combinatorics that

(nm)+(nm−1)=(n+1m).

Therefore, the sum of these two terms can be written as

[(nm)+(nm−1)]⋅u(n−m+1)v(m)=(n+1m)u(n+1−m)v(m).

It is clear that when m changes from 1 to n this combination will cover all terms of both sums except the term for i=0 in the first sum equal to

(n0)u(n−0+1)v(0)=u(n+1)v(0),

and the term for i=n in the second sum equal to

(nn)u(n−n)v(n+1)=u(0)v(n+1).

As a result, the derivative of (n+1)th order of the product of functions uv is represented in the form

y(n+1)=u(n+1)v(0)+n∑m=1(n+1m)u(n+1−m)v(m)+u(0)v(n+1)=n+1∑m=0(n+1m)u(n+1−m)v(m).

As can be seen, the expression for y(n+1) has a similar form as for the derivative y(n). Only now the upper limit of summation is equal to n+1 instead of n. Thus, the Leibniz formula is proved for an arbitrary natural number n.

Answer 2

Here is your answer....
Hope this helps you.....✌^_^