How to Apply Leibnitz Theorem: Formula, Proof, and Examples
In Mathematics, the Leibnitz theorem or Leibniz integral rule for derivation comes under the integral sign. It is named after the famous scientist Gottfried Leibniz. Thus, the theorem is basically designed for the derivative of the antiderivative. Basically, the Leibnitz theorem is used to generalise the product rule of differentiation. It states that if there are two functions let them be a(x) and b(x) and if they both are differentiable individually, then their product a(x). b(x) is also n times differentiable.
This theorem basically refers to the process through which one can find the derivative of an antiderivative. It is also known as successive differentiation. According to the proposition, the derivative on the nth order of the product of two functions can be expressed with the help of a formula. The formula for the above-mentioned theorem is as follows:
\[ (uv)^{n} = \sum_{i=0}^{n} \left(\begin{array}{c}n\\ i\end{array}\right) u^{n-i} v^{i} \]
In the above expression, \[ \left(\begin{array}{c}n\\ i\end{array}\right) \] represents the total number of i-combinations.
Hence, the Leibnitz theorem/formula for the nth derivative has been mentioned above.
Derivation of the Leibnitz Theorem/Formula
If A and B are the functions of x1, then dn(AB)/dxn = [nC0.An.B] + [nC1.An-1.B1] + [nC2.An-2.B2] + [nCr.An-r.Br] +...........+ [nCn.A.Bn]. Thus, the theorem will be proved by induction.
Step 1: By actual differentiation, we already know that
(AB)1 = A1.B + A.B1 (AB)2
= (A2.B + A1.B1) + (A1.B1 + A.B2)
= A2.B + 2A1.B1 + A.B2
= A2.B + 2A1.B1+ A.B2
= A2.B + 2C1 A1.B1 + A.B2
Hence, the theorem holds true for n = 1,2.
Step 2: We assume that the theorem is true for a particular value of n say k, so we have:
(AB)k = [Ak.B] + [kC1 Ak−1.B1] + [kC2 Ak−2.B2] + ...+ [kCr1Ak−r+1.Br−1] + [kCr Ak−r.Br + kCk A.Bk]
Differentiating both sides, we get
(AB)k+1 = [Ak+1.B + Ak.B1] + [kC1 Ak.B1 + kC1 . Ak−1.B2] +...+ [kC2 Ak−1.B2 + kC2 Ak−2.B3] + ...+ [kCr−1 Ak−r+2.Br−1] + [kCr−1 Ak−r+1.Br] + [kCr Ak−r+1.Br + kCr Ak−r.Br+1] +...[kCk A1.Bk + kCk A.Bk+1 (AB)k+1]
= [Ak+1.B +(1 + kC1) Ak.B1] + (kC1 + kC2) Ak−1.B2 +...+ ( kCr−1+ kCr ) Ak−r+1.Br +...+ kCk A.Bk+1 (kCr−1 + kCr = (k+1)Cr) (AB)k+1.
= Ak+1.B + (k+1)C1 Ak.B1 + (k+1)C2Ak−1.B2 +...+ (k+1)Cr Ak−r+1.Br +...+ (k+1)Ck+1 A.Bk+1
Thus, the theorem is true for n = k+1, i.e., it also holds true for the next higher integral value of k. Given that, we have observed that the theorem is true for n = 2, therefore, the theorem is true for (n = 2 + 1), i.e., n = 3, and, therefore, further true for n = 4 and so on. Hence, the theorem is true for all positive real values of n.
Solved Examples
Let's do some sample question solving:
Q1: If y = x3 eax, find yn , using Leibnitz theorem.
Let \[ u = e^{ax}\] , \[v = x^{3}\] . Now, \[ u_{n} = a^{n}e^{ax} \]
By Leibnitz’s Theorem ,
\[ y_{n} = a^{n} e^{ax} x^{3} + \left(\begin{array}{c}n\\ 1\end{array}\right) a^{n-1}e^{ax} 3x^{2} + \left(\begin{array}{c}n\\ 2\end{array}\right) a^{n-2}e^{ax} 6x + \left(\begin{array}{c}n\\ 3\end{array}\right) a^{n-3}e^{ax} 6 \]
= \[ e^{ax} a^{n-2} (a^{3}x^{3} + 3na^{2}x^{2} + 3n (n-1) ax + n(n-1) (n-2) ) \]
Consider a function in two variables x and y, i.e.,
z=f(x,y)
Let us consider the integral of z with respect to x, from a to b, i.e.,
\[I = \int_{a}^{b} f (x,y)dx \]
For this integration, the variable is only x and not y. y is essentially a constant for the integration process. Therefore, after we have evaluated the definite integral and put in the integration limits, y will still remain in the expression of I. This means that I is a function of y.
\[I (y) = \int_{a}^{b} f (x,y) dx \]… (1)
This relation (1) can be differentiated with respect to y as follows:
\[ I’ (y) = \frac{d}{dy} \left ( \int_{a}^{b} f(x,y)dx\right ) \]
=\[ \int_{a}^{b} \frac{\partial f(x,y)}{\partial{y}} dx \]
Where
\[ \frac{\partial f(x,y)}{\partial{y}} \]
stands for the partial derivative of f(x,y) with respect to y, that is, the derivative of f(x,y) w.r.t. y, treating x as a constant.
So as per the rules if you want to find a derivative in the nth order of the product of two functions, then this formula of Leibnitz is going to help you. Sometimes we refer to it as the differential under the integral sign.
FAQs on Leibnitz Theorem Made Simple
1. What is the Leibnitz Theorem in simple terms?
In simple terms, the Leibnitz Theorem is a powerful shortcut in calculus that provides a formula to find the nth derivative of a product of two functions. It is a generalisation of the standard product rule. If you have two functions, u(x) and v(x), that can be differentiated 'n' times, this theorem allows you to calculate the nth derivative of their product, u(x)v(x), directly without having to differentiate one step at a time.
2. What is the formula for the Leibnitz Theorem?
The formula for the Leibnitz Theorem for finding the nth derivative of the product of two functions, u and v, is given by:
(uv)n = nC0unv + nC1un-1v1 + nC2un-2v2 + ... + nCrun-rvr + ... + nCnuvn
Where:
- un and vn represent the nth derivatives of the functions u and v, respectively.
- nCr are the binomial coefficients.
3. What is the main difference between the Product Rule and the Leibnitz Theorem?
The main difference lies in their application for derivatives. The Product Rule is used to find only the first derivative of a product of two functions. The Leibnitz Theorem is a more general formula used to find any higher-order derivative (like the 2nd, 3rd, or nth derivative) of that same product. In fact, the Product Rule is just a special case of the Leibnitz Theorem where n=1.
4. How is the Leibnitz Theorem applied in solving problems? Can you provide an example?
The Leibnitz Theorem is applied by identifying the two functions (u and v) in a product and then systematically finding their successive derivatives. For example, to find the second derivative of y = x² log(x):
- Let u = x² and v = log(x).
- Find the successive derivatives: u₁ = 2x, u₂ = 2; and v₁ = 1/x, v₂ = -1/x².
- Apply the Leibnitz formula for n=2: (uv)₂ = 2C0u₂v + 2C1u₁v₁ + 2C2uv₂.
- Substitute the values: (uv)₂ = 1(2)log(x) + 2(2x)(1/x) + 1(x²)(-1/x²).
- Simplify to get the answer: 2log(x) + 4 - 1 = 2log(x) + 3.
5. Why does the formula for the Leibnitz Theorem look so similar to the Binomial Theorem?
This is an excellent observation. The structure of the Leibnitz Theorem formula is analogous to the Binomial Theorem because the process of successive differentiation of a product follows a similar pattern to the expansion of a binomial power. The binomial coefficients (nCr) appear in both, and the order of derivatives (one function's derivative order decreases while the other's increases) mirrors how the powers of the two terms behave in a binomial expansion.
6. How is the Leibnitz Theorem proven?
The Leibnitz Theorem is formally proven using the Principle of Mathematical Induction. The proof involves three main steps:
- Base Case: Show that the formula is true for n=1. This step demonstrates that the formula correctly simplifies to the standard Product Rule.
- Inductive Hypothesis: Assume that the formula is true for some positive integer n=k.
- Inductive Step: Differentiate the expression for n=k and, by algebraic manipulation, show that the formula also holds true for n=k+1.
This process confirms that if the theorem is true for one derivative, it must be true for the next, and therefore for all positive integers n.
7. Where is the Leibnitz Theorem most useful in higher mathematics?
The Leibnitz Theorem is particularly useful in several advanced areas of mathematics and science. Its primary applications include:
- Finding the nth derivative required for Taylor or Maclaurin series expansions of functions that are products.
- Solving certain types of linear differential equations where repeated differentiation of product terms is necessary.
- Applications in physics and engineering, especially in problems involving mechanics and wave theory where higher-order rates of change of combined quantities are studied.
