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?
The Leibnitz theorem, also known as the generalized product rule, is a formula used to find the $n$th derivative of the product of two functions. If $f(x)$ and $g(x)$ are functions that are $n$ times differentiable, the theorem states:
$$ \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^{n} \binom{n}{k} \cdot f^{(n-k)}(x) \cdot g^{(k)}(x) $$
This theorem is useful for simplifying the process of differentiating products multiple times, and is widely used in advanced calculus taught at Vedantu.
2. What is Leibniz' formula?
The Leibniz formula is another term for the generalized product rule in differentiation. It gives an explicit expression for the $n$th derivative of the product of two functions. The formula is:
$$ \frac{d^n}{dx^n}[u(x)v(x)] = \sum_{k=0}^{n} \binom{n}{k} \cdot \frac{d^{n-k}u(x)}{dx^{n-k}} \cdot \frac{d^k v(x)}{dx^k} $$
At Vedantu, this formula is used to teach students how to efficiently compute higher-order derivatives in calculus problems.
3. What is the Leibniz rule for limits?
The Leibniz rule for limits relates to interchanging the order of differentiation and integration under certain conditions. If $f(x, t)$ and its partial derivative with respect to $t$ are continuous, then:
$$ \frac{d}{dt}\int_{a}^{b} f(x, t)\,dx = \int_{a}^{b} \frac{\partial}{\partial t} f(x, t)\,dx $$
This result simplifies many calculations and is a key concept in advanced calculus, as explained thoroughly in Vedantu's learning sessions.
4. Is Leibnitz theorem in JEE Mains?
Yes, Leibnitz theorem is included in the JEE Mains syllabus and is particularly important for questions involving higher-order derivatives and product rules. Vedantu's JEE preparatory courses cover this theorem in detail, providing numerous examples and guided practice to ensure mastery for the exam.
5. How is Leibnitz theorem used to find higher order derivatives?
The Leibnitz theorem is especially powerful for calculating higher order derivatives of products of functions. The process involves:
- Identifying the two functions to differentiate (say, $f(x)$ and $g(x)$).
- Applying the formula: $$ \frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x) $$
- Computing the required derivatives for each combination and summing all the terms.
6. What is the difference between Leibnitz theorem and product rule?
The main difference is:
- The product rule provides the formula for the first derivative of the product of two functions: $$ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) $$
- The Leibnitz theorem generalizes this to find the $n$th derivative: $$ \frac{d^n}{dx^n}[u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} u^{(n-k)}(x) v^{(k)}(x) $$
7. Can you provide a solved example using Leibnitz theorem?
Certainly. Let's find the second derivative of $f(x) = x^2 e^x$ using the Leibnitz theorem:
$f(x) = x^2$ and $g(x) = e^x$
We need $ \frac{d^2}{dx^2} [x^2 e^x] $:
Using Leibnitz theorem for $n = 2$:
$$\sum_{k=0}^2 \binom{2}{k} f^{(2 - k)}(x) g^{(k)}(x)$$
Calculate derivates:
- $f''(x) = 2$, $f'(x) = 2x$, $f(x) = x^2$
- $g''(x) = e^x$, $g'(x) = e^x$, $g(x) = e^x$
$$2 \cdot e^x + 2 \cdot 2x e^x + 1 \cdot x^2 e^x = (x^2 + 4x + 2) e^x$$
Vedantu provides step-by-step examples like this in live classes for a strong conceptual understanding.
8. Why is understanding the Leibnitz theorem important for calculus students?
Understanding the Leibnitz theorem is crucial because it:
- Enables students to tackle complex differentiation problems involving products of functions.
- Is foundational for solving challenging questions in engineering and entrance exams like JEE.
- Improves conceptual clarity on higher order derivatives.
9. Are there any tips for remembering Leibnitz theorem for exams?
Yes, here are some tips to remember the Leibnitz theorem:
- Memorize the format: It is a summation over all $k$ from 0 to $n$ using binomial coefficients.
- Use mnemonic devices or write out the pattern with small $n$ values to observe the structure.
- Practice multiple examples to identify repeating elements in derivatives.
- Rely on Vedantu’s revision notes and dedicated practice problems for reinforcement before exams.
