## Leibnitz Theorem For nth Derivative

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 d^{n}(AB)/dx^{n} = [nC_{0}.A_{n}.B] + [nC_{1}.A_{n-1}.B_{1}] + [nC_{2}.A_{n}-_{2}.B_{2}] + [nC_{r}.A_{n-r}.B_{r}] +...........+ [nC_{n}.A.B_{n}]. Thus, the theorem will be proved by induction.

Step 1: By actual differentiation, we already know that

(AB)_{1} = A_{1}.B + A.B_{1} (AB)_{2}

= (A_{2}.B + A_{1}.B_{1}) + (A_{1}.B_{1} + A.B_{2})

= A_{2}.B + 2A_{1}.B_{1} + A.B_{2}

= A_{2}.B + 2A_{1}.B_{1}+ A.B_{2}

= A_{2}.B + 2C_{1} A_{1}.B_{1} + A.B_{2}

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

**1. What is a Leibnitz theorem?**

In mathematics, a Leibnitz theorem is actually the rule of the Leibnitz which is defined for derivatives of the antiderivative. According to the stated proposition, the derivative on the nth order of the product of two functions can be demonstrated using a formula. The functions that could possibly have rendered function as a derivative are called as antiderivatives (or primitive) of the function. The formula that provides all these antiderivatives is known as the indefinite integral of the function, and such a process of determining the antiderivatives is what we call integration.

**2. What is the Leibnitz theorem for differentiation of an integral state?**

In the Leibnitz theorem, there is an integral sign for differentiation. You can differentiate the product of two functions upto nth order and this can be expressed with the help of the formula. This theorem was given by Gottfried Leibnitz. He describes that for integral of the form, the derivative of integral is always expressible. This theorem has its uses in both calculus and arithmetics. This theorem is going to help you in solving a number of problems directly.