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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:

\[(uv)^{n} = \sum_{i=0}^{n}\binom{n}{i} u^{n-i}v^{i}\]

In the above expression, \[\binom{n}{i}\] represents the total number of i-combinations.

Hence, the Leibnitz theorem/formula for the nth derivative has been mentioned above.

If A and B are the functions of x_{1} 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} = [A_{k}.B] + [kC_{1} A_{k-1}.B_{1}] + [kC_{2} A_{k-2}.B_{2}] + ...+ [kC_{r-1}A_{k-r+1}.B_{r-1}] + [kC_{r} A_{k-r}.B_{r} + kC_{k} A.B_{k}]

Differentiating both sides, we get (AB)_{k+1} = [A_{k+1}.B + Ak.B_{1}] + [kC_{1} A_{k}.B_{1} + kC_{1} . A_{k-1}.B_{2}] +...+ [kC_{2} A_{k-1}.B_{2} + kC_{2} A_{k-2}.B_{3}] + ...+ [kC_{r-1} A_{k-r+2}.B_{r-1}] + [kC_{r-1} A_{k-r+1}.B_{r}] + [kC_{r} A_{k-r+1}.B_{r} + kC_{r} A_{k-r}.B_{r+1}] +...[kC_{k} A_{1}.B_{k} + kC_{k} A.B_{k+1} (AB)_{k+1}]

= [A_{k+1}.B +(1 + kC_{1}) A_{k}.B_{1}] + (kC_{1} + kC_{2}) A_{k-1}.B_{2} +...+ ( kC_{r-1}+ kC_{r} ) A_{k-r+1}.B_{r} +...+ kC_{k} A.B_{k+1} ( kC_{r-1} + kC_{r} = (k+1)C_{r}) (AB)_{k+1}.

= A_{k+1}.B + (k+1)C_{1} A_{k}.B_{1} + (k+1)C_{2}A_{k-1}.B_{2} +...+ (k+1)C_{r} A_{k-r+1}.B_{r} +...+ (k+1)C_{k+1} A.B_{k+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.

Let's do some sample question solving:

Q1: If y = x^{3} e^{ax}, find y_{n} , using Leibnitz theorem.

Let u = e^{ax} ; v = x^{3}

D_{n}u = ane^{ax} & Dv = 3x^{2}

D_{n-1}u = a_{n-1}e^{ax} & D^{2}v = 6x

D_{n-2}u = a_{n-2}e^{ax} & D^{3} = 6

Now by LEIBNITZ’s THEOREM, we have

= D_{n}(x^{3}e^{ax}) = D_{n}(e^{ax})x^{3} + nc_{1} D_{n-1}(e^{ax}) . Dx^{3} + nc_{2} D_{n-2}(e^{ax}) . D_{2}x^{3} + nc_{3} D_{n-3} (e^{ax}) . D_{3}x^{3} = a_{n}e^{ax} x^{3} + n a_{n-1} e^{ax}. 3x^{2} + n(n-1)/2 a_{n-2} e^{ax} . 6x^{3} + n(n-1)(n-2)/6 . a_{n-3} e^{ax} .6 = ane^{ax} x^{3} + n a_{n-1} e^{ax}. 3x^{2} + n(n-1) a_{n-2} e^{ax} . 3x^{3} + n(n-1)(n-2) . a_{n-3} e^{ax} = e^{ax}{a_{n}x^{3} = 3_{n}a_{n-1} x^{2} + 3n(n-1)a_{n-2}x + n(n-1)(n-2)a_{n-3}}

This section includes the types of questions you will be getting from the chapter to solve. These questions include:

Find the nth differential coefficient of :-

Sin3x

Sin x cos 3x

e ax cos2x sin x

x

^{2}(x + 2)(2x + 3)

If y = e

^{ax}sin bx, prove that y^{2}– 2ay_{1}+ (a_{2}+ b_{2})y = 0.If y = cos (msin

^{-1}x) prove that (1 - x^{2})y^{2}– xy_{1}+ m^{2}y = 0 & (1 - x^{2}) y_{n+2}– (2n+1)xy_{n+1}+ (m_{2 }- n_{2})y_{n }= 0.If y = (sin

^{-1}x)^{2}, prove that (1 - x^{2})y^{2}- xy_{1}- 2 = 0 and deduce that (1 - x^{2}) y_{n+2}- (2n + 1)xy_{n+1}- n_{2}y_{n}= 0.If y = e tan

^{-1}x, prove that (1 + x^{2})y_{n+2}+ {2(n + 1)x^{-1}}y_{n+1}+ n(n+1)y_{n}= 0.

The Leibniz integral rule provides a designated formula for differentiation of a definite integral whose limits are functions of the differential variable.

\[\frac{d}{dx} \int_{sinx}^{x^{3}} \sqrt{t^{2} + 1} dt = \sqrt{(x^{3})^{2} + 1 . 3x^{2}} - \sqrt{(sinx)^{2} + 1 . cosx}\]

\[= 3x^{2} \sqrt{x^{6} + 1} - cosx \sqrt{sin^{2}x + 1}\]

It is basically known as differentiation under the integral sign. This rule can be applied to assess some unusual definite integrals such as:

\[\frac{d}{dt} \int_{a}^{b} f(x, t)dx = \int_{a}^{b} \delta_{t} f(x, t) dx\]

FAQ (Frequently Asked Questions)

Q1. What is a Leibnitz Theorem?

Answer: 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.

Q2. What is Leibnitz Theorem for Differentiation of an Integral State?

Answer: Under the integral sign for differentiation, the Leibniz's rule named after Gottfried Leibniz, pronounces that for an integral of the form, the derivative of this integral is expressible.

Q3. What is Leibnitz Theorem for the Calculus of an Integral State?

Answer: A link between the derivative of an integral and the integral of a derivative, that is,

d/dt[∫_{S1(t)}^{S2(t)} A(t, s)dx] = ∫_{S1(t)}^{S2(t)} [∂A(t,s)/∂t]ds + A(t, S_{2}) dS_{2}/dt - A(t,S_{1}) dS_{1}/dt

where,

S_{1} and S_{2} = the limits of integration,

s = space variable or dummy distance such as height z,

t = time,

A = some meteorological quantity, like temperature or humidity, that is a function of both time and space.