The concept of the derivative in its current form was first developed by Newton in 1666, and a few years after him, by Gottfried Wilhelm Leibniz independently of each other. Leibniz, the greatest comprehensive genius of the 17th century, was Newtons rival in the inventing calculus. The general opinion today is that each of them discovered calculus independently of the other. Although Newtons discovery was made earlier, Leib-niz published the results earlier. He has extracted many of the rules of derivation, which a student learns in the beginning of an introductory course in calculus. The derivative of function f can also be represented by f^. This symbol emphasizes that f^ is a new fun-ction obtained by deriving from function f and its value at x is given by f^ (x). This symbol was used by Joseph Louis Lagrange in 1770 AD. Derivatives of higher orders are shown as the f^(first derivative), f^  (second derivative), and f^  (third derivative), f^((4))  (fourth derivative) and .......〖 f〗^((n))  (nth derivative). The rule for finding the derivative of the product of two functions is still called Leibnizs rule.

A generalization of a mathematical theorem is a theo-rem that gives a wider result with the same premise of the theorem. So that the generalized theorem is obtain-ned from it. Through the generalization of a theorem, it is possible to look at that theorem from a broader angle. Most mathematical works, even the results of outstanding mathematicians, are usually generaliz-ations of existing works and concepts. There are two types of the patterns for generalization: "result-based generalization" and the "process-based generalization". Generalization based on the result is a generalization that is obtained by recognizing a general pattern of the result itself. Process-based generalization is a general-ization resulting from a process that ends in a result, that is, a way that contains a specific chain of steps dependent on previous results (Niazai et al., 2022).

Leibnizs rule and its generalization are used to obtain the derivative of higher order in the product of two or more functions. For example, in functions such asu=sinx, v=lnx, and w=x^2, if we want to obtain the derivative of higher orders in their product (y=uvw), we use the generalization of Leibnizs rule for the product of three functions. This method makes us go through a less process in obtaining the derivative of higher orders, especially when the product of several functions cannot be written as a sum of several terms.

The derivative of the product of two or more functions and derivatives of the first to nth order:

Suppose u and v are two functions of x and we have: y=u.v then the derivative of the function y will be: (first order derivative): y^=〖(uv)〗^

Leibnizs symbolization method:

(d(u.v))/dx=u.dv/dx+v.du/dx

We assume that u and v are continuous. The deriva-tive of their product will be:  y^=u^ v+uv^

We also have several functions:

y=uvw

y^=〖(uvw)〗^→y^=u^ vw+uv^ w+uvw^

If the function is continuous and differentiable, we have: y^=f^ (x), if y^=f^ (x) is also differentiable, the second derivative of the original function y=f(x)  can be found:

y^=f^ (x)=〖(f^ (x))〗^            y^=〖(y^)〗^

Similarly, if y^=f^ (x) is differentiable, the third derivative of the function f(x)  is as follows:

y^=〖(f^ (x))〗^=〖(y^)〗^

Derivatives of higher order can be defined as follows:

y^((n))=〖(y^((n-1) ))〗^=(d^n y)/(dx^n )

Leibniz rule (derivative):

If y=u.v, where u and v are functions of x. we can say that the nth order derivative of y with respect to x is obtained from the following equation:

(I) y^((n))=uv^((n))+nu^ v^(n(n-1))+(n(n-1))/2! u^ v^((n-2))+(n(n-1)(n-2))/3! u^ v^((n-3))+⋯+u^((n)) v

The relationship (I) is also written as follows:

y^((n))=∑_(k=0)^n▒〖(n¦k) u^((k)) v^((n-k)) 〗

The sum of derivative orders of functions u and v in each term is equal to (n):

(k)+(n-k)=n

The coefficient of each term in relation (I) is equal to: n!/(k!(n-k)!)

The general expression of the relation (I) is as follows:  Tp=n!/(k!(n-k)!) u^((k)) v^((n-k))

Generalization of Liebnitzs rule for n function:

Suppose u_(i )is a function of x, we have u_i=f_i (x) and

y=∏_(i=1)^n▒u_i                       i∈N

y=u_1 u_2 u_3…u_n

If the functions u_1 u_2 u_3…u_n are continuous and differentiable, we have:

y^=〖(u_1 u_2 u_3…u_n)〗^

y^=u_1^ u_2 u_3…u_n+u_1 u_2^ u_3…u_n+u_1 u_2 u_3^…u_n+⋯+u_1 u_2 u_3…u_n^

y^=(2u_1^ u_2^ u_3 〖…u〗_(n-1)+2u_1^ u_3^ u_2…u_n+2u_1^ u_4^ u_2 u_3 〖…u〗_n+⋯+2u_1^ u_n^ u_2 u_3…u_(n-1) )+(u_1^ u_2 u_3…u_n+u_1 u_2^ u_3…u_n+u_1 u_2 u_3^…u_n+⋯+u_1 u_2 u_3…u_n^ )+(2u_1 u_2^ u_3^…u_n+2u_1 u_2^ u_4^ u_3…u_n+2u_1 u_2^ u_5^ u_4 u_3…u_n+⋯+2u_1 u_2^ u_n^ u_3 u_((n-1) ) )+(2u_1 u_2 u_3^ u_4^ u_5 u_6…u_n+2u_1 u_2 u_3^ u_5^ u_4 u_6…u_n+⋯+2u_n^ u_3^ u_1 u_2…u_(n-1) )+⋯(2u_n^ u_1^ u_2…u_(n-1)+2u_n^ u_2^ u_1 u_3…u_(n-1)+2u_n^ u_3^ u_1 u_2…u_(n-1)+⋯+2u_n^ u_(n-1)^ u_1 u_2 u_3…u_(n-2))

We have for the derivative of the nth order of function y=∏_(i=1)^n▒u_i : (i∈N)

(II) y^((n))=(u_1 u_2 u_3…u_n^((n) )+u_1 u_2…u_(n-1)^((n) ) u_n+⋯+u_1 u_2^((n) ) u_3…u_n+u_1^((n) ) u_2 u_3…u_n )+u_1^((n)) u_2 u_3…u_n+n(u_1 u_2…u_(n-1)^ u_n^((n-1) ) )+n(u_1 u_2…u_(n-2)^ u_(n-1) u_n^((n-1) ) )+⋯+n(u_1^ u_2…u_(n-1) u_n^((n-1) ) )+n(n-1) u_1^ u_2^ u_3…u_n^((n-2) )+n(n-1) u_1^ u_3^ u_2…u_n^((n-2))+⋯+n(n-1) u_1^ u_(n-1)^ u_2 u_3…u_n^((n-2))+⋯

From the expansion sentences y^( )  and the relation (II), the general sentence can be concluded:

Tp=n!/(a_1 〖!a〗_2 〖!a〗_3 !…a_n !)×u_1^(〖(a〗_1)) u_2^(〖(a〗_2)) u_3^(〖(a〗_3))…u_n^(〖(a〗_n))

j∈N,a_j∈Z            a_1+a_2+a_3+⋯+a_n=n

a_j≥0

a_j is the derivative of functions u_i:

Order of the derivative of the function a_1:u_1

Order of the derivative of the function a_2:u_2

Order of the derivative of the function a_3:u_3

.................................................. 

...................................................

Order of the derivative of the function a_n:u_n

-The zeroth order of the derivative of a function is that function itself.

{█(〖u_i〗^((0))=u_i@a_j=0)┤

(III)Tp=n!/(∏_(j=1)^n▒aj!)×∏_█(i=1@j=1)^n▒〖u_i〗^((aj))               

The general sentence of relation (II)

- Note (1): The general sentence of relation (II) is concluded through process-based generalization.

Examples:

Leibnizs rule can be concluded from relation (II) (from its general sentence formula):

If we omit u_3 u_4…u_n  functions in the general sentence of relation (II), we will have:

Tp=n!/(a_1 !a_2 !)×u_1^((a_1)) u_2^((a_2))

Tp=n!/(k!(n-k)!) u_1^((k)) u_2^((n-k))                      k≥0

{█(a_1 〖+a〗_2=n@a_1=k)→a_2 ┤=n-k

→y^((n))=∑_(k=0)^n▒〖(n¦k) u_1^((k)) u_2^((n-k)) 〗

With a little precision, we can see that the relationship between the coefficients and the order of the derivative of the functions in the expansion of Leibnizs rule is the same as the relationship between the coefficients and powers in the expansion of Newtons binomial. Also, the relationship between the derivative order of fun-ctions and coefficients in the generalization of Leibnizs derivative rule in the general sentence is the same as the relationship between the coefficient and power in the expansion of polynomials (general sentence):

〖(x_1 x_2 x_3 〖+⋯+x〗_n)〗^n

Tp=n!/(a_1 !a_2 !a_3 !…a_n !) x_1^(a_1 ) x_2^(a_2 ) x_3^(a_3 )…x_n^(a_n ) (IV)

a_1+a_2+a_3+⋯+a_n=n

Example: if the functions u, v and w are functions of the variable x and we have    y=uvw, the expan-sion of the third derivative of the function y is equal to

y^((3))=(〖uvw)〗^

y=u_1 u_2→y^((n))=(〖u_1 u_2)〗^((n))

y^((3))=u^ vw+uv^ w+uvw^+6u^ v^ w^+3u^ v^ w+3u^ v^ w+3uv^ w^+3u^ vw^+3u^ vw^+3v^ uw^

Note (2): It may be possible to conclude the general sentence of the relation (II) through the generalization based on the result. The relation (I) is the same as Newtons binomial expansion. The general sentence of relation (II) is also the same as the general sentence of polynomial expansion, that is, relation (IV). Of course, we are more interested in process-based generalization here.