Vedic Sutra #9:     Differential Calculus

We want to differentiate   y = (x + a1)(x + a2)(x + a3)...

Note: Given a polynomial, we first factor it as noted above ... if we can !

Example #1:
    y = x² + 3x + 2 = (x+1)(x+2)
Then     dy/dx = (x+1) + (x+2) ... the sum of different factor-products, 1 at a time

Example #2:
    y = x⁴ + 10x³ + 35x² + 50x + 24 = (x+1)(x+2)(x+3)(x+4) ... factor the polynomial, if we can !
    dy/dx = (x+2)(x+3)(x+4) + (x+1)(x+3)(x+4) + (x+1)(x+2)(x+4) + (x+1)(x+2)(x+3)
    ... the sum of different factor-products, 3 at a time
    d²y/dx² = 2{(x+3)(x+4) +(x+2)(x+4) + (x+2)(x+3) + (x+1)(x+4) + (x+1)(x+3) + (x+1)(x+2)}
    ... the sum of different factor-products, 2 at a time
    d³y/dx³ = 6{(x+1) + (x+2) + (x+3) + (x+4)2}
    ... the sum of different factor-products, 1 at a time

Note:
The numbers in blue are related to the degree of the polynomial, eh?

If y = (x + a₁)(x + a₂)(x + a₃)... (x + a_n)   for n factors
then
[1]   log(y) = log(x + a₁) + log(x + a₂) + log(x + a₃) + ...
then (differentiating)
[2]   (1/y) dy/dx = 1/(x + a₁) + 1/(x + a₂) + 1/(x + a₃) + ...
then
[3]   dy/dx = y/(x + a₁) + y/(x + a₂) + y/(x + a₃) + ...
and, replacing y by (x + a₁)(x + a₂)(x + a₃)...
[4]   dy/dx is the sum of factor-products, taken n-1 at a time.

Note:
There are n terms in dy/dx, each a product of n-1 factors, where the first term is missing (x + a₁), the second is missing (x + a₂) etc.

Differentiating again, to obtain d²y/dx², we use the same ritual [1] - [4], obtaining for each term a sum of n-1 terms.
Each is a sum of factor-products with n-2 factors ... but some factor-products are repeated !

Consider
    y = (x + a₁)(x + a₂)(x + a₃)... (x + a_n) = xⁿ + (a₁+a₂+...+a_n)x^n-1 + ...
then n-1 differentiations would yield:
    d^n-1y/dx^n-1 = [n!]x + [(n-1)!](a₁+a₂+...+a_n) = [(n-1)!] {(x+a₁)+(x+a₂)+...+(x+a_n)}
which is (n-1)! times the sum of factor-products taken 1 at a time

In general, for the higher derivatives, we have:
    d^my/dx^m = m!{sum of factor-products taken n-m at a time}