## ABSTRACT

The result we now describe provides a generalization of the derivative of a

product. Let F(x )/f (x )g (x ), where f (x ) and g (x) are functions which may both be differentiated n times. Applying the rule for the differentiation of a

product to F (x ) gives

dF

dx

d[ fg]

dx f

dg

dx g

df

dx :

A further differentiation gives

d2F

dx2

d2[ fg]

dx2 f

d2g

dx2 2

df

dx

dg

dx g

d2f

dx2 ,

and one more differentiation gives

d3[ fg]

dx3 f

d3g

dx3 3

df

dx

d2g

dx2 3

d2f

dx2

dg

dx g

d3f

dx2 :

Continuing in this manner it is not difficult to show (by induction) that, in general,

dn[ fg]

dxn f

dng

dxn n

df

dx

dn1g

dxn1

n(n 1)

d2f

dx2

dn2g

dxn2

n(n 1)(n 2)

d3f

dx3

dn3g

dxn3 . . .n

dn1f

dxn1

dg

dx g

dnf

dxn :

This result is called Leibniz’s formula, and the coefficients n , n (n/1)/2!,

n (n/1)(n/2)/3! are seen to be the binomial coefficients n

k

: On account

of this, Leibniz’s formula may be written more concisely as

dn

dxn [ f (x) g(x)]

n

k

f (k)(x) g(nk) (x),

with the convention that f (0)/f(x ) and g(0)/g (x ). In general the n th derivative dn [ f(x ) g (x)]/dxn will contain n/1 terms, but

if f(x) (or g (x )) is a polynomial of degree m , f (n ) (x )/0 (or g(n ) (x )/0) for n n

and a simple expression can be found for dng /dxn .