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 .
