Leibnitz theorem

Let f and g be two functions defined over an interval I of R having nth order derivatives on I.
Then the product of the two functions fg also has nth derivative on I.
It is given by (where x ∈ I)

(fg)(n)(x) =
= f(n)(x).g(x) + nC1 f(n – 1)(x) g'(x) + ..... + nCr f(n – r)(x) g(r)(x) +.....+ f(x) g(n)(x)

Note that g(0)(x) means g(x).


If we use u = f(x) and v = g(x) and the Leibnitz theorem can be stated as

(uv)n = nC0 un v0 + nC1 un – 1 v1 + ..... + nCr un – r vr + ..... + nCn u0 vn

OR

Dn(uv) = nC0 Dn(u). v + nC1 Dn – 1(u) D(v) + ..... + nCr Dn – r(u) Dr(v) + ..... + nCn u.Dn (v)

(We know nC0 = nCn = 1)