Can the product rule be extended to the product of three functions?
1 Answer
Mar 22, 2017
We have:
F(x)=f(x)g(x)h(x) F(x)=f(x)g(x)h(x)
Let
u(x)=f(x)g(x) => F(x) = u(x)h(x) u(x)=f(x)g(x)⇒F(x)=u(x)h(x)
We can differentiate using the product rule:
F'(x) = u(x)h'(x) + u'(x)h(x) ..... [1]
And as
u'(x)=f(x)g'(x) + f'(x)g(x)
Substituting this result, along with the definition of
F'(x) = f(x)g(x)h'(x) + {f(x)g'(x) + f'(x)g(x)} \ h(x)
" " = f(x)g(x)h'(x) + f(x)g'(x)h(x) + f'(x)g(x)h(x)
QED