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) we can similarly apply the product rule to get:

u'(x)=f(x)g'(x) + f'(x)g(x)

Substituting this result, along with the definition of u(x) into [1] we get:

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