so that
For a function , we have by the chain rule,
and
Let and . This substitution is made just to make the application of the chain rule clearer.
Differentiating again wrt gives
By the chain rule,
and our substitution shows that, for instance,
and so
Similarly, we find
Putting everything together, we have
and we can similarly find that
Adding together these derivatives, we see the mixed partials cancel, and recalling that , we end up with
as required.