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I've seen this multiple times in questions of this form

Suppose $z=f(x,y)$ has continuous second order partial derivatives and $x=g(s,t)$, $y=h(s,t)$, find $\dfrac{\partial^2 f}{\partial t^2}$ (or find something similar).

Typically, $g(x,t)$ and $h(s,t)$ are given. And the answer typically involves some terms of second partial derivatives like $\dfrac{\partial^2 f}{\partial x^2}$ and $\dfrac{\partial^2 f}{\partial y^2}$.

It's quite obvious that $f(x,y)$ must have second order partial derivatives.

But why continuous?

kel c.
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  • I think you mean $g(s,t)$. Also, I don't really understand your question. Why is it obvious that $f(x,y)$ must have second order partial derivatives? The question as it stands guarantees the rule you obtain will be valid for all $\mathbb{R}^2$ (or an open set in it). Assuming the second order partial derivatives aren't continuous makes things get harder. – Mark Fantini Nov 16 '13 at 07:00
  • if $f(x,y)$ doesn't have second order derivatives, then $\dfrac{\partial^2 f}{\partial t^2}$ doesn't exist. To rephrase this question from a different angle, what difference does it make if the second partial derivatives of $f(x,y)$ are not continuous? – kel c. Nov 16 '13 at 21:41

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