The first one - I have no idea...
the second - I guess we can use Cauchy's mean value theorem, but how??
the third - Taylor's theorem?
I do not really like the notation,
and .
It makes me want to vomit.
Let us assume,
has an inverse on some open interval
is the inverse function.
Further, assume is twice differenciable on the interval.
Then, is twice differenciable on the interval.
Thus,
throughout the interval.
Take derivative (chain rule),
Thus,
*)
Take derivative again, use chain on left use quotient on right.
Thus, (I assume, but am lazy to check that).
Thus,
In your notation,
*)Note at some point. There are two possibilities. Either zero throught the interval in that case, but then cannot exists, because it is one-to-one map. And it cannot happen that it is zero at some point but not all, that will lead to non-differenciability. Thus, we can divide.
This is wrong, probably because you trusted the problem statement to be correct.
Suppose y is a function of x and you want to switch the roles of y and x, i.e. make y the independent variable of the function x. It is safer to introduce new variables, let me show it for Phoebe83 in particular. Let:
Then:
Of course, y' means that y as a function of x is being differentiated with respect to x, while x' means that x as a function of y is being differentiated with respect to y. Taking the second derivative, which I'll take with respect to t again, using the chain rule:
In the last step, I explicitly wrote the quotient rule. This simplifies to:
So there should be a cube instead of a square in the denominator.