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,
$\displaystyle y=y(x)$ and $\displaystyle x=x(y)$.
It makes me want to vomit.
Let us assume,
$\displaystyle y=f(x)$ has an inverse on some open interval
$\displaystyle y^{-1}=f^{-1}(x)$ is the inverse function.
Further, assume $\displaystyle y$ is twice differenciable on the interval.
Then, $\displaystyle y^{-1}$ is twice differenciable on the interval.
Thus,
$\displaystyle f(f^{-1}(x))=x$ throughout the interval.
Take derivative (chain rule),
$\displaystyle (f^{-1}(x))'f'(f^{-1}(x))=1$
Thus,
$\displaystyle f'(f^{-1}(x))=\frac{1}{(f^{-1}(x))'}$*)
Take derivative again, use chain on left use quotient on right.
$\displaystyle (f^{-1}(x))'f''(f^{-1}(x))=-\frac{(f^{-1}(x))''}{f^{-1}(x))')^2}$
Thus, (I assume, but am lazy to check that).
$\displaystyle (f^{-1}(x))'f''(f^{-1}(x))=f''(x)$
Thus,
$\displaystyle f''(x)=-\frac{(f^{-1}(x))''}{(f^{-1}(x))')^2}$
In your notation,
$\displaystyle \frac{d^2 y}{dx^2}=-\frac{ \frac{d^2 x}{dy^2} }{\left( \frac{ dx }{dy}\right)^2}$
*)Note $\displaystyle (f^{-1}(x))'\not = 0$ at some point. There are two possibilities. Either zero throught the interval in that case, $\displaystyle f^{-1}(x)=C$ but then $\displaystyle f(x)$ 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.
Yes! Use the Extended Mean Value theorem.
Work on the interval,
$\displaystyle [0,x]$.
Both are differenciable on,
$\displaystyle I=(0,x)$
Both are continous on,
$\displaystyle [0,x]$.
Now,
$\displaystyle f(x)=\ln (1+x)$ thus, $\displaystyle f'(x)=\frac{1}{1+x}$
And,
$\displaystyle g(x)=\sin^{-1} (x)$ thus, $\displaystyle g'(x)=\frac{1}{ \sqrt{1-x^2} }$
Where, $\displaystyle g'(x)\not = 0 \forall x\in I$.
Thus, $\displaystyle \exists c\in I$
$\displaystyle \frac{\sqrt{1-c^2}}{1+c}=\frac{\ln (1+x)}{\sin^{-1} x}$
But, $\displaystyle 0<c<x$.
Thus,
$\displaystyle \sqrt{\frac{1-x}{1+x}}<\sqrt{\frac{1-c}{1+c}}=\frac{\ln (1+x)}{\sin^{-1} x}$
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:
$\displaystyle
\left\{ \begin{array}{l}
x = u \\
y = t \\
\end{array} \right.
$
Then:
$\displaystyle
y' = \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dy}}}} = \frac{1}{{x'}}
$
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:
$\displaystyle
y'' = \frac{{dy'}}{{dx}} = \frac{{\frac{{dy'}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dt}}}}\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{1}{{\frac{{dx}}{{dt}}}}\frac{d}{{dt}}\left( {\frac{1}{{\frac{{dx}}{{dt}}}}} \right)\frac{1}{{\frac{{dx}}{{dt}}}}\left( {\frac{{0 \cdot \frac{{dx}}{{dt}} - 1 \cdot \frac{{d^2 x}}{{dt^2 }}}}{{\left( {\frac{{dx}}{{dt}}} \right)^2 }}} \right)
$
In the last step, I explicitly wrote the quotient rule. This simplifies to:
$\displaystyle
y'' = - \frac{{\frac{{d^2 x}}{{dt^2 }}}}{{\left( {\frac{{dx}}{{dt}}} \right)^3 }} = - \frac{{x''}}{{x'^3 }}
$
So there should be a cube instead of a square in the denominator.