# Thread: Help with some questions...

1. ## Help with some questions...

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?

2. I do not really like the notation,
$y=y(x)$ and $x=x(y)$.
It makes me want to vomit.

Let us assume,
$y=f(x)$ has an inverse on some open interval
$y^{-1}=f^{-1}(x)$ is the inverse function.
Further, assume $y$ is twice differenciable on the interval.
Then, $y^{-1}$ is twice differenciable on the interval.

Thus,
$f(f^{-1}(x))=x$ throughout the interval.
Take derivative (chain rule),
$(f^{-1}(x))'f'(f^{-1}(x))=1$
Thus,
$f'(f^{-1}(x))=\frac{1}{(f^{-1}(x))'}$*)
Take derivative again, use chain on left use quotient on right.
$(f^{-1}(x))'f''(f^{-1}(x))=-\frac{(f^{-1}(x))''}{f^{-1}(x))')^2}$
Thus, (I assume, but am lazy to check that).
$(f^{-1}(x))'f''(f^{-1}(x))=f''(x)$
Thus,
$f''(x)=-\frac{(f^{-1}(x))''}{(f^{-1}(x))')^2}$
$\frac{d^2 y}{dx^2}=-\frac{ \frac{d^2 x}{dy^2} }{\left( \frac{ dx }{dy}\right)^2}$

*)Note $(f^{-1}(x))'\not = 0$ at some point. There are two possibilities. Either zero throught the interval in that case, $f^{-1}(x)=C$ but then $f(x)$ 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.

3. Yes! Use the Extended Mean Value theorem.

Work on the interval,
$[0,x]$.
Both are differenciable on,
$I=(0,x)$
Both are continous on,
$[0,x]$.
Now,
$f(x)=\ln (1+x)$ thus, $f'(x)=\frac{1}{1+x}$
And,
$g(x)=\sin^{-1} (x)$ thus, $g'(x)=\frac{1}{ \sqrt{1-x^2} }$
Where, $g'(x)\not = 0 \forall x\in I$.
Thus, $\exists c\in I$
$\frac{\sqrt{1-c^2}}{1+c}=\frac{\ln (1+x)}{\sin^{-1} x}$
But, $0.
Thus,
$\sqrt{\frac{1-x}{1+x}}<\sqrt{\frac{1-c}{1+c}}=\frac{\ln (1+x)}{\sin^{-1} x}$

4. ## ?

thank you very much!

i didn't understand one thing... (in the image)
can you explain?

5. Originally Posted by ThePerfectHacker
Thus, (I assume, but am lazy to check that).
$(f^{-1}(x))'f''(f^{-1}(x))=f''(x)$
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:

$
\left\{ \begin{array}{l}
x = u \\
y = t \\
\end{array} \right.
$

Then:

$
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:

$
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:

$
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.

6. How does that happen then?
If these expressions are not equal then the original problem is not true*)

*)Unless $(f^{-1}(x))'f''(f^{-1}(x))-f''(x)=C$
They differ by some constant throught the interval.

7. That's why I said:

"This is wrong, probably because you trusted the problem statement to be correct."

The intial statement isn't correct, at least I think it's incorrect.