Differentiating the derivative of an inverse function I'm having some trouble with differentiation of the derivative of an inverse function.
Let $v(x)$ be a function, and $v^{-1}(x)$ is its inverse function. $\frac{d}{dx} v(x)$ is denoted as $v'(x)$.
I was trying to do $\frac{d}{dx} \ (v')^{-1} (x)$
According to the implicit function theorem, I know $(v^{-1})'(x) = \frac{1}{v'(v^{-1}(x))}$
So should the answer be $$\frac{(v')^{-1}(x) \ \cdot \ v''(v^{-1}(x))}{v'(v^{-1}(x))^2}$$ or $$\frac{1}{v''((v')^{-1}(x))}?$$
The first one is my answer, and the latter is suggested by my friend.
Thank you for the help!
 A: From $v^{-1}(v(y))=y$ by differentiation $\left(v^{-1}(v(y))\right)'.v'(y)=1$ or
$$\left(v^{-1}(x)\right)'=\dfrac{1}{v'(v^{-1}(x))}$$
where $x=v(y)$, by differentiation 
\begin{align}
\left(v^{-1}(x)\right)''
&=\dfrac{0-\frac{d}{dx}v'(v^{-1}(x))\times1}{\left(v'(v^{-1}(x))\right)^2}\\
&=-\dfrac{1}{\left(v'(v^{-1}(x))\right)^2}\left(\frac{d}{dx}v'(v^{-1}(x))\right)\\
&=-\dfrac{1}{\left(v'(v^{-1}(x))\right)^2}\left(v''(v^{-1}(x))\left(v^{-1}(x)\right)'\right)\\
&=-\dfrac{v''(v^{-1}(x))}{\left(v'(v^{-1}(x))\right)^2}\dfrac{1}{v'(v^{-1}(x))}\\
&=-\dfrac{v''(v^{-1}(x))}{\left(v'(v^{-1}(x))\right)^3}
\end{align}
A: Long but necessary.
This calculus madness gets very confusing very fast, unless you have it very clear what objects you're dealing with (functions), how they can interplay with each other, and the precise statements of the theorems that you're going to use to make even a single step forward in the calculations. Do this, and you'll avoid most common mistakes.

What we're talking about. First of all, it's not completely correct to say that "$v(x)$ is a function", because it is a number, and that's a fundamentally different beast. The object you have in mind is actually $v$, which takes in a number $x$ and outputs a number $v(x)$. Since we're in the context of simple one-dimensional calculus, these two numbers $x$ and $v(x)$ will both be real numbers, and this fact is summarized in the notation $v : \mathbb R \to \mathbb R$ ("$v$ is a function from the set real numbers to the set of real numbers").
Another object is the differential operator, which is usually explained in calculus classes as an operator that takes in a function, say $u$, and outputs another function, its derivative, $u'$. Thing is, you can't do this to any function – you need your input function to satisfy a certain condition called differentiability. Not all functions are differentiable, so you'll need theorems that tell you if the ones you're working with actually are.
Of course, if the derivative $u'$ turns out to also be differentiable, you can differentiate it (i.e. apply the differential operator again) to find yet another function $u''$ that is called the second derivative of $u$. And on and on it goes.
If you think this way about things, then your question (from what I understand) becomes

Given an invertible and differentiable function $v : \mathbb R \to \mathbb R$, how can I express the second derivative of its inverse function $v^{-1}$ in terms of $v^{-1}$ itself, $v$, $v'$, $v''$, etc.?


Useful information. Let's get to the theorems we need. The first one is the so-called "chain rule":

Chain rule. Let $p : \mathbb R \to \mathbb R$ and $q : \mathbb R \to \mathbb R$ be differentiable. Then their composition $q \circ p$ is also differentiable, and for all real numbers $x$ the following identity holds:
  $$(q\circ p)'(x) = q'(p(x))\ p'(x) \tag{1}$$

Nice. And even more nicely, this turns out to be true even if more than two functions are involved, i.e., if the "chain is longer". So if the chain consists of three differentiable functions, say $a,b,c$, the theorem states that $(c \circ b \circ a)$ is differentiable and we have
$$(c \circ b \circ a)'(x) = c'(b(a(x)))\ b'(a(x))\ a'(x) \tag{1a}$$
But the theorem we really need is the one you've stated yourself,

Inverse function theorem. Let $h : \mathbb R \to \mathbb R$ be an invertible, differentiable function. Furthermore suppose that its inverse $h^{-1}$ is continuous, and that for all real numbers $x$ we have $h'(x) \neq 0$. Then $h^{-1}$ is also differentiable, and for all real numbers $y$ the following formula holds:
  $$(h^{-1})'(y) = \frac{1}{h'(h^{-1}(y))} \tag{2}$$

(If you're asking yourself why it is required that $h'$ never be zero, look at the formula: if it were to vanish, then you'd be dividing by zero! On the other hand, the requirement that $h^{-1}$ be continuous is a bit more subtle.)
Notice that I've highlighted two specific sentences in the statements above: that's because the theorems are not about the formulas included in them, they're really about establishing the differentiability of a function. In the first case such function is $(p \circ q)$; in the second case it is $(h^{-1})$. 

The answer, step by step. Now we can tackle your question. Suppose $v$ satisfies all the conditions stated in the inverse function theorem, that is


*

*It is invertible;

*Its inverse $v^{-1}$ is continuous;

*It is differentiable;

*Its derivative $v'$ is never zero.


Then $v$ satisfies formula $(2)$ above: for all real numbers $y$
$$(v^{-1})'(y) = \frac{1}{v'(v^{-1}(y))} \tag{3}$$
We want the derivative of the function $(v^{-1})'$, but we need to understand if it makes sense to even talk about such a thing. Let's first understand how this function builds its outputs from its inputs. When it gets given an input $x$, first it asks $v^{-1}$ to take $x$ as an input and build the number $v^{-1}(x)$, then it asks $v'$ to take that as an input and build the output $v'(v^{-1}(x))$, and then it takes the reciprocal of it. We can think of the last action as passing the output of $v'$ as an input to another function $r$, the "reciprocal" function that takes any (non-zero! cf. above) real number $t$ and outputs its reciprocal $t^{-1}$.
This is good: it means we can re-write formula $(3)$ as
$$(v^{-1})'(x) = (r \circ v' \circ v^{-1})(x)$$
In this situation, if we can make sure that all functions $v^{-1}$, $v'$ and $r$ are differentiable, then the chain rule would grant us that $(v^{-1})' = (r \circ v' \circ v^{-1})$ is too, and moreover, it would give us a formula for its derivative.
Well, but we've already established that $v^{-1}$ is differentiable, and we know how its derivative is defined through formula $(3)$. Furthermore, $r$ is famously differentiable (but again remember where it is defined...) and we also know that the derivative is given by the formula $r'(t) = - t^{-2}$. Sadly there is no theorem that tells us that $v'$ is differentiable only relying on the assumptions we've made, so in order to proceed we'll need to add it to the requirements on $v$ at the beginning of our journey:


*Its derivative $v'$ is differentiable.


Good: so we're all set to apply the chain rule. We can write:
$$\begin{split}
(v^{-1})''(y) &= (r \circ v' \circ v^{-1})'(y) \\
&= r'(v'(v^{-1}(y)))\ v''(v^{-1}(y))\ (v^{-1})'(y) \qquad && \mathrm{apply\ (1a)}\\
&= - \frac{1}{v'(v^{-1}(y))^2}\ v''(v^{-1}(y))\ \frac{1}{v'(v^{-1}(y))} \qquad && \mathrm{apply\ (3)\ to\ last\ factor} \\
&= - \frac{v''(v^{-1}(y))}{v'(v^{-1}(y))^3} \qquad && \mathrm{simplify}
\end{split}$$
There you go.
