Decomposing higher order derivatives of composite functions I'm new to calculus, and just a little hazy on the skeleton of higher order derivatives when the chain rule is involved. I'm given a table of function and first derivative values, and I need to solve for the second derivative at a specific point. On an abstract level, that's fairly straightforward:
(1)$$k(x) = f(g(x))$$
(2)$$\frac{d}{dx}k(x) = \frac{d}{dx}f(g(x))\frac{d}{dx}g(x)$$
(3)$$\frac{d^2}{dx^2}k(x)=\frac{d^2}{dx^2}f(g(x))\left[\frac{d}{dx}g(x)\right]^2+\frac{d}{dx}f(g(x))\frac{d^2}{dx^2}g(x)$$
But I don't know anything about the second derivative, so it's difficult to solve for a specific value of x (I could graph the function, but I'm looking for an exact solution).
I'm curious, if I take $\frac{d^2}{dx^2}f(g(x))$ for example, is this the equivalent of $\frac{d}{dx}\left[\frac{d}{dx}f(g(x))\right]$? Of course, had I started with this expression, I would apply the chain rule, so I would find myself back at (3). In terms of one of the second-order derivatives in (3), however, since the chain rule was already applied, is it fair to assume $$\frac{d^2}{dx^2}f(g(A))=\frac{d^2}{dx^2}f(B)=\frac{d}{dx}\left[\frac{d}{dx}f(B)\right]=\frac{d}{dx}[C]$$
Edit: The point I'm trying to make here is breaking up the expression into first-order derivatives without applying the chain rule, since g(x) is essentially a constant in this context (to my understanding).
 A: The short answer is, if you don't have any information about $f''$ or $g''$, you can't evaluate $k''$. Your step (3) is correct, and that's as good as it gets.
Yes, $\frac{d^2}{dx^2}f(g(x)) = \frac{d}{dx}\left[\frac{d}{dx}f(g(x))\right]$, that's exactly what a second derivative represents, and that's why the fancy bit you do with $A$, $B$, and $C$ works, it's just that $C = \frac{d}{dx}f(g(x))$ that you have in step (2), so it's not particularly helpful. A second derivative can be written as a first derivative of a different function, where that different function is just the first derivative of the original function.
The notation is a little bit tricky here, so it's important to remember this: when the $g(x)$ appears inside of another function, like $f$ or $f'$ or $f''$, all it means is we're evaluating that outer function at $g(x)$ rather than at $x$.
It doesn't mean that $g(x)$ is a constant, but it tells us that in order to figure out where we want to evaluate a function or a derivative, we first have to evaluate $g(x)$, then plug that into the function.
Let's write out what you actually want: the second derivative of $f$ with respect to $x$ evaluated at $g(x)$.
I'm going to use prime notation, because that's easier. So, we have
$$ k(x) = f(g(x)) $$
$$ k'(x) = f'(g(x))\cdot g'(x) $$
$$ k''(x) = \underbrace{[f''(g(x))\cdot g'(x)]}_{\frac{d}{dx}f'(g(x))}\cdot \underbrace{g'(x)}_{g'(x)} + \underbrace{f'(g(x))}_{f'(g(x))}\cdot \underbrace{g''(x)}_{\frac{d}{dx}g'(x)} $$
We can simplify a bit to
$$ k''(x) = f''(g(x))\cdot [g'(x)]^2 + f'(g(x))\cdot g''(x) $$
This is the same thing you got in step (3), and there's no way to simplify it any further without more information about $f$ or $g$.
