Stuck on Some Derivative in Derivation of Faà di Bruno's Formula

Question

I have no problem with the underlying idea of Faà di Bruno's Formula as generalization of composite function derivatives of single-variable functions. Where I am not sure that there is some recursive operation related to chain rule. Here I will directly delve into the part I am stuck on:

Let $f(u)=y$ and $g(x)=u$ be single-variable functions of class $C^n$ defined on $\mathbb{R}$. We can write the first derivative of the composite function $f(g(x))$ with respect to $x$, which is obvious, as such: $$\frac{dy}{dx}= \frac{du}{dx} \frac{dy}{du}$$

The problematic part shown in the box below comes up when trying to take the second derivative as follows: $$\frac{d^2y}{dx^2}= \frac{d^2u}{dx^2} \frac{dy}{du}+ \frac{du}{dx} \fbox{$\frac{d}{dx} \left( \frac{dy}{du} \right)$}$$

I did not know how to take derivative of $\frac{dy}{du}$ with respect to $x$ at first. Then noting to use $\frac{dy}{dx}= \frac{du}{dx} \frac{dy}{du}$ as a chain rule operator where we can plug other functions between parentheses $($ and $)$: $$\frac{d \left(\ \right)}{dx}= \frac{du}{dx} \frac{d \left(\ \right)}{du}$$

However, even though it works out to give the correct result, which is as shown: $$\frac{d^2y}{dx^2}= \frac{d^2u}{dx^2} \frac{dy}{du}+ \left(\frac{du}{dx} \right)^2 \frac{d^2y}{du^2}$$

And something feels wrong about such a symbolic manipulation to me when taking $\frac{dy}{du}$ as some other function, say $h(x)=z$, of which we know that it contains $g(x)$ function itself so that we can plug it in the chain rule, otherwise no need to apply chain rule in the first place. It is also obvious that we cannot represent $h(x)$ contains $g(x)$ as $h(g(x))$. Thus I end up here in some kind of circular argument. If all make sense, where do I fail to comprehend? Is it mathematically valid of my treatment of chain rule in such an operator fashion?

Answer 1

The derivation of $\frac{d^2 y}{dx^2}$ is sound.

We obtain \begin{align*} \color{blue}{\frac{d^2 y}{dx^2}} &=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ &=\frac{d}{dx}\left(\frac{dy}{du}\,\frac{du}{dx}\right)\\ &=\left(\frac{d}{dx}\left(\frac{dy}{du}\right)\right)\frac{du}{dx}+\frac{dy}{du}\,\left(\frac{d}{dx}\left(\frac{du}{dx}\right)\right)\\ &=\left(\frac{d^2y}{du^2}\,\frac{du}{dx}\right)\frac{du}{dx}+\frac{dy}{du}\,\frac{d^2u}{dx^2}\\ &\,\,\color{blue}{=\frac{d^2y}{du^2}\,\left(\frac{du}{dx}\right)^2+\frac{dy}{du}\,\frac{d^2u}{dx^2}}\tag{1} \end{align*} in accordance with OPs derivation.

It is convenient to make a plausibility check by calculating a specific example in two ways. We consider \begin{align*} y&=f(u)=3\sin (u)+u^2+3\\ u&=g(x)=4x^2+1\\ \end{align*} We have according to (1) \begin{align*} \frac{d^2 y}{dx^2}=&\left(\frac{d^2}{du^2}\left(3\sin(u)+u^2+3\right)\right)\,\left(\frac{d}{dx}\left(4x^2+1\right)\right)^2\\ &\quad+\left(\frac{d}{du}\left(3\sin(u)+u^2+3\right)\right)\,\left(\frac{d^2}{dx^2}\left(4x^2+1\right)\right)\\ &=\left(\frac{d}{du}\left(3\cos(u)+2u\right)\right)\left(8x\right)^2+(3\cos(u)+2u)\frac{d}{dx}\left(8x\right)\\ &=\left(-3\sin(u)+2\right)64x^2+(3\cos(u)+2u)8\\ &=\left(-3\sin\left(4x^2+1\right)+2\right)64x^2+\left(3\cos\left(4x^2+1\right)+2\left(4x^2+1\right)\right)8\\ &\,\,\color{blue}{=-192x^2\sin\left(4x^2+1\right)+24\cos\left(4x^2+1\right)+192x^2+16}\tag{2} \end{align*}

On the other hand we obtain \begin{align*} \frac{d^2 y}{dx^2}=&\frac{d^2}{dx^2}\left(3\sin\left(4x^2+1\right)+\left(4x^2+1\right)^2+3\right)\\ &=\frac{d}{dx}\left(8x\cdot 3\cos\left(4x^2+1\right)+2\left(4x^2+1\right)8x\right)\\ &=\frac{d}{dx}\left(24x\cos\left(4x^2+1\right)+64x^2+16x\right)\\ &=\left(24\cos\left(4x^2+1\right)+24x\left(-\sin\left(4x^2+1\right)\right)8x\right)+192x^2+16\\ &\,\,\color{blue}{=-192x^2\sin\left(4x^2+1\right)+24\cos\left(4x^2+1\right)+192x^2+16} \end{align*} in accordance with (2).

Answer 2

Here's a way of writing it without the dee-stuff; I think that it may help you see that you are not really trapped in a circular argument.

We're talking about derivatives of the function $f\circ g:x\mapsto f(g(x))$.

The Chain Rule is this: $(f\circ g)'=(f'\circ g)g'$.

The Product Rule is this: $ (\phi\psi)'=\phi' \psi +\phi\psi'$.

We will then have $$ \begin{eqnarray*} (f\circ g)''&=&((f\circ g)')'&\\ &=&((f'\circ g)g')'&\text{Chain Rule}\\ &=&(f'\circ g)'g'+(f'\circ g)g''&\text{Product Rule}\\ &=&(f''\circ g)(g')^2+(f'\circ g)g''&\text{Chain Rule.}\\ \end{eqnarray*} $$