Why is this the second derivative here?

Question

Let $F\colon\mathbb{R}^{2}\to\mathbb{R}$ and $f\colon\mathbb{R}\to\mathbb{R}$ be two $\mathcal{C}^{1}$ functions. Lets say we have a neighborhood of a point $\left(x_{0},y_{0}\right)$ such that for every $\left(x,y\right)$ we have $F\left(x,y\right)=0$ if and only if $y=f\left(x\right).$ So we can say that in this neighborhood we have $F\left(x,f\left(x\right)\right)=0$ and if we let $g\colon\mathbb{R}\to\mathbb{R}^{2}$ be defined by $g\left(x\right)=\left(x,f\left(x\right)\right)$ then $F\left(g\left(x\right)\right)=0$ and therefore \begin{align*} 0 & =\left[F\left(g\left(x\right)\right)\right]'=D_{F\circ g}\left(x\right)=D_{F}\left(g\left(x\right)\right)\cdot D_{g}\left(x\right)=\\ & =\left(\frac{\partial F}{\partial x}\left(x,f\left(x\right)\right),\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)\right)\cdot\begin{pmatrix}1\\ f'\left(x\right) \end{pmatrix}=\\ & =\frac{\partial F}{\partial x}\left(x,f\left(x\right)\right)+\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)\cdot f'\left(x\right) \end{align*} $$ \Rightarrow\qquad f'\left(x\right)=-\frac{\frac{\partial F}{\partial x}\left(x,f\left(x\right)\right)}{\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)} $$ Im trying to compute $f''\left(x\right)$. I was thinking it would be something like $$ f''\left(x\right)=-\frac{\frac{\partial^{2}F}{\partial x^{2}}\left(x,f\left(x\right)\right)\cdot\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)-\frac{\partial F}{\partial x}\left(x,f\left(x\right)\right)\cdot\frac{\partial^{2}F}{\partial x\partial y}\left(x,f\left(x\right)\right)}{\left[\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)\right]^{2}} $$ but I know it is wrong. The correct answer is $$ f''\left(x\right)=-\frac{\left[\frac{\partial^{2}F}{\partial x^{2}}\left(x,f\left(x\right)\right)+\frac{\partial^{2}F}{\partial x\partial y}\left(x,f\left(x\right)\right)\cdot f'\left(x\right)\right]\cdot\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)-\frac{\partial F}{\partial x}\left(x,f\left(x\right)\right)\cdot\left[\frac{\partial^{2}F}{\partial x\partial y}\left(x,f\left(x\right)\right)+\frac{\partial^{2}F}{\partial y^{2}}\left(x,f\left(x\right)\right)\cdot f'\left(x\right)\right]}{\left[\frac{\partial F}{\partial y}\left(x,f\left(x\right)\right)\right]^{2}} $$ and I don't understand why.

Any help please?

Answer 1

The point is that $$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}(x,f(x))\right) \color{red}{\neq}\frac{\partial^2F}{\partial x^2}(x,f(x)), $$but$$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}(x,f(x))\right) =\frac{\partial^2F}{\partial x^2}(x,f(x))\cdot 1+\frac{\partial^2F}{\partial x\partial y}(x,f(x))\cdot f'(x) $$instead. Similar for the other derivative.

Answer 2

To answer the question in the comments (and indirectly, the problem):

Consider the functions $$G: \mathbb{R}^2 \to \mathbb{R}, \ G: (x,y)^T\mapsto G(x,y) \\ H:\mathbb{R}\to\mathbb{R}^2, \ H: x\mapsto (H_1(x),H_2(x))^T$$

Then, $G\circ H: \mathbb{R}\to\mathbb{R}, \ G\circ H: x\mapsto G(H_1(x),H_2(x))$. The chain rule$^1$ states that

$$ (G\circ H)'(x) = (G'\circ H)(x)\cdot H'(x) $$

where $(\cdot) '$ is the $m\times n$ matrix of partial derivatives $\partial_j(\cdot)_i$ of a function $(\cdot):\mathbb{R}^n\to\mathbb{R}^m$ and $(\cdot)' = \partial(\cdot)$ for $n = m = 1$.

So, $$G'(x) = ( \partial_1G(x,y),\partial_2G(x,y) ), \\ (G'\circ H)(x) = (\partial_1 G(H_1(x),H_2(x)),\partial_2G(H_1(x),H_2(x)), \\ H'(x) = (\partial H_1(x),\partial H_2(x))^T$$ Multiplying, we see that

$$ \partial (G\circ H)(x) = \partial_1 G(H_1(x),H_2(x))\partial H_1(x) + \partial_2 G(H_1(x),H_2(x))\partial H_2(x)$$

Let $H_1 = E$ (the identity function), $H_2 = f$, and $G = \partial_1 F$. Then,

$$\partial_k G = \partial_k \partial_1 F \\ \partial H_1 = 1 \\ \partial H_2 = \partial f$$

and so

$$ \frac{\partial}{\partial x} \left(\frac{\partial F}{\partial x}(x,f(x))\right) = \\ \partial(\partial_1 F\circ (E,f))(x) = (\partial_1\partial_1 F\circ (E,f))(x) + (\partial_2\partial_1 F\circ (E,f))(x)\cdot\partial f(x) \\ = \frac{\partial^2F}{\partial x^2}(x,f(x)) + \frac{\partial^2 F}{\partial y\partial x}(x,f(x))\cdot f'(x)$$

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1. Spivak, Calculus on Manifolds pp. 19, "2-2 Theorem (Chain Rule)"