directional derivative of an expression

Question

let $f:\mathbb{R}^n \times\mathbb{R}^m \rightarrow \mathbb{R}^m$ and $f(x,y)=0$ where $x\in \mathbb{R}^n $ and $y\in \mathbb{R}^m$. then suppose that we can write $y$ as a function of $x$ like $g(x)$ (in a small interval)

so $h(x)=f(x,g(x))=0$.

Now I see a statement in this book thatI can't understandt:

$D_xh = 0 $

$Dh = D_xf + D_yf.D_xg = 0$

my first question: what is the meaning of the first line $D_xh = 0 $ (what is the meaning of $D_xh$) and how he concluded it?

second question: how he concluded the second statement $Dh = D_xf + D_yf.D_xg = 0$ ?

Answer 1

First question

$D_x h = 0$ means that the partial derivative $\frac{\partial}{\partial x}$ of $h$ is always vanishing. This is the consequence that $h$ is supposed to be always vanishing. By the way $h$ is a function of $x$ only so the partial derivative is equal to the partial derivative.

Second question

The (Fréchet) derivative of a map $(x,y) \mapsto f(x,y)$ is $$Df.(k,l) = D_x f.k+D_y f.l$$

You then get the formula $Dh = D_xf + D_yf.D_xg = 0$ using the chain rule for derivatives.

Answer 2

Hint.

The directional derivative can be taken as

$$ Df_u=<\nabla f, \vec u > $$

Now $\vec u$ is tangent to $f(x,y)=0$ because is taken over $(x,g(x))$ and $\nabla f$ is normal to the tangent plane to $f(x,y)=0$ so obviously

$$ <\nabla f, \vec u > = 0 $$