Differentiating a function Let $f(x_1,\dots,x_n)$ be a $C^{\infty}$ real valued function.
I want to prove the following.
$$\frac{d(f(tx)}{dt}=\sum_{i=1}^n\frac{df}{dx_i}(tx_1, \dots, tx_n) \cdot x_i$$.
How can I prove this? What kind of theorem do I need to use?
I checked this with one variable and calculated for several $f$ but I don't know how to prove it.
Thank you.
( I don't know what is the good title for this question.)
 A: You have a function $f:\Bbb R^n\to\Bbb R$; and the function $g:\Bbb R\to\Bbb R^n$ defined by $t\mapsto tx$. Note that $f\circ g:\Bbb R\to\Bbb R$. The chain rule says that $$\frac{d}{dt}(f\circ g)'(t)=\nabla f(tx)\cdot \frac{d}{dt}(tx)=\nabla f(tx)\cdot x$$
But $$\nabla f(tx)=\sum_{i=1}^nD_if(tx)e_i\;\;,\;\;x=\sum_{i=1}^n x_ie_i$$ so $$\nabla f(tx)\cdot x=\sum_{i=1}^n D_if(tx)x_i$$

Recall that the chain rule is the statement relating the total derivatives of $f\circ g$, $f$ and $g$. That is, if $g$ is differentiable at $a$, $f$ is differentiable at $g(a)$ then $f\circ g$ is differentiable at $a$ and $$D(f\circ g)(a)=Df(g(a))\circ Dg(a)$$
 (or what amounts  to multiplying the corresponding matrices)
In particular, when $f$ is a scalar function on $\Bbb R^n$, $Df(a)=\nabla f(a)(\;\cdot \;)$.
A: I think you're looking for something more than "applying chain rule". 
Here is a draft of the proof in dimension 2:
Fix $x$ $y$ and also $h$.
$$
\frac{f(tx+ hx,ty+hy)-f(tx,ty)}{h}=\\= \frac{f(tx+ hx,ty+hy)-f(tx+ hx,ty)+f(tx+hx,ty)-f(tx,ty)}{h}
$$
Define $F(s)=f(tx+hx,s)$ and $G(r)=f(r,ty)$ (observe that $F'=f_y$ and $G'=f_x$).
Using Barrow's rule:
$$
\frac{f(tx+ hx,ty+hy)-f(tx+ hx,ty)}{h}=\frac{F(ty+hy)-F(ty)}{h}=\frac{1}{h}\int_0^1F'(ty+shy)\,hy \,ds \\\int_0^1f_y(tx+hx,ty+shy)\,y\,ds
$$
and
$$
\frac{f(tx+ hx,ty)+f(tx+hx,ty)-f(tx,ty)}{h}=\frac{G(tx+hx)-G(tx)}{h}=\frac{1}{h}\int_0^1G'(tx+rhx)\,hx\,dr \\ =\int_0^1f_x(tx+rhx,ty)\,x\,dr
$$
Now let h tend to zero.
