# Directional derivative question

Let $f: \mathbb{R}^n \to \mathbb{R}$. For $x \in \mathbb{R}^n$, the limit $$\lim_{s \to 0} \frac{f(a + sx) - f(a)}{s}$$ if it exists is called the directional derivative of $f$ at $a$ in the direction $x$ and is denoted $D_x f(a)$. I want to show that $D_{tx}f(a) = tD_xf(a)$. Help is appreciated.

-
You seem to be using the variable $t$ in two different contexts. It would be good to have another variable. –  user17762 Jun 13 '12 at 3:50
That actually makes things a lot clearer. I was misinterpreting this problem. –  Lambert Jun 13 '12 at 3:59
\begin{align*}D_{tx}f(a)&=\lim_{s\to 0}\frac{f(a+stx)-f(a)}s\\\\ &=t\lim_{s\to 0}\frac{f(a+stx)-f(a)}{st}\\\\ &=t\lim_{st\to 0}\frac{f(a+stx)-f(a)}{st}\\\\ &=tD_xf(a)\;. \end{align*}
Added: Note that this makes sense only if $t\ne 0$. If $t=0$, the difference quotient is identically $0$, since $f(a+stx)=f(a)$, and its limit is therefore $0$, which is indeed $tD_{tx}f(a)$.
+1. A nitpick though. You might want to argue the case $t=0$ separately. –  user17762 Jun 13 '12 at 3:53