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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.

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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

1 Answer 1

I’d prefer to give a hint, but I can’t think of one that doesn’t give the game away. It’s a straightforward computation: by definition

$$\begin{align*}D_{tx}f(a)&=\lim_{s\to 0}\frac{f(a+stx)-f(a)}s\\\\ &=t\lim_{s\to 0}\frac{f(a+stx)-f(a)}{st}\\\\ &=t\lim_{st\to 0}\frac{f(a+stx)-f(a)}{st}\\\\ &=tD_xf(a)\;. \end{align*}$$

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 $tD_{tx}f(a)$.

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+1. A nitpick though. You might want to argue the case $t=0$ separately. –  user17762 Jun 13 '12 at 3:53
    
@Marvis: You’re absolutely right; thanks. –  Brian M. Scott Jun 13 '12 at 3:55

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