Simple question about directional derivative

**slevvio** · October 10th 2011, 03:15 PM

Hello, I just have a question about something that was left as an exercise in my notes. It seems really obvious to me but I can't work out how to prove it for sure. Can anybody give me a hand.

Let $f,g: U \rightarrow \mathbb{R}$ where $U \subseteq \mathbb{R}^n$ is open. Then let $p \in U$ . Then define the directional derivative of $f$ at $p$ to be

$D_{v,p}f = \operatorname{\frac{d}{dt}}\bigg|_{t=0}f(p+tv)$ .

So my question is, why is it true that if there exists an open set $W \subseteq U$ such that $p\in W$ and $f|_W = g|_W$ , then $D_{v,p}f = D_{v,p}g$ ?

**Drexel28** · October 10th 2011, 03:24 PM

Originally Posted by slevvio

Hello, I just have a question about something that was left as an exercise in my notes. It seems really obvious to me but I can't work out how to prove it for sure. Can anybody give me a hand.

Let $f,g: U \rightarrow \mathbb{R}$ where $U \subseteq \mathbb{R}^n$ is open. Then let $p \in U$ . Then define the directional derivative of $f$ at $p$ to be

$D_{v,p}f = \operatorname{\frac{d}{dt}}\bigg|_{t=0}f(p+tv)$ .

So my question is, why is it true that if there exists an open set $W \subseteq U$ such that $p\in W$ and $f|_W = g|_W$ , then $D_{v,p}f = D_{v,p}g$ ?

Intuitively because derivatives are local operations. If two functions look locally the same then they should have the same derivative. We may assume without loss of generality that $W$ is some ball $B_\delta(p)$ . Now, choose $\varepsilon>0$ be be arbitrary, then choosing $t$ so small that $p+tv\in B_\delta$ we see that $f(p+tv)=g(p+tv)$ but we may also choose $t$ so small that $|D_v f(p)-\Delta_t f(p)\|,\|D_v g(p)-\Delta_t g(p)\|<\varepsilon$ (where $\Delta_t f(p)$ is the difference qutotient at $p$ with increment $t$ ) we see then that choosing $t$ to be small enough we have that

$\begin{aligned}\|D_v f(p)-D_v g(p)\| &\leqslant \|D_v f(p)-\Delta_t f(p)\|+\|D_v g(p)-\Delta_t f(p)\|\\ &=\|D_v f(p)-\Delta_t f(p)\|+\|D_v g(p)-\Delta_t g(p)\|\\ &<2\varepsilon\end{aligned}$

Where we used the fact that how we chose $t$ we have that $\Delta_t f(p)=\Delta_t g(p)$ since they agree on $W$ . But, since $\varepsilon$ was arbitrary, equality follows.

**slevvio** · October 10th 2011, 03:37 PM

I thought about this problem a bit more and we are really asking to show that

$\operatorname{\frac{d}{dt}}(f\circ F)(0) = \operatorname{\frac{d}{dt}}(g\circ F)(0)$

where $F: t \mapsto p + tv$ . That is,

$\displaystyle\lim_{h \rightarrow 0} \frac{f\circ F(h) - f\circ F(0)}{h} = \displaystyle\lim_{h \rightarrow 0} \frac{g\circ F(h) - g\circ F(0)}{h}$

But since we care about the limit we can simply choose $h$ small enough so that $p+hv$ is contained in some ball contained in $W$ and hence these two limits are equal

**slevvio** · October 10th 2011, 03:38 PM

haha looks what I just wrote is a derigorised version of what you've put

**Drexel28** · October 10th 2011, 03:39 PM

Originally Posted by slevvio

haha looks what I just wrote is a derigorised version of what you've put

Haha, right!

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