# Thread: Simple question about directional derivative

1. ## Simple question about directional derivative

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

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

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

2. ## Re: Simple question about directional derivative

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

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

So my question is, why is it true that if there exists an open set $\displaystyle W \subseteq U$ such that $\displaystyle p\in W$ and $\displaystyle f|_W = g|_W$, then $\displaystyle D_{v,p}f = D_{v,p}g$ ?
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 $\displaystyle W$ is some ball $\displaystyle B_\delta(p)$. Now, choose $\displaystyle \varepsilon>0$ be be arbitrary, then choosing $\displaystyle t$ so small that $\displaystyle p+tv\in B_\delta$ we see that $\displaystyle f(p+tv)=g(p+tv)$ but we may also choose $\displaystyle t$ so small that $\displaystyle |D_v f(p)-\Delta_t f(p)\|,\|D_v g(p)-\Delta_t g(p)\|<\varepsilon$ (where $\displaystyle \Delta_t f(p)$ is the difference qutotient at $\displaystyle p$ with increment $\displaystyle t$) we see then that choosing $\displaystyle t$ to be small enough we have that

\displaystyle \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 $\displaystyle t$ we have that $\displaystyle \Delta_t f(p)=\Delta_t g(p)$ since they agree on $\displaystyle W$. But, since $\displaystyle \varepsilon$ was arbitrary, equality follows.

3. ## Re: Simple question about directional derivative

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

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

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

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

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

4. ## Re: Simple question about directional derivative

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

5. ## Re: Simple question about directional derivative

Originally Posted by slevvio
haha looks what I just wrote is a derigorised version of what you've put
Haha, right!