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Math Help - Simple question about directional derivative

  1. #1
    Senior Member slevvio's Avatar
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    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 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 ?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Simple question about directional derivative

    Quote Originally Posted by slevvio View Post
    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.
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  3. #3
    Senior Member slevvio's Avatar
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    Re: Simple question about directional derivative

    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
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  4. #4
    Senior Member slevvio's Avatar
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    Re: Simple question about directional derivative

    haha looks what I just wrote is a derigorised version of what you've put
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Re: Simple question about directional derivative

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