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How would you show that for the directional derivative $D_vf(p)$ of $f$ at location $p$ with respect to $v$ the following formula holds for $c \in \mathbb{R}$ $D_{cv}f(p) = cD_vf(p)\, ?$

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    @AlexP: That is only true if the gradient of $f$ at $p$ exists.2012-06-18

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The definition of the directional derivative is: $ D_vf(p)=\lim_{h\rightarrow 0} \frac{f(p+hv)-f(p)}{h}. $ Then we have: $ D_{cv}f(p)=\lim_{h\rightarrow 0} \frac{f(p+hcv)-f(p)}{h}=\lim_{h\rightarrow 0} c\frac{f(p+hcv)-f(p)}{ch} $ $ =c\lim_{h'\rightarrow 0}\frac{f(p+h'v)-f(p)}{h'}=cD_vf(p) $ Where $h'=ch$ and because $h\rightarrow 0$ iff $ch\rightarrow 0$.

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    Ah of course! Silly me2012-05-17