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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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As a hint: consider that the directional derivative is the dot product of v with the gradient of f evaluated at p, and then think about what properties of the dot product you know. It may help to write out an explicit example in order to visualize it. – Alex Petzke May 17 '12 at 12:06
@AlexP: That is only true if the gradient of $f$ at $p$ exists. – Jonas Meyer Jun 18 '12 at 4:05

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, thank you but what about showing $ D_{v+w}f(p)= D_{v}f(p) + D_{w}f(p)$? I get $\lim_{h\rightarrow 0} \frac{f(p+h(v+w))-f(p)}{h} =\lim_{h\rightarrow 0} \frac{f(p+hv+hw))-f(p)}{h} =\lim_{h\rightarrow 0} \frac{f(p+hv+hw))-f(p)}{2h} +\lim_{h\rightarrow 0} \frac{f(p+hv+hw))-f(p)}{2h}$ but then I get stuck? – user26069 May 17 '12 at 12:26
Try instead to write $f(p+h(v+w))-f(p)$ as $f(p+h(v+w))-f(p+hv)+f(p+hv)-f(p)$, as use the fact that the sum of the limits is the limit of the sum (when they exist!). – guaraqe May 17 '12 at 12:31
Wait, but how are they equal if the function isn't linear? ($f:\mathbb{R}^n \to \mathbb{R}^m$) – user26069 May 17 '12 at 12:40
Its just a sum and a subtraction of the same term, it does not depend on linearity! – guaraqe May 17 '12 at 13:39
Ah of course! Silly me – user26069 May 17 '12 at 13:43

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