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I want to show that if $f:\mathbb{R}^n\to \mathbb{R}$ and $df_a$ is the derivative of the function at $a$ then $df_a(v)=\displaystyle\frac{\partial f}{\partial v}(a)$. I saw a few proofs of this proposition, but all of them were pretty much the same (using the same definition for derivability) and I wanted do a proof using another definition.

I have a sketch, but I believe it may have some troubles. Instead of proving that $df_a(a)=\displaystyle\frac{\partial f}{\partial v}(a)$ I show that $df_a(tv)=\displaystyle\frac{\partial f}{\partial tv}(a)$ considering the limit

$$\lim_{tv\to 0}\frac{|f(a+tv)-f(a)-\frac{\partial f}{\partial tv}(a)|}{|tv|}\\=\lim_{t\to 0}\frac{|f(a+tv)-f(a)-t\frac{\partial f}{\partial v}(a)|}{|t||v|} \\\leq\lim_{t\to 0}\frac{|f(a+tv)-f(a)-t\frac{\partial f}{\partial v}(a)|}{|t|}\\=\lim_{t\to 0}\left|\frac{f(a+tv)-f(a)-t\frac{\partial f}{\partial v}(a)}{t}\right|\\=\left|\lim_{t\to 0}\frac{f(a+tv)-f(a)}{t}-\frac{\partial f}{\partial v}(a)\right|=0.$$

Follows that $df_a(tv)=\displaystyle\frac{\partial f}{\partial tv}$ and since $df_a(tv)=tdf_a(v)$ and $\displaystyle\frac{\partial f}{\partial tv} = t \displaystyle\frac{\partial f}{\partial v}$ is $df_a(v)=\displaystyle\frac{\partial f}{\partial v}$. Now I'm unsure about the second step, whether is legitimate to change $tv\to 0$ by $t\to 0$, which would make no differencefor $f(a+tv)$ but I have some doubts if this remains true for the partial derivative as well.

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Is $f$ a function of one variable? – 5xum Jul 16 '14 at 14:06
@5xum No, $f:\mathbb{R}^n\to \mathbb{R}$ – Cure Jul 16 '14 at 15:59
up vote 2 down vote accepted

I suppose you want to show the directional derivative can be defined as follows: $$ \frac{\partial}{\partial v}f(x)=\lim_{t\rightarrow 0}\frac{f(x+tv)-f(x)}{t} $$ where $v$ is usually taken to be a unit vector. The above limit can be computed via some textbook tricks like $$ f(x+x',y+y')-f(x,y+y')+f(x,y+y')-f(x,y)\approx x'\frac{\partial}{\partial x}f(x,y+y')+y'\frac{\partial}{\partial y}f(x,y)+o(x'^{2}+y'^{2}) $$for the two dimensional case. For the general case it is similar. For this to work we need to assume that all the first order derivatives are continuous, so when the vector $(x',y')$ is small enough the above approximates $x'\frac{\partial}{\partial x}f(x,y)+y'\frac{\partial}{\partial y}f(x,y)$ as desired. I think there must be better ways to do this, though.

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