# Multivariable Derivative and Partial Derivatives

If $L_p(x)=c+\sum_{i=1}^n b_ix^i$ is a derivative of $f$ at $p$, then $\displaystyle b_i=\frac{\partial f}{\partial x^i}(p),$ $1\leq i\leq n$. In particular, if $f$ is differentiable at $p$, these partial derivatives exist and the derivative $L_p$ is unique.

Proof:
Let $h=x-p$, then $\lim_{x\to p}\frac{f(x)-L_p(x)}{\|x-p\|}=0$ becomes $\displaystyle \lim_{h\to 0}\frac{f(p+h)-f(p)}{\|h\|}-\lim_{h\to 0}\frac{L_p(h)-c}{\|h\|}=0$ since $L_p(p+h)=L_p(p)+L_p(h)-c=f(p)+L_p(h)-c$.

By choosing $h=(0,\dots,h^i,\dots, 0)$ (all zeroes except in $i$th position), then as $h\to 0$, $\displaystyle \lim_{h\to 0}\frac{f(p+h)-f(p)}{\|h\|}=\frac{\partial f}{\partial x^i}(p)$ and $\displaystyle \lim_{h\to 0}\frac{L_p(h)-c}{\|h\|}=\lim_{h^i\to 0}\frac{b_ih^i}{h^i}=b_i.$ So $b_i=\frac{\partial f}{\partial x^i}(p)$. ## Author: mathtuition88

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