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 ith 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).

Advertisements

About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s