- #1

joypav

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$\frac{f(a+h)-f(a)-Bh}{\left| h \right|}\rightarrow 0$ as $h\rightarrow 0$

($B$ is called the derivative of $f$ at $a$)The definition seems a-ok, but I am trying to just work a simple problem so that I can see it in action and I am stuck.

Say $f(x,y)=(xy, x^2)$. Obviously, $f: R^2 \rightarrow R^2$.

And let $B = \left[\begin{array}{c} a & b \\ c & d \end{array}\right]$

We are in $R^2$ so $h = \left[\begin{array}{c}h_1 \\ h_2 \end{array}\right]$.

$f(x+h_1, y+h_2) - f(x,y) = \left[\begin{array}{c}(x+h_1)(y+h_2) \\ (x+h_1)^2 \end{array}\right]-\left[\begin{array}{c}xy \\ x^2 \end{array}\right]=\left[\begin{array}{c}xh_2+yh_1+h_1h_2 \\ 2xh_1+h_1^2 \end{array}\right]$

$B\cdot h = \left[\begin{array}{c} a & b \\ c & d \end{array}\right]\cdot \left[\begin{array}{c}h_1 \\ h_2 \end{array}\right] = \left[\begin{array}{c} ah_1+bh_2 \\ ch_1+dh_2 \end{array}\right]$

$f(x+h_1, y+h_2) - f(x,y) - B\cdot h = \left[\begin{array}{c}xh_2+yh_1+h_1h_2-ah_1-bh_2 \\ 2xh_1+h_1^2-ch_1-dh_2 \end{array}\right]$

We need to choose $a, b, c, d$ so that,

$\frac{1}{\sqrt{h_1^2+h_2^2}} \cdot \left[\begin{array}{c}xh_2+yh_1+h_1h_2-ah_1-bh_2 \\ 2xh_1+h_1^2-ch_1-dh_2 \end{array}\right] \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$

Meaning, we need,

$\frac{xh_2+yh_1+h_1h_2-ah_1-bh_2}{\sqrt{h_1^2+h_2^2}} \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$

and

$\frac{2xh_1+h_1^2-ch_1-dh_2}{\sqrt{h_1^2+h_2^2}} \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$

**OKAY**, now that that's all out there, I need help! First of all, is there a simple way to figure this out from what I have?

From previous things I've worked on, I believe I can just find $B$ with partial derivatives. So it

**should be**

$B = \left[\begin{array}{c} y & x \\ 2x & 0 \end{array}\right]$

How do I go about getting this? Or do I just need to show that this $B$ will work?