MHB Derivatives in Higher Dimensions

[joypav]

Looking at Munkres "Analysis on Manifolds", it says for $A\subset R^n, f: A \rightarrow R^m$ suppose that $A$ contains a neighborhood of $a$. Then $f$ is differentiable at $a$ if there exists an $n$ by $m$ matrix $B$ such that,

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

 

[I like Serena]

Hey joypav,

We can write:
$$\lim_{(h_1,h_2)\to 0}\frac{xh_2+yh_1+h_1h_2-ah_1-bh_2}{\sqrt{h_1^2+h_2^2}}
=\lim_{(h_1,h_2)\to 0}\frac{(x-b)h_2+(y-a)h_1+h_1h_2}{\sqrt{h_1^2+h_2^2}} = 0$$
For the limit to be zero, it needs to be zero on any path that $(h_1,h_2)$ can take to zero.
So for instance along the path where $h_1=0$ and $h_2\to 0^+$.
Suppose $x-b\ne 0$. Then along that path, the limit is $x-b\ne 0$, which is a contradiction. Therefore we must have $b=x$.
Similarly we must have $a=y$.
We can prove that the remaining expression has indeed the limit $0$.

The same reasoning applies to the other limit.And yes, one might also show that the $B$ you found with partial derivatives just works.
Do note that it's possible that a partial derivative does not exist, even though we can still find the derivative of $f$ at a point.
And a matrix with partial derivatives is also not guaranteed to work in all cases, even if it exists.
In other words, the definition that Munkres gives is certainly relevant.Btw, I hope you don't mind, but I'd classify this as Calculus, so I've moved the thread there.

 

[joypav]

That makes a lot more sense. Thank you!