1. ## [SOLVED] Partial derivative

The problem is:
Prove that the mixed partial derivative f xy (0,0) is not equal to the mixed partial derivative f yx (0,0) ,
given the function:
f(x,y) = x.y when |x|>=|y|
= -x.y when |x|<|y|

Friends, I have tried a few times but am always landing up by proving the opposite, ie, f xy (0,0) = f yx (0,0), which is not right.
One has to use the first principles to find these derivatives at the point (0,0).
If one uses the differentiation formulae then one lands up with 0/0 undefined form.

Kindly help.

2. Originally Posted by fanofandrew
Prove that the mixed partial derivative f xy (0,0) is not equal to the mixed partial derivative f yx (0,0) ,
given the function:
f(x,y) = x.y when |x|>=|y|
= -x.y when |x|<|y|
For starters, we have to calculate the first partial derivtives:
$f_x(x,y) = \begin{cases}y&\text{if }|x|>|y|,\\-y&\text{if }|x|<|y|;\end{cases}\qquad\quad f_y(x,y) = \begin{cases}x&\text{if }|x|>|y|,\\-x&\text{if }|x|<|y|.\end{cases}$

(If |x| = |y| ≠ 0 then f(x,y) is discontinuous and so the partial derivatives are not defined.) At the origin, we have to calculate the derivatives from first principles: $f_x(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = 0,$ and similarly $f_y(0,0)=0.$

For the second partial derivatives at the origin, we again use first principles. Notice that $f_x(0,k) = -k$ for all k, and $f_y(h,0) = h$ for all h. Therefore $f_{xy}(0,0) = \lim_{k\to0}\frac{f_x(0,k)-f_x(0,0)}k = \lim_{k\to0}\frac{-k}k = -1.$ But $f_{yx}(0,0) = \lim_{h\to0}\frac{f_y(h,0)-f_y(0,0)}h = \lim_{h\to0}\frac hh = 1.$

3. ## sincere thanks

Dear sir (Opalg),
Thank you very much for the reply.
Your explanation is lucid as it stands. There is but one point which I cant understand. Why is it, that sometimes the first principle method helps to avoid 0/0 undefined form,(hence skirts inconclusiveness to give a definite derivative) but at other times it is the using of derivative formulae which gives a definite answer (as you have done here: you have used both: derivation formula for the first derivative and first principle for the second derivative) and the first principle does not help at all? For here, fx (0,k) by first principle will give 0 as the answer, while by formula it is -k as you have pointed out.
Why this dichotomy?
Yours truly,
fanofandrew

4. Originally Posted by fanofandrew
Why is it, that sometimes the first principle method helps to avoid 0/0 undefined form,(hence skirts inconclusiveness to give a definite derivative) but at other times it is the using of derivative formulae which gives a definite answer (as you have done here: you have used both: derivation formula for the first derivative and first principle for the second derivative) and the first principle does not help at all?
The rules for differentiating formulas apply at any point where the function is given by that formula throughout some neighbourhood of the point. But if the point is at the boundary of the region where the function is defined by that formula, the rule for differentiating it cannot be applied, and you have to argue from first principles.

For the function given by $f(x,y) = \begin{cases}xy&\text{if }|x|\geqslant|y|\\-xy&\text{if }|x|<|y|,\end{cases}$ you have to calculate derivatives from first principles at any point (x,y) for which |x|=|y|, because such points are on the boundary between the two domains of definition.

Originally Posted by fanofandrew
For here, fx (0,k) by first principle will give 0 as the answer, while by formula it is -k as you have pointed out.
Not true. Both methods will give the same result $f_x(0,k) = -k$ provided that k≠0. When k=0 you have to argue from first principles.

5. But if the point is at the boundary of the region where the function is defined by that formula, the rule for differentiating it cannot be applied, and you have to argue from first principles.
Dear sir,
Thank you for this insight which is not available in the undergraduate books I am consulting. May I make so bold as to ask why this is so? Is it from the consideration of left and right handed limits? Since, any point on the boundary of the region of definition of the function may not answer to any one of the left or right handed limit, and consequently fail to have a limiting value. Because,
for $lim_{x\to{a}}f(x)=l$to be true, it must be
$lim_{x\to{a-0}}f(x)=l_1=l=l_2=lim_{x\to{a+0}}f(x)$

If that is so, then how does first principle deliver the goods because there too the $lim_{h\to0}$ implies both $lim_{h\to0+}$ and $lim_{h\to0-}$?

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### partial derivatives from the first principle

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