Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to do a problem which wants me to compute the directional derivatives at $(0, 0)$ of $$f(x, y) = \frac{xy}{\sqrt{x^2 + y^2}}, \quad f(0, 0) = 0.$$

There are two equations I know for computing the directional derivative, and them seem to be inconsistent for some reason.

Here are the two formulas I have for directional derivatives, $(a, b)$ a unit vector $$D_{(a, b)}f(x, y) = \lim_{t \to 0} \frac{f(x + ta, y + tb) - f(x, y)}{t}.$$

$$D_{(a, b)}f(x, y) = aD_xf(x, y) + bD_yf(x, y).$$ where $D_x$ and $D_y$ are the partial derivatives.

Computing the partials derivatives I get $$D_x(0, 0) = \lim_{t \to 0} \frac{f(t, 0) - f(0, 0)}{t} = \lim_{t \to 0} \frac{\displaystyle\frac{t\cdot 0}{\sqrt{t^2 + 0^2}} - 0}{t} = 0$$

$$D_y(x, 0) = \lim_{t \to 0} \frac{f(0, t) - f(0, 0)}{t} = \lim_{t \to 0} \frac{\displaystyle\frac{0\cdot t}{\sqrt{0^2 + t^2}} - 0}{t} = 0$$

Which in turn should imply $D_{(a, b)}(0, 0) = a\cdot 0 + b\cdot 0 = 0$.

But computing with the limit formula $$D_{(a, b)}f(x, y) = \lim_{t \to 0} \frac{f(ta, tb) - f(0, 0)}{t} = \lim_{t \to 0}\frac{\displaystyle\frac{(ta)(tb)}{\sqrt{(ta)^2 + (tb)^2}} - 0}{t} = \lim_{t \to 0}\frac{t^2 ab}{t^2\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}} = ab \neq 0$$ if $a \neq 0$ and $b \neq 0$.

I imagine I am doing something silly, or I have a formula wrong or something. But I haven't been able to correct myself yet. Can anyone tell me where my reasoning is going wrong?

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

For $u=(a,b) \ne 0$ and $t \ne 0$ one has $$ \frac{f(0+tu)-f(0)}{t}=\frac{f(tu)}{t}=\frac{t^2ab}{t\sqrt{t^2(a^2+b^2)}}=\frac{t}{|t|}f(u). $$ If $ab=0$, then $D_uf(0)=0$. If $ab \ne 0$, since the limit of $t/|t|$ as $t$ tends to 0 doesn't exist, then the directional derivative does not exist.

share|improve this answer
    
Ah thank you, that clarifies where my issue was. $\sqrt{t^2} = |t|$. –  Danikar Jun 30 '12 at 0:07
add comment

The directional derivatives exist but this function "puckers" at the origin. The actual derivative does not exist. Since it's homogeneous of degree 0, it cannot be continuous at the origin.

share|improve this answer
    
So when the derivative doesn't exist the second formula, $D_{(a, b)}(x, y) = aD_x(x, y) + bD_y(x, y)$ is not correct? I believe that is the one that uses differentiability to derive. –  Danikar Jun 30 '12 at 0:03
1  
If the derivative does not exist, you can't write $D_{(a,b)}(x,y)$, 'cause it does not exist. –  ncmathsadist Jun 30 '12 at 0:08
    
Well certainly it still exists in the case where $(a, b) = (1, 0)$ and $(a, b) = (0, 1)$ since those just coincide with $D_x$ and $D_y$. –  Danikar Jun 30 '12 at 0:14
    
Antecedent problem. All of the directional derivatives exist but the function is discontinuous at the origin. This is a classic counterexample. –  ncmathsadist Jun 30 '12 at 0:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.