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Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ if $f(x,y) = (x^2 +y^2)\text{log}(x^2+y^2)$.

Ok, to my understanding, $\frac{\partial f}{\partial x}$ will just be the derivative keeping all terms besides $x$ fixed.

I use the logarithm base-change law: $\text{log}(x^2+y^2) = \frac{\text{ln}(x^2+y^2)}{\text{ln}(10)}$.

Then $\frac{\partial f}{\partial x} = \frac{(2x+y^2)\text{ln}(x^2+y^2)}{\text{ln}(10)} + \frac{1}{\text{ln}(10)}(\frac{2x+y^2}{x^2+y^2})(x^2+y^2) = \frac{2x+y^2}{\text{ln}(10)}[\text{ln}(x^2+y^2)+1]$.

Similarly, $\frac{\partial f}{\partial y} = \frac{2y+x^2}{\text{ln}(10)}[\text{ln}(x^2+y^2)+1]$. Please tell me if I did this correctly.

My real question is if $f$ is differentiable, which is not required for my homework, but I am curious about this case. At $(x,y)=(0,0)$, the partials are discontinuous. A theorem in my book states:

Let $f:U \subset \mathbb{R}^n \to \mathbb{R}^m$. Suppose the partial derivatives $\partial f_i / \partial x_j$ of $f$ all exist and are continuous in a neighborhood of a point $\textbf x \in U$. Then $f$ is differentiable at $\textbf x$.

If the partials of $f$ are discontinuous at $(x,y) = (0,0)$, does this imply that $f$ is not differentiable as a whole? If we state the condition $(x,y) \neq (0,0)$, does $f$ become differentiable? Please explain (preferably in terms of neighborhoods).

Bobby Lee
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2 Answers2

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I am afraid that your derivatives are wrong.

They should be $$\frac{\partial f}{\partial x}=\frac{2 x \left(\log \left(x^2+y^2\right)+1\right)}{\log (10)}$$ and $$\frac{\partial f}{\partial y}=\frac{2 y \left(\log \left(x^2+y^2\right)+1\right)}{\log (10)}$$ I hope this will allow you to continue.

Claude Leibovici
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  • Wow, I'm dumb. When I read about the partial derivative I didn't realize "fixed" meant "constant", so obviously $y^2$ gets eliminated with $\frac{\partial f}{\partial x}$... – Bobby Lee Feb 07 '14 at 17:50
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In your example $f$ is not defined in $(0, 0)$ and thus it is not differentiable in $(0, 0)$.

dani_s
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  • A function might be undefined at a point but still extend differentiably to this point. For example $\sin(x)/x$. But not in the our case though. – Gil Bor Feb 07 '14 at 18:06
  • @GilBor It is still not differentiable, by definition you need $f(x)$ to be defined for $f$ to be differentiable at $x$. – dani_s Feb 07 '14 at 18:32
  • Correct, but when a function is not defined at some point, we me be able, sometimes, to extend its definition in a differentiable manner. For example, the formula $f(x)=\sin(x)/x$ does not define $f$ at $x=0$, but by defining $f(0)=1$ the new function is differentiable at $x=0$. This may or may not be possible to do for a given function undefined at a point. It turns out that for the function given by the OP it is impoosible to extend it diffrentiably (or even continiously) to $(0,0)$. – Gil Bor Feb 07 '14 at 19:02