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I am working on a question which asks when the fuction $f: \mathbb{R}^2 \rightarrow \mathbb{R}: (x,y) \mapsto x \exp(|y|) $ is differentiable, and where it is differentiable.

I can quite easily see that it is not differentiable in the points $(a,0)$ where $a\neq0$, as it would be the function would be of the form $a\exp(|0|)$ and $\exp(|0|)$ is not differentiable according to: Is $e^{|x|}$ differentiable?.

My issue comes with the point $(0,0)$. Can anyone help with how I would show it is differentiable or not in this point?

MathsBBB
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You can just use the definition... In order for $f$ to be differentiable at $(0,0)$ you must have $$ \lim_{(x,y)\to (0,0)}\dfrac{f(x,y)-f(0,0)-f'_x(0,0) x - f'_y(0,0) y}{\sqrt{x^2+y^2}} = 0. $$

Since, $f(0,0)=0$, $f'_x(0,0) = 1$ and $f'_y(0,0)=0$, you just need to check if $$ \lim_{(x,y)\to (0,0)}\frac{f(x,y)-x}{\sqrt{x^2+y^2}} = 0. $$

The above limit is fact 0, because

$$ \left| \frac{f(x,y)-x}{\sqrt{x^2+y^2}}\right|=\left|\frac{x(e^{|y|}-1)}{\sqrt{x^2+y^2}}\right|\leq e^{|y|}-1 \to 0 , \quad (x,y \to 0) $$

and so you conclude that $f$ is differentiable at $(0,0)$.

PierreCarre
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