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Here is my question: Find all the points $\left ( x,y \right )$ in $\mathbb{R}^{2}$ where the following function is differentiable: $f\left ( x,y \right )=\left | e^{x}-e^{y} \right |.\left ( x+y-2 \right )$

The only thing that came to my mind is to distinguish three cases:$x> y$, $x and $x=y$. In each case the function $f$ can be written differently. But, no idea how to find the points where the function is differentiable. Any one can help?

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    You know that $f$ is differentiable on the open set $\mathcal U=\{(x,y)\in\mathbb R^2, x\neq y\}$. So you have to look at the differentiability at $(x_0,x_0)$. Compute $\lim_{h\to 0^+}f(x_0+h,x_0)$ and $\lim_{h\to 0^-}f(x_0+h,x_0)$.2012-01-29
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    @Davide Giraudo: I agree with the first part that $f$ is differentiable on the open set $U$. As for the limits, they are both equal to zero. Can you please elaborate more why do you have to compute these limits and how ...?2012-01-29
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    Sorry I meant $\lim_{h\to 0^+}\frac{f(x_0+h,x_0)-f(x_0,x_0)}h$, $\lim_{h\to 0^-}\frac{f(x_0+h,x_0)-f(x_0,x_0)}h$ in order to see whether the partial derivative exist.2012-01-29
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    @Davide Giraudo The first limit is equal to: $e^{x_{0}}.(2x_{0}-2)$ and the second one is equal to $-e^{x_{0}}.(2x_{0}-2)$. The only case where these two are equal is at $x_{0}=1$. What does this mean? Does it mean that $f$ is only differentiable at the point $(1,1)$?2012-01-29
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    It means that $f$ has partial derivatives at $(x,y)=(1,1)$ (since the computation of the derivative in $y$ will give the same result). Now you have to check weather $f$ is differentiable at $(1,1)$.2012-01-29

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