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I have to find out if the function $f: \mathbb{R^2} \to \mathbb{R}$ with $f(x,y) :=$ min{x,y} and the function $g:\mathbb{R^n} \to \mathbb{R}$ with $g(x) := \left\lVert x \right\rVert_2$ are partially differentiable. Furthermore, I have to find out the gradients $\nabla f(a)$ and $\nabla g(a)$ for all points $a$, where they exist.

I know that $\min \text {{x,y}} = \frac{1}{2} (x+y-|x-y|)$.

$\left\lVert x \right\rVert_2$ = $\sqrt {\sum_{k=1}^n x_i^2 }$

So the partial derivate $\frac{\partial g}{\partial x_i}$ for a generic $x_i$ would be

$$\frac{\partial f}{\partial x_i} = \frac{1}{2\sqrt{\sum_{k=1}^n x_i^2}}\cdot2x_i = \frac{x_i}{\left\lVert x \right\rVert}$$ and the gradient of g is $\nabla g= \frac{1}{\left\lVert x \right\rVert}\cdot x$

Though I think that a proof requires more...

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$f$ has partial derivative w.r.t. $x$ at all points $(x,y)$ with $x \neq y$ , it has partial derivative w.r.t. $y$ at all points $(x,y)$ with $x \neq y$, and $g$ has no partial derivative at $0$. At other points your calculations are OK. Justification: for a fixed real number $a$ the function $\max \{x,a\}$ is differentiable at all point except $a$. It is not differentiable at $a$ because the right hand derivative at $$ is $1$ and the left hand derivative is $0$. The rest should be clear.

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  • $\begingroup$ A point here is a point $(x,y)$. The question is, at which points $(x,y)\in{\mathbb R}^2$ the function $(x,y)\mapsto f(x,y):=\min\{x,y\}$ is partially differentiable. $\endgroup$ – Christian Blatter Nov 8 '18 at 17:34
  • $\begingroup$ @ChristianBlatter I have edited my answer. $\endgroup$ – Kavi Rama Murthy Nov 8 '18 at 23:23

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