# Are $f(x,y) :=$ min{x,y} and $g:\mathbb{R^n} \to \mathbb{R}$ with $g(x) := \left\lVert x \right\rVert_2$ partially differentiable?

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...

$$f$$ 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.
• 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. – Christian Blatter Nov 8 '18 at 17:34