In the $f:\mathbb{R^2}\rightarrow \mathbb{R}$ case it is helpful to consider the direction vector $u=(\cos\theta, \sin\theta)$ in order to quickly check all directional derivatives by simply adjusting for the angle. for example, in the function $f(x,y)=\dfrac{x^2y} {\sqrt{x^2+y^2}}$, switching to polar coordinate gives us $f(\cos\theta,\sin\theta)=\cos^2\theta\sin\theta$ How can we generalize this to higher dimensions? I had the question of determining differentiablity of the function I just gave as an example and this simplification made the problem a lot easier, so I ask because in case where I have to prove that a function isn't differentiable I would like to look for contradictions to the theorem "If $f$ is differentiable the the directional derivative at point $a$ in direction $v$ is equal to the gradient computed at $a$ multiplyed by $v$".


In two variables it looks ok to use polar coordinates $(sin \theta, cos\theta)$ .You can also manage three dimensional polar co-ordinates but higher (for n- sphere) things get messy. But for higher variables you can to use definiton of derivative. That implies

$$‖\frac{f(a+h,b+k)-f(a,b)-\nabla{f}.‖(h,k)‖}{‖(h,k)‖}‖\to 0$$This is for two variables (Checking differentiability at $(a,b)$ and $(h,k)$ is a point on the ball close to $(a,b)$ )but you can generalize .

To prove that a function is not differentiable your observation is right .

Further there is a theorem that "If all partial derivatives exist and continuos in a neighborhood of $x$ , then $f$ is differentiable at $x$."


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