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Math Help - Directional derivatives, not differentiable function

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    MHF Contributor arbolis's Avatar
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    Directional derivatives, not differentiable function

    Let f(x,y)= \begin{cases} \frac{x|y|}{\sqrt{x^2+y^2}} \hspace{2cm} \text{if (x,y)} \neq (0,0) \\  0 \hspace{3.1cm} \text{if (x,y) =0}   \end{cases}.
    Demonstrate that f has directional derivatives in all directions in (0,0) but that f is not differentiable in (0,0).
    My attempt :
    I notice I can't use the theorem that states that if f is differentiable in a point, then it has directional derivatives in all directions in this point.
    I believe f is continuous in (0,0)... I didn't prove it via the definition but I've failed at showing that it is not continuous. I realize that it would have been easy if f wasn't continuous because f could not be differentiable but this is not the case.
    I'm stuck at starting the exercise.
    I guess I have to use the definition to show that f is not differentiable, i.e. that there does not exist a linear function \varphi such that \lim _{\vec h \to 0} \frac{f(\vec {x_0} + \vec h)-f(\vec x_0)- \varphi (\vec h)}{\| h \|}=0.
    Seems quite hard, I don't know how to start.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Let f(x,y)= \begin{cases} \frac{x|y|}{\sqrt{x^2+y^2}} \hspace{2cm} \text{if (x,y)} \neq (0,0) \\ 0 \hspace{3.1cm} \text{if (x,y) =0} \end{cases}.
    Demonstrate that f has directional derivatives in all directions in (0,0) but that f is not differentiable in (0,0).
    My attempt :
    I notice I can't use the theorem that states that if f is differentiable in a point, then it has directional derivatives in all directions in this point.
    I believe f is continuous in (0,0)... I didn't prove it via the definition but I've failed at showing that it is not continuous. I realize that it would have been easy if f wasn't continuous because f could not be differentiable but this is not the case.
    I'm stuck at starting the exercise.
    I guess I have to use the definition to show that f is not differentiable, i.e. that there does not exist a linear function \varphi such that \lim _{\vec h \to 0} \frac{f(\vec {x_0} + \vec h)-f(\vec x_0)- \varphi (\vec h)}{\| h \|}=0.
    Seems quite hard, I don't know how to start.
    the first part of the problem is straightforward. for the second part, suppose f is differentiable at the origin. then there must exist functions g,h : \mathbb{R}^2 \longrightarrow \mathbb{R}, which are defined in some punctured

    neighbourhood of the origin, \lim_{(x,y)\to(0,0)}g(x,y)= \lim_{(x,y)\to(0,0)}h(x,y)=0, and for all (x,y) sufficiently close to (0,0) we have: f(x,y)=f(0,0) + xf_x(0,0) + yf_y(0,0) + xg(x,y)+yh(x,y). since

    f(0,0)=f_x(0,0)=f_y(0,0)=0, we get: \frac{x|y|}{\sqrt{x^2+y^2}}=f(x,y)=xg(x,y)+yh(x,y)  . put y=x to get \frac{1}{\sqrt{2}}=g(x,x)+h(x,x), which is a contradiction because \lim_{x\to0}g(x,x)=\lim_{x\to0}h(x,x)=0.
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