Recall. Definition. Let $A \subset \mathbf{R}^{m} ;$ let $f: A \rightarrow \mathbf{R}^{n} .$ Suppose $A$ contains a neighborhood of a. Given $\mathbf{u} \in \mathbf{R}^{m}$ with $\mathbf{u} \neq 0$, define $$ f^{\prime}(\mathbf{a} ; \mathbf{u})=\lim _{t \rightarrow 0} \frac{f(\mathbf{a}+t \mathbf{u})-f(\mathbf{a})}{t} $$

provided the limit exists.

  • Let $f: \mathbf{R}^{2} \rightarrow \mathbf{R}$ be defined by setting $f(\mathbf{0})=0$ and $$ f(x, y)=x y /\left(x^{2}+y^{2}\right) \text { if }(x, y) \neq 0 $$

(a) For which vectors $\mathbf{u} \neq 0$ does $f^{\prime}(0 ; \mathbf{u})$ exist? Evaluate it when it exists.

(b) Do $D_{1} f$ and $D_{2} f$ exist at $0$ ?

(c) Is $f$ differentiable at $0$ ?

(d) Is $f$ continuous at $0$ ?

My Attempt.

$a)$. Let $u=(u_1,u_2)$. Then

$$ f^{\prime}(\mathbf{0} ; \mathbf{u})=\lim _{t \rightarrow 0} \frac{f(\mathbf{0}+t \mathbf{u})-f(\mathbf{0})}{t} =\lim _{t \rightarrow 0}\dfrac {u_1u_2} {t(u_1^2+u_2^2)} $$

So since the limit does not exist, then $f^{\prime}(0 ; \mathbf{u})$ does not exist.

$b)$ $\frac{d f}{d x}=\frac{y\left(-x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}$ and $\frac{d f}{d y}=\frac{x\left(x^{2}-y^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}.$


$$\frac{d f}{d x}(0,0)=\lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x,y)}{h}=\lim _{h \rightarrow 0} \frac{f\left(h,0\right)-f\left(0,0\right)}{h}=\lim _{h \rightarrow 0} \frac{0}{h}=0$$

Similarly, $$\frac{d f}{d y}(0,0)=0.$$

c) Since partial derivatives exists at $0$ and continous at $0$, then $f$ is differentaiable at $0$.

d) Yes, $f$ is continuous at $0$.

  1. Please, may you check my attemp? Thanks...
  2. If you take $u$ to be $u=\left(u_{1}, 0\right)$ with $u_{1} \neq 0$ then the directional derivative seems to exist in (a)? Thanks @Willy.k for the question.
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    $\begingroup$ In (a), what about $u = (0,u_{2})$ or $(u_{1},0)$? $\endgroup$ – IamWill Nov 5 '19 at 21:17
  • $\begingroup$ @Willy.K Sorry ? I couldn't understand. $\endgroup$ – user295645 Nov 5 '19 at 21:21
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    $\begingroup$ I mean, if you take $u$ to be $u=(u_{1},0)$ with $u_{1}\neq 0$ then the directional derivative seems to exist in (a), right? $\endgroup$ – IamWill Nov 5 '19 at 21:24
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    $\begingroup$ I think too! Besides, note that the partial derivatives are nothing but directional derivatives, with $u = (1,0)$ and $u=(0,1)$, so this shows both partial derivatives exits (this proves (b)). $\endgroup$ – IamWill Nov 5 '19 at 21:29
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    $\begingroup$ Remember that the computations of the derivative are well-defined only if we know the these derivatives exist. So, you should prove the existence of the partial derivatives before calculating them. $\endgroup$ – IamWill Nov 5 '19 at 21:32

(a) As pointed out in a comment, $f'(0,u)=0$ if $u_1=0$ or $u_2=0.$ Your idea for other $u$ is correct.

(b) You shouldn't write $\dfrac{f(x+h,y)-f(x,y)}{h}.$ What are $x,y?$ The basic argument is correct however.

(c) This is not correct. Why do you think the partial derivatives are continuous at $0?.$

d) No, not continuous at $0.$ Proof: $f(x,x) =1/2$ for $x\ne 0,$ but $f(0) =0.$ There's no hope.

  • $\begingroup$ For $a),c)$ and $d)$ you are right, yess. Thanks. But for $b)$, this is just definiton of partial derivative, that is: Let $f(x, y)$ be a function of two variables. Then we define the partial derivatives as $$ f_{x}=\frac{\partial f}{\partial x}=\lim _{h \rightarrow 0} \frac{f(x+h, y)-f(x, y)}{h} $$ $f_{y}=\frac{\partial f}{\partial y}=\lim _{h \rightarrow 0} \frac{f(x, y+h)-f(x, y)}{h}$ if these limits exist. $\endgroup$ – user295645 Nov 5 '19 at 21:54
  • $\begingroup$ @JamesEnsor but we are looking at $(0,0).$ Should be $(f(0+h,0)-f(0,0))/h$ Also the best argument here is to just note $f=0$ on both axes. $\endgroup$ – zhw. Jul 17 '20 at 17:18

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