# A question From Munkres' Analysis on Manifolds (P. 48, E1)

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

So

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

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.
• In (a), what about $u = (0,u_{2})$ or $(u_{1},0)$? – IamWill Nov 5 '19 at 21:17
• @Willy.K Sorry ? I couldn't understand. – user295645 Nov 5 '19 at 21:21
• 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? – IamWill Nov 5 '19 at 21:24
• 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)). – IamWill Nov 5 '19 at 21:29
• 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. – 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.
• 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. – user295645 Nov 5 '19 at 21:54
• @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. – zhw. Jul 17 '20 at 17:18