Consider the function $f(x,y)=\begin{cases}     \frac{xy}{x^2+y^2}, &amp; \text{if $(x,y)\ne(0,0)$}\\     0, &amp; \text{otherwise}   \end{cases}$  Question: Is it differentiable everywhere? Does it have partial derivatives everywhere?  My attempt:  $f$ is differentiable on $\Bbb R^2\backslash\{(0,0)\}$ because it is the ratio between two differentiable functions and the denominator doesn't vanish. Is it still the case in $(0,0)$?  If we write $x=r\cos\theta$ and $y=r\sin\theta$ then the function becomes   $\begin{cases}     \frac{r^2\cos\theta\sin\theta}{r^2} &amp; \text{if } r\ne0 &amp; \\      0 &amp;\text{if } r=0 \end{cases}$ = $\begin{cases} \cos\theta\sin\theta &amp;  \end{cases}$   We realize that $f$ doesn't depend on the radius, so the fact that $f(1,1)=1/2\ne-1/2=f(-1,1)$ so we can approach zero by decreasing the radius and preserving the values $1/2$ and one hand and $-1/2$ on the other. So the function is discontinuous at zero, so not differentiable.  Does it have partial derivatives everywhere?  $$\frac{\partial f}{\partial x}=\lim_{h\to\ 0}\frac{f(h,0)}{h}=0$$ $$\frac{\partial f}{\partial y}=\lim_{h\to\ 0}\frac{f(0,h)}{h}=0$$ So the partial derivatives exist in zero and elsewhere.
54805
Let $F:\mathbb{R^2}\rightarrow \mathbb{R}$ be a function, such as : $$F(x,y)=\begin{cases}        0 &amp; (x,y)=(0,0) \\       \frac{xy}{x^2+y^2} &amp; (x,y)\neq (0,0)    \end{cases}$$      Is $F$ differentiable at $(0,0)$?   My attempt :  Consider the limit, $\lim_{(x,y)\rightarrow (0,0)} F(x,y)$ along the line $y=x$. The limit equals to $\frac{1}{2}$. However, $F(0,0)=0$. Hence $F$ is not continuous at $(0,0)$ which implies that $F$ isn't differentiable as well.  Is this approach okay ?
12544
Let  $$f(x,y)=\frac{x^3-y^3}{x^2+y^2}$$  My solution manual says that this function is not diffb. in $(0,0)$ because it is not linear. Well my problem is that I don't see why this function is linear, and I also don't see why that would imply that $f$ is not differentiable.
35296
Show that $f(x,y)=\dfrac{xy^2}{x^2+y^2}$ (with $(x,y)\not=(0,0)$ and $f(0,0)=0$) is continuous but not differentiable at $(0,0)$.  I tried to show continuity with an $\epsilon -\delta$ argument but I don't know how to factorize the expression so that I can have something useful.   For differentiability I think I should show that the partials are not continuous at $(0,0)$. But finding the partials is also painful.
29487
Problem : Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable?   For the sake of generality, let's assume that it is unknown to us that $|x|$ is not differentiable at $x = 0$  Attempted Solution :  Using the definition of differentiability, a function is differentiable over an interval $I$ $\text{iff}$  $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \ \ \forall x\in I$$  Now implicitly this definition of differentiability requires $$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}$$  Therefore $f$ will be differentiable only when  $$\lim_{h\to 0^+} \frac{(x + h + |x+h|)-(x + |x|)}{h} = \lim_{h\to 0^-} \frac{(x - h + |x-h|)-(x + |x|)}{h}$$  But it is unclear what to do next as $x$ is just arbitrary $$ \begin{equation} \begin{aligned} |x+h| &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq -h\\ -x-h &amp; \text{if} &amp; x &lt; -h \end{cases} \\ &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq 0\\ -x-h &amp; \text{if} &amp; x &lt; 0 \end{cases} \end{aligned} \end{equation}$$    A Wrong Solution :  We could take the derivative of $f$, (and evaluate the domain of the derivative $f'(x)$ to values for which it is defined), using the rules of differentiation, but we would get a wrong answer.  $$f'(x) = \frac{|x| + x}{|x|}$$ $$\implies f'(x) = \begin{cases} 2 &amp; \text{if} &amp; x \geq 0 \\ 0 &amp; \text{if} &amp; x &lt; 0 \end{cases}$$  This implies the derivative $f'(0)$ exists, when it does not as per the definition of differentiability. Why is that so?    Questions:   How can a solution be found using the definition of differentiablity? Why does taking the derivative of $f$, $f'(x)$ and evaluating it's domain, not give the correct values of $x$ for which $f$ is differentiable?
77539
Is $f(x,y) = |xy|$ differentiable in $(0,0)$? I have no idea how to approach this problem.
8592
I had an exam today, and I was asked about the inverse function theorem, and the exact conditions and statement (as stated in Mathematical Analysis by VA Zorich): Let $X, Y \subset \mathbb{R}$ be open sets and let the functions $f: X \to Y$ and $f^{-1}: Y \to X$ be mutually inverse and continuous at points $x_{0} \in X$ and $y_{0}=f(x_{0})$, respectively. If $f$ is differentiable at $x_{0}$ and $f'(x_{0}) \neq 0$, then $f^{-1}$ is also differentiable at $y_{0}$  and its derivative is $$(f^{-1})'(y_{0})=(f'(x_{0}))^{-1}=\frac{1}{f'(x_{0})}.$$  Then I was asked to come up with an example which would show that the condition that $f^{-1}$ be continuous at $y_{0}$ is not redundant, i.e. that there exists a differentiable bijection $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, but whose inverse $f^{-1}$ is non-differentiable at some point $y_{0} \in \mathbb{R}$ (my professor stated that both the domain and range are $\mathbb{R}$, but I'd be open to any example whose domain/range are open subsets of $\mathbb{R}$). I couldn't come up with an answer on the spot, so I was left with a homework assignment, which I'm shamelessly asking for help with here.  Here are some examples that almost, but don't quite fit the bill:   $f(x)=x^3, f:\mathbb{R} \to \mathbb{R}$, whose inverse $f^{-1}(y)=\sqrt[3]y$ is not differentiable at $0$, but the problem is that $f'(0)=0$. If $f'(x) \neq 0$ were removed as a conditions, many counterexamples could be easily found, because any bijection $f$ whose derivative is zero at $x$ implies that $f^{-1}$ is not differentiable at $f(x)$. An example which is possibly closer to what I'm looking for is the one found in the answer of , which fits the bill completely except for the domain/range, because $f^{-1}$ is not continuous at 1, let alone differentiable, but its domain is not an open or connected set, so it wouldn't be easy to extend its domain to $\mathbb{R}$ and still keep all of its properties.   I'm aware that the inverse $f^{-1}$ of a continuous bijection $f$ defined on an interval ($\mathbb{R}$ in this case) is also continuous, so the "pathological inverse" that I'm looking for is continuous (unlike example 2), but not differentiable (i.e. "spiky") at a point, even though $f$ isn't "spiky" anywhere.
72711
If f: $\mathbb{R}^n \rightarrow \mathbb{R}$ is a convex function, i heard that f is almost everywhere differentiable. Is it true? I can't find a proof (n-dimentional).  Thank you for any help
83617
I've taken multivariate calculus and am wondering if I can see a specific function where the order of taking the partial derivative matters. I've been told that there are some exceptions where $ \dfrac{\partial ^2 f}{\partial x \partial y} \ne \dfrac{\partial ^2 f}{\partial y \partial x} $, so I'm curious to see what this looks like.  EDIT: And why would this true?
111053
Is there a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) = 0$ when $x &lt; 0$ and $f'(x) = 1$ when $x \geq 0$?   Attempt: I was trying to think of an example. First thing that came to mind was $$ f(x) =  \begin{cases} x &amp; x \geq 0 \\ 0 &amp; x &lt; 0 \end{cases}. $$ I wanted to know if it is differentiable at $x = 0$. Since $$ \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = 0 = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x} $$ the derivative at $x = 0$ exists and equals $0$. But according to the problem, the derivative should be $1$, so I guess this is not a correct example. Does anyone have an example, or is such a function not possible at all?
24804
I want to show that $f(x,y) = \sqrt{|xy|}$ is not differentiable at $0$.  So my idea is to show that $g(x,y) = |xy|$ is not differentiable, and then argue that if $f$ were differentiable, then so would $g$ which is the composition of differentiable functions $\cdot^2$ and $g$.  But I'm stuck as to how to do this.  In the one variable case, to show that $q(x) = |x|$ is not differentiable, I can calculate the limit $\frac{|x + h| - |x|}{h}$ as $h\to 0^+$ and $h\to 0^-$, show that the two one-sided limits are distinct, and conclude that the limit $$\lim_{h\to 0}\frac{|x + h| - |x|}{h}$$ does not exist.  The reason this is easier is that I do not have to have in mind the derivative of the function $q$ in order to calculate it.  But in the case of $g(x,y) = |xy|$, to show that $g$ is not differentiable at $0$, I would need to show that there does not exist a linear transformation $\lambda:\mathbb{R}^{2}\to\mathbb{R}$ such that  $$\lim_{(h,k)\to (0,0)}\frac{\left||hk| - \lambda(h,k)\right|}{|(h,k)|} = 0$$  I thought of assuming that I had such a $\lambda$, and letting $(h,k)\to (0,0)$ along both $(\sqrt{t},\sqrt{t})$ and $(-\sqrt{t},-\sqrt{t})$ as $t\to 0^{+}$, but this didn't seem to go anywhere constructive.
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Let $f$ be a function defined on the interval $(-1,1)$ such that for all $x,y\in(-1,1)$,  $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$.  Suppose that $f$ is differentiable at $x=0$.  Show that $f$ is differentiable on $(-1,1)$.  For me I can see that $\tan(x)$ is obviously a case of $f$, but I cannot find a general proof for this $f$. Could any kind soul help?
89844
Consider a function: $$f:\mathbb R ^2\to\mathbb R $$ when does $$\dfrac {\partial f(x,t)}{\partial t\partial x}=\dfrac {\partial f(x,t)}{\partial x\partial t}$$  Thinking about it in terms of the limit definition of derivative,  and thinking about it as taking slices of surfaces and measuring the slope on the edge, seems to be giving me the feeling that answer is always.  However I have some memories of there being requirements like piecewise continuous, and smooth.   What is the full set of conditions?
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Suppose the function $f:[0,\delta) \to \mathbb{R}$ is continuous, differentiable in $(0,\delta)$ and $f(0)=0$.  If the limit $\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L$ exists, then is it always the case that $L = 0$.  This seems to be true both for functions with well-behaved derivatives such as $f(x) = x,$  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,$$  as well as functions with "bad" derivatives such as   $$f(x) = \begin{cases}x \ln x &amp;\mbox{if }x&gt;0 ,\\0  &amp;\mbox{if } x=0,  \end{cases},$$  where  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$  Is there a simple proof or counterexample?
31835
Consider the function $$f(x)=\left\{\begin{array}{cc} x^2-4 &amp; \text{ if }x\leq 2\\4x+3&amp;\text{ if } x\gt 2\end{array}\right.$$ This function is differentiable at $x=2$ since $\lim_{h\to 0^{\pm}}\frac{f(2+h)-f(2)}{h}=4$ (EDIT: this actually isn't true but it is true that $\lim_{x\to 2^-}f^\prime (x)=4=\lim_{x\to 2^+}f^\prime(x)$); however, it's not continuous at $x=2$. How is that possible, doesn't differentiability at $x=a$ imply continuity at $x=a$? This question came up when I tried to answer the question of finding $a$ and $b$ such that the function $$f(x)=\left\{\begin{array}{cc} ax^2-b &amp; \text{ if }x\leq 2\\bx+3&amp;\text{ if } x\gt 2\end{array}\right.$$ The solution is achieved by finding conditions on $a$ and $b$ such that it's continuous, and also such that the left/right derivates exist. The left/right derivative question gives $4a=b$. With the condition of continuity you get the additional condition that $4a-b=2b+3$, giving a unique solution. But doesn't differentiability imply continuity? What's wrong with just solving $4a=b$ like in the first example above?
33120
The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^
50786
I was reading a solution when I came across this statement.     So $$\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}.$$ Then there exists a differentiable function $g$ on $\mathbb{R}^n$ such that $\frac{\partial g}{\partial x_i}=f_i$.   Why is this true?
11555
Let $f(x,y)=     \begin{cases}       \frac{2x^2y}{x^2+y^2} &amp; if &amp; (x,y) \neq (0,0) \\       0       &amp; if &amp; (x,y)=(0,0) .  \end{cases}$  (i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.  (ii)¿Is $f$ continuous on (0,0)?  For (i) I took $u=(u_{1},u_{2}) \in \mathbb{R^{2}}$ such $\|u\|=1$ and $0=(0,0)$. So $$\lim_{t \to 0} \frac{f(0+ t u)-f(0)}{t}=\lim_{t \to 0}\frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=\lim_{t \to 0}\frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}.$$ But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?  For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $\lbrace (\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}$ and $\lbrace (\frac{1}{k},0) \rbrace_{k \in \mathbb{N}}$ which are two different sequences in $\mathbb{R}^{2}$ converging to $(0,0)$. However, $\lbrace f(\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}=\lbrace \frac{1}{k^{2}} \rbrace_{n \in \mathbb{N}} \to 0$ and $\lbrace f(\frac{1}{k},0) \rbrace_{k \in \mathbb{N}} \to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?