Studio funzione in 2 variabili
Ciao a tutti! 
Volevo proporvi il seguente esercizio:
Data la funzione $ f(x,y)={ ( xysen(1/(xy)); xy!=0),( 0;xy=0 ):} $
Se ne determinino i punti di continuità, derivabilità, differenziabilità.
Chiaro che lo studio va fatto in $(0,0)$
Ora, per la continuità pensavo che, qualunque sia l'argomento, $ |sen(x)|<=|x| $, da cui discende in modo immediato la continuità.
Per la derivabilità: $ { ( phi(x) = f(x,0) = 0 ),( psi(y)=f(0,y)=0 ):} $
$ phi'(0) = psi'(0)= \partial_(x)f(0,0) = \partial_(y)f(0,0)$
Da cui discende la derivabilità di f.
Per la differenziabilità, essendo $f(0,0)=0$, ottengo:
$ lim_((x,y)rarr(0,0)) (xysen(1/(xy)))/sqrt(x^2+y^2) $
che per il discorso del seno di prima diventa: $|(xysen(1/(xy)))/sqrt(x^2+y^2)|<=|(xy)/sqrt(x^2+y^2)|$
ed in coordinate polari: $|rhosenthetacostheta|<=rhorarr0$ ovviamente per $rhorarr0$
Il che implica la differenziabilità.
Volevo il vostro parere, perchè mi sembra troppo facile così
Nota: Come si mettono gli spazi bianchi in ASCII?
Volevo proporvi il seguente esercizio:
Data la funzione $ f(x,y)={ ( xysen(1/(xy)); xy!=0),( 0;xy=0 ):} $
Se ne determinino i punti di continuità, derivabilità, differenziabilità.
Chiaro che lo studio va fatto in $(0,0)$
Ora, per la continuità pensavo che, qualunque sia l'argomento, $ |sen(x)|<=|x| $, da cui discende in modo immediato la continuità.
Per la derivabilità: $ { ( phi(x) = f(x,0) = 0 ),( psi(y)=f(0,y)=0 ):} $
$ phi'(0) = psi'(0)= \partial_(x)f(0,0) = \partial_(y)f(0,0)$
Da cui discende la derivabilità di f.
Per la differenziabilità, essendo $f(0,0)=0$, ottengo:
$ lim_((x,y)rarr(0,0)) (xysen(1/(xy)))/sqrt(x^2+y^2) $
che per il discorso del seno di prima diventa: $|(xysen(1/(xy)))/sqrt(x^2+y^2)|<=|(xy)/sqrt(x^2+y^2)|$
ed in coordinate polari: $|rhosenthetacostheta|<=rhorarr0$ ovviamente per $rhorarr0$
Il che implica la differenziabilità.
Volevo il vostro parere, perchè mi sembra troppo facile così
Nota: Come si mettono gli spazi bianchi in ASCII?
Risposte
Niente? 
The continuity doesn't follow from $|\sin(x)|\leq |x|$ but from $|x\sin(1/x)|\leq |x|$.
On the other hand, if you want to study the differentiability of the function $f$ on $(0,0)$ you have to find one (the only one) linear application $L$ such that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L[(x,y)-(0,0)]}{\|(x,y)\|}=0.$$
I'll show you the step-by-step way to do it:
This $L$ is completely determined by the partial derivatives $\partial_x f(0,0)$ and $\partial_y f(0,0)$ in the following way: $L(x,y)=\partial_xf(0,0)x+\partial_yf(0,0)y$. So the only thing you have to do is to compute these derivatives.
Now, $$\partial_x f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0,$$
and the same happens with $\partial_yf(0,0)$. Then, the only candidate to be the differential of $f$ at $(0,0)$ is the null function $L(x,y)=0$.
Let's check it. First, let us observe that the candidate to be the limit is $0$ because, when $x=0$ or $y=0$, the function is identically $0$ and so is the limit of the definition of differentiability. So, we have to check if it is true that
$$0=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{\|(x,y)\|}=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$
even when we approach the point $(0,0)$ with points $(x,y)$ such that $xy\ne 0$.
But, if $xy\ne 0$,
$$\left|\frac{f(x,y)}{\|(x,y)\|}\right|=\left|\frac{xy\sin\left(\frac{1}{xy}\right)}{\|(x,y)\|}\right|\leq \frac{|xy|}{\|(x,y)\|}\leq \frac{|x|\|(x,y)\|}{\|(x,y)\|}=|x|,$$ which goes to $0$ when $(x,y)\to 0$. When $xy=0$, the function is identically $0$, so for these points is also clear that the limit is $0$. Then, the function is differentiable at $(0,0)$ and its differential is $L(x,y)=0$.
This is the step-by-step way to do it. If you check your final argument, you'll realise that what you have is that
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\left|\frac{|\rho^2\cos\theta\sin\theta|}{\rho}\right|\leq \rho\to0.$$
Probably you wanted to say that thing. Anyways, you have to be careful with this kind of inequalities.
On the other hand, if you want to study the differentiability of the function $f$ on $(0,0)$ you have to find one (the only one) linear application $L$ such that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L[(x,y)-(0,0)]}{\|(x,y)\|}=0.$$
I'll show you the step-by-step way to do it:
This $L$ is completely determined by the partial derivatives $\partial_x f(0,0)$ and $\partial_y f(0,0)$ in the following way: $L(x,y)=\partial_xf(0,0)x+\partial_yf(0,0)y$. So the only thing you have to do is to compute these derivatives.
Now, $$\partial_x f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0,$$
and the same happens with $\partial_yf(0,0)$. Then, the only candidate to be the differential of $f$ at $(0,0)$ is the null function $L(x,y)=0$.
Let's check it. First, let us observe that the candidate to be the limit is $0$ because, when $x=0$ or $y=0$, the function is identically $0$ and so is the limit of the definition of differentiability. So, we have to check if it is true that
$$0=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{\|(x,y)\|}=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$
even when we approach the point $(0,0)$ with points $(x,y)$ such that $xy\ne 0$.
But, if $xy\ne 0$,
$$\left|\frac{f(x,y)}{\|(x,y)\|}\right|=\left|\frac{xy\sin\left(\frac{1}{xy}\right)}{\|(x,y)\|}\right|\leq \frac{|xy|}{\|(x,y)\|}\leq \frac{|x|\|(x,y)\|}{\|(x,y)\|}=|x|,$$ which goes to $0$ when $(x,y)\to 0$. When $xy=0$, the function is identically $0$, so for these points is also clear that the limit is $0$. Then, the function is differentiable at $(0,0)$ and its differential is $L(x,y)=0$.
This is the step-by-step way to do it. If you check your final argument, you'll realise that what you have is that
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\left|\frac{|\rho^2\cos\theta\sin\theta|}{\rho}\right|\leq \rho\to0.$$
Probably you wanted to say that thing. Anyways, you have to be careful with this kind of inequalities.
Wow, amazing answer! Thanks a lot and yes, those inequalities are so tricky!
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