$$f(x,y)=\begin{cases} \frac{\tan x}{x}+y, & 0<|x|<1 \\ 1+y,& x=0 \\ \end{cases}$$

Prove that it is differentiable on $(-1,1) \times \mathbb R$. I use the Frechet definition of differentiability.

I run into trouble with this type of assignment because the usual route that is taken is the following:

1.) Finding the partial derivatives on the main part of the domain(in this case $x \neq 0$). Seeing if the partial derivatives are continuous, in which case they are here.Conclusion: It is differentiable on that part of the domain.

2.) I find the partial derivative on the point of the domain that is not already evaluated in 1.) like so: $$ \frac{\partial f}{\partial x}(0,y)= \lim_{h \to0} \frac{f(h,y)-f(0,y)}{h}=\lim_{h\to0}\frac{\frac{\tan h}{h}+y-(1+y)}{h}=\lim_{h\to 0}\frac{\tan h -h}{h^2}=????$$

$$ \frac{\partial f}{\partial x}(0,y)= \lim_{h \to0} \frac{f(0,y+h)-f(0,y)}{h}=\lim_{h\to0}\frac{1+y+h-(1+y)}{h}=\lim_{h\to 0}1=1$$

3.)Then from there I would traditionally do:

$$f((0,y)+(h_1,h_2))-f(0,y)= \frac{\partial f}{\partial x}(0,y)h_1+ \frac{\partial f}{\partial y}(0,y)h_2+ R(h),$$ then when plugging in the limits I found I have to sort out $R(h)$ proving that $\frac{R(h)}{\|h\|}\to 0.$ As you can tell, it's not going to plan, as I do it this way, what am I not seeing?

Definition of differenciability:

Let $X$ and $Y$ be normed vector spaces upon the same field $\mathbb R$ or $\mathbb C$ and $U$ an open set in $X$. For a function $f:U \to Y$ it is said to be differentiable in point $x \in U$ if there exists a continuous linear map $A_x:X \to Y$ such that: $$f(x+h)-f(x)=A_xh+R(h)$$ where $$\lim_{h \to 0}\frac{R(h)}{\|h\|}=0. \text{ or } R(h)=o(h)$$

  • $\begingroup$ showing $f(x, 0)$ is differentiable is sufficient, as $f(x,y) = f(x,0) + y$. As for the differentiability of $f(x,0)$ use $\tan(x) = \sin(x) / \cos(x)$, or l'hospital. $\endgroup$ – user251257 Aug 25 '15 at 22:37
  • $\begingroup$ Thats exactly what Im having trouble with, putting it all ttogether $\endgroup$ – Bozo Vulicevic Aug 25 '15 at 22:40

To prove that $f$ is differentiable at zero use Maclaurin $$ \tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\ldots\text{(higher order terms)}. $$ Then $f(x,y)=1+y+g(x,y)$ where $$ g(x,y)=\left\{ \begin{array}{ll} \frac{x^2}{3}+\frac{2x^4}{15}+\ldots\quad&\text{if } 0<|x|<1,\\ 0\qquad&\text{if } x=0. \end{array} \right. $$ Since $g(x,y)=o(\|(x,y)\|)$, the function $f$ is differentiable at zero by definition.

UPDATE: We check $g(x,y)=o(\|(x,y)\|)$ by definition $$ \left|\frac{g(x,y)}{\sqrt{x^2+y^2}}\right|\le \left|\frac{g(x,y)}{x}\right|\le \left|\frac{x}{3}+\frac{2x^3}{15}+\ldots\right|\to 0,\qquad\text{when } x\to 0. $$ Here and above everywhere $\ldots$ means the higher order terms in $x$.

P.S. Here $g(x,y)$ (actually depends only on $x$) plays the role of $R(h)$, and the formula $f(x,y)=1+y+g(x,y)$ after changing $(x,y)$ to $(h_1,h_2)$ looks like $$ f(h_1,h_2)=f(0,0)+h_2+g(h_1,h_2) $$ which means that the Frechet derivative at zero is $A_0=[0\ 1]$ or $$ A_0h=h_2=\left[\matrix{0 & 1}\right]\left[\matrix{h_1\\h_2}\right]. $$

  • $\begingroup$ Can you just state how and why, maybe prove that $g(x,y)=o(\|(x,y)\|)$ $\endgroup$ – Bozo Vulicevic Aug 25 '15 at 22:58
  • $\begingroup$ @BozoVulicevic Sure. Adding some more to the answer. $\endgroup$ – A.Γ. Aug 25 '15 at 23:07

It is sufficient to show that $x\mapsto f(x,0)$ is differentiable, as $f(x,y)=f(x,0)+y$ and sum of differentiable functions is differentiable again.

As for the differentiability of $f(x,0)$: $$ \frac{\partial}{\partial x} f(0,0) = \lim_{x\to 0} \frac{\tan(x) - x}{x^2} = \lim_{x\to 0} \frac{1 + \tan^2(x) - 1}{2x} = \lim_{x\to 0} \frac{2\tan(x)(1+\tan^2(x))}{2} = 0. $$ Further, $f(x,0)$ is a composition of differentiable functions on $(-1,1)\setminus\{ 0 \}$ and thus differentiable.

In summary, $f$ is differentiable on $(-1,1)\times\mathbb R$.


We did this already, didn't we? If we define $g(t) = (\tan t)/t, 0< |t|< 1, g(0)= 1,$ then $g \in C^\infty(-1,1).$ Clearly $f(x,y) = g(x) + y, (x,y) \in (-1,1)\times \mathbb {R}.$ Thus $f\in C^\infty((-1,1)\times \mathbb {R}).$ So certainly $f$ is Frechet differentiable in this domain.

  • $\begingroup$ we did a different one, but I lost track of it. :D Thanks again ! $\endgroup$ – Bozo Vulicevic Aug 25 '15 at 23:17
  • $\begingroup$ I really am having troubles with these assignment, im gonna ask another question, same subject, I hope you can make the time to see it. $\endgroup$ – Bozo Vulicevic Aug 25 '15 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.