Prove that the following functions is differentiable on $(-1,1) \times \mathbb R$

Question

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

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 $$\lim_{h \to 0}\frac{R(h)}{\|h\|}=0. \text{ or } R(h)=o(h)$$

Answer 1

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 $$ A_0h=h_2=\left[\matrix{0 & 1}\right]\left[\matrix{h_1\\h_2}\right]. $$

Answer 2

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 $(-1,1)\times\mathbb R$.

Answer 3

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.