Determine the differentiability of $$f(x,y)=\begin{cases} x+ \frac{\sin y}{y}, & \text{if $y\neq 0;$ } \\x+1, & \text{if $y=0;$ } \end{cases}$$

I am using the Frechet derivative as my definition of differentiability. Since the partial derivatives of $ x+ \frac{\sin y}{y}, \text{if $y\neq 0;$ }$ are continuous function we know that $f(x,y)$ is differentiable $y\neq 0.$

But how about when $y=0$. Usually, at least with my other assignments, the second case was $(0,0)$ and I would do $f((0,0)+(h_1,h_2))-f(0,0)= \frac{\partial f}{\partial x}(0,0)h_1 + \frac{\partial f}{\partial y}(0,0)h_2 + R(h)$(usually these partial derivatives would be zero) after that the function $R(h)$ would be a function of $h_1, h_2$ meaning what is left to prove is that $$\lim_{h_1 \to 0, h_2 \to 0}\frac{R(h)}{\|(h_1,h_2)\|}$$ thats how differentiability is proved, but here we have $f((x,0)+(h_1,h_2))...$ which is obviously much different, any suggestions, as to what to do?


Let $g(t) = (\sin t)/t, t\ne 0, g(0) = 1.$ Then $g \in C^\infty (\mathbb {R}).$ (In fact, $g(t) = 1-t^2/3!+t^4/5! - \cdots,$ a power series that converges everywhere in $\mathbb {R}.$) Check that $f(x,y) = x+g(y)$ for all $(x,y).$ It follows that $f \in C^\infty(\mathbb {R}^2).$ This implies $f$ is Frechet differentiable everywhere.

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  • $\begingroup$ Is there a theory that would imply(could you send a link with a possible proof) that $f(x,y)=x+g(y)$ implies $f \in C^{\infty}\mathbb{(R^2)}$ because $g(t) \in C^{\infty}(\mathbb R)$ and $u(x)=x \in C^{\infty}(\mathbb R)$, because I haven't done this theory in class, so I don't know whether the professor would approve of this. This seems intuitively understandable (the technique you used), but formally how would I justify it ? $\endgroup$ – Bozo Vulicevic Aug 16 '15 at 15:00
  • $\begingroup$ Sums, products, compositions of smooth functions are smooth. So for example $h(x,y) = y$ is smooth, and since $g$ is smooth, so is $(g\circ h) (x,y) = g(y).$ $\endgroup$ – zhw. Aug 16 '15 at 16:19

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