3
$\begingroup$

I have to prove if this function is differentiable.

$$f(x,y)= \begin{cases} \frac{\cos x-\cos y}{x-y} \iff x \neq y \\-\sin x \iff x=y \end{cases}$$

if $x \neq y$ it is continuous, but i want to see if it is continuous in x=y too.

i can rewrite f as $$ f(x,y)= \begin{cases} \frac{g(x)-g(y)}{x-y} \iff x \neq y \\ g'(x)=g'(y) \iff x=y \end{cases}$$

and see that $lim_{xy \to xx} g(x,y)=g'(x)$. THus, it is continuous. Also, the partial derivatives exist: $$f_x(x,y)=\begin{cases} \frac{-\sin x(x-y)-\cos x+\cos y}{(x-y)^2} \\ -\cos(x) \end{cases}$$ $$f_y(x,y)=\begin{cases} \frac{\sin y(x-y)+\cos x-\cos y}{(x-y)^2} \\ 0 \end{cases}$$ If I proved that they are continuous, too, for the theorem of the total differential, the function would be differentiable. Still, I'm not sure this is the right way of reasoning.

$\endgroup$
1
$\begingroup$

$\cos x-\cos y=-2\sin\frac{x+y}2\,\sin\frac{x-y}2$ and $\operatorname {sinc}u=\frac{\sin u}{u}$ is a known analytical function. So $f(x,y)=-\sin\frac{x+y}2\operatorname {sinc}\frac{x-y}2$.

$\endgroup$
0
$\begingroup$

This seems like a perfectly OK way to do your work.

ALternatively, you might look at the function

$$ h(u, y) = f(u+y, y). $$

Writing that out, you get $$ h(u, y) = f(u+y,y)= \begin{cases} \frac{\cos(u+y)-\cos(y)}{u} \iff u+y \neq y \\ -\sin(y) \iff u = 0 \end{cases} $$

Which you can rewrite as $$ h(u, y) = f(u+y,y)= \begin{cases} \frac{\cos(u+y)-\cos(y)}{u} \iff u \neq 0 \\ -\sin(y) \iff u = 0 \end{cases} $$

In these rotated coordinates, it might be a little easier to prove things.

$\endgroup$
0
$\begingroup$

One has $$f(x,y)=-\int_0^1\sin\bigl((1-t)y+t\,x\bigr)\ dt\qquad\forall\ (x,y)\in{\mathbb R}^2\ .$$ This shows that $f\in C^\infty({\mathbb R}^2)$.

$\endgroup$

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.