How can one prove, that this function is totally differentiable on $\mathbb{R^2}$ and not continuous partially differentiable on $\mathbb{R^2}$?

$$ f(x, y) := \begin{cases} (x^4-y^4)\cos\left(\dfrac{1}{\|(x,y)\|^3_2}\right), & (x, y) \neq (0,0); \\ 0, & (x, y) = (0, 0). \end{cases} $$

I know that to prove the total derivative one first has to check, whether this function is continuous and partially derivable, or not.

I also know that I can use the following formula to prove that a function is totally differentiable:

$$ \dfrac{f(x, y) - f(0, 0) - \left(\left(\dfrac{\partial f}{\partial x}\right)(0, 0)\left(\dfrac{\partial f}{\partial y}\right)(0, 0)\right)\cdot\left({x-0}\atop{y-0}\right)}{|(x, y) - (0, 0)|} $$

If this formula gives a $0$, the function is totally differentiable.

I am stuck though, since I can't even find out if the function is continuous, or what the total derivative would be.

Can I substitute $x^4 = a $ and $y^4 = b$ and then we could follow:

$$(a-b)\cos\left(\frac{1}{\left\|\sqrt{\sqrt{(x,y)}} \right\|^3_2}\right) =$$

$$ = (a-b)\cos\left(\frac{1}{\left\| \sqrt{(x,y)} \right\|^1_2}\right)$$

And then maybe l'Hospital (though I don't know how that would work for the denominator) and then calculating the limit for $n\to\infty$, which would be $0\cdot 1$ (I think, because $\cos\left(\frac{1}{x}\right)\to 1$ for $\lim\to\infty$), which gives us $0$, proving that the function is continuous.


1 Answer 1


First of the function is totally differentiable on $\mathbb{R}^*$

Next it is continuous around $(0,0)$ because $$|f(x,y)| < x^4+y^4$$

therefore $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$$

so f is continuous in $(0,0)$ and $f(0,0) = 0$

Next the gradient of $f$ is

$$\partial_xf = 4x^3\cos\left((x^2+y^2)^{-3/2}\right)-3(x^4-y^4)(x^2+y^2)^{-5/2}x\sin\left((x^2+y+2)^{-3/2}\right)$$

$$\partial_yf = -4y^3\cos\left((x^2+y^2)^{-3/2}\right)-3(x^4-y^4)(x^2+y^2)^{-5/2}y\sin\left((x^2+y+2)^{-3/2}\right)$$

so we expect that it is totally differentiable at $(0,0)$ and that $\nabla f(0,0) = (0,0)$

This is indeed true because

$$\left|\frac{f(x,y)}{(x^2+y^2)^{1/2}}\right| = \left|(x^2-y^2)(x^2+y^2)^{1/2}\cos\left((x^2+y^2)^{-3/2}\right)\right| \leq (x^2+y^2)^{3/2} \rightarrow_{(x,y)\rightarrow (0,0)} 0$$


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.