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Let $f:\mathbb{R}^2 \to \mathbb{R}$ defined by:

$f(x,y) = \begin{cases} \left( x^2 \right) \sin \left(\frac{y}{x^2} \right) & \text { if } x \neq 0\\ 0 & \text{ if } x = 0 \end{cases}$

Is the function $f$ differentiable or not? I don’t know whether my solution is right, so I’ll post it here.

My solution:

If $x \neq 0$, than $f(x,y) = x^2 \sin(y/x^2)$ is differentiable as it is the product of differentiable functions.

Now for $x=0$, let's consider the limit:

$$\begin{align*} \lim_{h \to 0} \frac{f(h,y) - f(0,y)}{h} &= \lim_{h \to 0} h^2\sin(\frac{y}{h^2})\\ &= \lim_{h \to 0} h^2\sin(yh^{-2})\\ &= \lim_{h \to 0} h^{2} = 0 \end{align*}$$

Therefore the function $f$ is differentiable.

MY NEW SOLUTION: I first calculated the partial derivatives with respect to x and y with limits at the point $(0,y)$.

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

$\frac{\partial f}{\partial y}(0,y) = \lim_{k \to 0} \frac{f(0,y+k)-f(0,y)}{k} = \lim_{k \to 0} \frac{0-0}{k} = 0$

Consequently, since $f(x, 0) = 0$ and the only candidate for a derivative matrix is $f'(x, 0) = (0, 0)$, we have:

$E(h,k):= f(0+h,y+k) = f(h,y+k) = h^2 \sin(\frac{y+k}{h^2})$

Then:

$0 \leq \frac{|E(h,k)|}{||(h,k)||} \leq \frac{h^2\sin(\frac{y+k}{h^2})}{\sqrt{h^2+k^2}} \leq |h| \cdot \frac{|h|}{\sqrt{h^2+k^2}} \leq |h|$

This converge to zero when $(h,k) \to 0$, therefore $f$ is differentiable at the point $(0,y)$.

EUEU
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  • First, you forgot that $h^2/h = h$. Second, you need to do more than compute a partial derivative to determine differentiability. (You also forgot to calculate the partial derivative with respect to $y$.) There are hundreds of similar questions posted on this site. – Ted Shifrin May 20 '22 at 17:29
  • @TedShifrin Hello, so if a calculate de partial derivatives with respect to x and y, and prove that they're continuous then the function f is differentiable, right? – EUEU May 20 '22 at 17:51
  • Yes, that's a theorem. However, you'll find that the partial derivatives of this function are not continuous. – Ted Shifrin May 20 '22 at 18:10

1 Answers1

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A function $f:\Bbb{R}^{2}\to\Bbb{R}$ is differentiable when $f(h,k)-f(0,0)=f_{x}(0,0)h+f_{y}(0,0)k + h\phi(h,k)+k\psi(h,k)$ where $\psi$ and $\phi$ are function such that $\displaystyle\lim_{(h,k)\to(0,0)}\phi(h,k)=\lim_{(h,k)\to(0,0)}\psi(h,k)=0$.

Then set $\phi(h,k)=h\sin(\frac{k}{h^{2}})$ and $\psi(h,k)=0$.

And you'll get differentiability.

More generally a function $f:\Bbb{R}^{n}\to\Bbb{R}^{m}$ is differentiable at a point $p$ if for the Jacobian matrix $J$ you have $f(p+h)-f(p)= Jh + ||h||\phi(h)$ such that $$\lim_{||h||\to 0}\phi(h)=0 $$

Mr.Gandalf Sauron
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  • Your first definition is a less standard, and often more awkward, definition. Better to just argue that $f(h,k)-f(0,0) - (f_x(0,0)h +f_y(0,0)k) = E(h,k)$ has the property that $E(h,k)/\sqrt{h^2+k^2}\to 0$ as $(h,k)\to (0,0)$. Your "more generally" is confusion, as these are — to the casual observer — quite different definitions. – Ted Shifrin May 20 '22 at 18:09
  • For functions from $\Bbb{R}^{2}\to \Bbb{R}$ this definition might be better suited for the op as the op might not be familiar with Linear transformations. Besides I find it useful and it is an equivalent condition anyways. But yeah I'll edit it in too !! – Mr.Gandalf Sauron May 20 '22 at 18:13
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    @Ted Shifrin. Property, that $E(h,k)=\varepsilon(h,k)\sqrt{h^2+k^2}$, where $\varepsilon(h,k)\to 0$, when $(h,k)\to 0$ is equivalent to property $E(h,k)=\varepsilon_1(h,k)\cdot h+\varepsilon_2(h,k)\cdot k$, $\sqrt{h^2+k^2}\ne 0$, where $\varepsilon_1(h,k)\to 0$, $\varepsilon_2(h,k)\to 0$ when $(h,k)\to 0$.

    And since they are equivalent, then what is the meaning of your words

    "less standard, and often more awkward"

    ?

     – zkutch May 20 '22 at 19:41