Let \(f:\mathbb R^2\to\mathbb R\) be defined by \[ f(x,y)=\begin{cases} x^{4/3}\sin(y/x) & \text{if } x\ne0,\\ 0 & \text{if } x=0. \end{cases} \] Determine all points at which \(f\) is differentiable.
Solution
1. The function \(f\) is differentiable at every point \((x,y)\) with \(x\neq0\).
On the open set \[ \{(x,y)\in\RR^2:x\neq0\}, \] the functions \(x\mapsto x^{4/3}\) and \((x,y)\mapsto y/x\) are differentiable, and so is their composition with \(\sin\). Hence \[ f(x,y)=x^{4/3}\sin(y/x) \] is differentiable there.
2. For every \(y_0\in\RR\), the function \(f\) is differentiable at \((0,y_0)\), with derivative equal to the zero linear map.
Since \[ f(0,y_0)=0, \] it suffices to prove \[ \frac{\abs{f(h,y_0+k)}}{\sqrt{h^2+k^2}} \longrightarrow0 \] as \((h,k)\to(0,0)\).
If \(h=0\), the numerator is \(0\). If \(h\neq0\), then \[ \begin{aligned} \frac{\abs{f(h,y_0+k)}}{\sqrt{h^2+k^2}} &= \frac{\abs{h}^{4/3}\abs{\sin((y_0+k)/h)}}{\sqrt{h^2+k^2}}\\ &\leq \frac{\abs{h}^{4/3}}{\abs{h}}\\ &= \abs{h}^{1/3}. \end{aligned} \] The last quantity tends to \(0\) as \((h,k)\to(0,0)\). Therefore \[ f(0+h,y_0+k)-f(0,y_0) = o\bigl(\sqrt{h^2+k^2}\bigr), \] which is exactly differentiability at \((0,y_0)\) with derivative \(0\).
3. The function is differentiable at\[ \boxed{\text{every point of }\RR^2}. \]
Step 1 covers every point with nonzero first coordinate, and step 2 covers every point with first coordinate zero.
4. Q.E.D.
Step 3 is the required determination.