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PrithiviRaj
PrithiviRaj
12:45
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Q: Check differentiability of $f(x,y)=y\sin\frac{1}{x}$ at $(0,0)$

MartinGiven , $f(x,y)=y\sin\frac{1}{x}$ when $x\ne0$ and $f(0,0)=0$ . Investigate differentiability at $(0,0)$ . I've found that it is continuous at $(0,0)$ and the partial derivatives $f_x=0$ $\forall (x,y)$ and $f_y=\begin{cases} \sin\frac1x & \text{ if } x\ne0\\ 1 & \text{ if } x= 0 \end{cases}.$...

multivariable-calculus derivatives
look the answer given by GNU supporter.
@simply

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