Consider the function f(x)={xsin(1/x)x≠00x=0f(x) = \begin{cases} x \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}f(x)={xsin(1/x)0x=0x=0. Is fff differentiable at x=0x=0x=0?
Yes, f′(0)=0f'(0)=0f′(0)=0
Yes, f′(0)=1f'(0)=1f′(0)=1
No, the derivative does not exist
No, the function is discontinuous