Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 3 / 1E / Solution

For $t\ne 0$, ordinary differentiation of the two smooth components givesAt zero, use the definition of the derivative in the Euclidean norm:Thus $f$ has differentiability everywhere and $\boxed{f^{\prime }(0)=(0,0)}$. The two vectors $(\sin (1/t),\cos (1/t))$ and $(-\cos (1/t),\sin (1/t))$ are orthogonal unit vectors, so the squared norm of the displayed derivative is $4t^2+1$. ConsequentlyThis discontinuous derivative of a differentiable plane curve also shows why componentwise scalar mean value theorem conclusions do not supply a single intermediate point reproducing an entire vector-valued difference quotient.