Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 3G / c / Solution

The function $g(t)=|t{|}^{3/2}$ is differentiable everywhere: for $t\ne 0$, its derivative is $(3/2)\mathrm{sgn}(t)\sqrt{|t|}$, while at zero the difference quotient is $\mathrm{sgn}(t)\sqrt{|t|}\to 0$.

The function $k(t)=|t{|}^{1/2}$ is differentiable for $t\ne 0$, with derivative $\mathrm{sgn}(t)/(2\sqrt{|t|})$. At zero its difference quotient is $\mathrm{sgn}(t)/\sqrt{|t|}$, which has no finite limit. Applying the coordinatewise-sum argument givesAt those points,where the first coefficient is zero at $x=0$. At any point with $y=0$, the restriction to the vertical coordinate line already fails to be differentiable.