Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 10F / Solution

Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Solved by gpt-5.6-sol high.