Differentials — Question 10

Question 10 – Solution

Strategy. Existing partial derivatives propose a linear map, but differentiability requires the remainder divided by distance to vanish along every path.

Step 1: Partials $$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)}h =\lim_{h\to 0}\frac{h}{h}=1,$$ $$f_y(0,0)=\lim_{k\to 0}\frac{f(0,k)}k=0.$$ The candidate differential is therefore $L(dx,dy)=dx$.

Step 2: Remainder test Differentiability would require $$\frac{f(x,y)-x}{\sqrt{x^2+y^2}}\longrightarrow 0.$$ Along $y=x$, $$f(x,x)=\frac{x^3}{2x^2}=\frac x2,$$ so $$\frac{f(x,x)-x}{\sqrt{2x^2}} =\frac{-x/2}{\sqrt 2|x|}=-\frac{\operatorname{sgn}(x)}{2\sqrt 2},$$ which does not tend to zero.

Conclusion Although both partials exist, $f$ is not differentiable at the origin, so $$\boxed{df=dx\text{ is not a valid total first-order approximation there}.}$$