← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q4b — Solution

15 marks · Section A

Question

If $f(x, y) = \begin{cases} xy\dfrac{x^2 - y^2}{x^2 + y^2}, & \text{when } (x, y) \neq (0, 0) \\ 0, & \text{when } (x, y) = (0, 0), \end{cases}$ then find $f_{xy}(0, 0)$ and $f_{yx}(0, 0)$.

Technique

Compute the mixed second-order partials at the origin from the limit definition, working through the first-order partials $f_x(0,y)$ and $f_y(x,0)$. This is the classic example where $f_{xy}(0,0)\neq f_{yx}(0,0)$, because the mixed partials are not continuous at the origin and Clairaut’s theorem fails.

Solution

Step 1 — $f_x(0,y)$ for arbitrary $y$.

By definition, $f_x(0,y)=\lim_{h\to0}\frac{f(h,y)-f(0,y)}{h}.$ For $(h,y)\neq(0,0)$, $f(h,y)=hy\dfrac{h^2-y^2}{h^2+y^2}$, and $f(0,y)=0$. Thus $f_x(0,y)=\lim_{h\to0}\frac{hy\frac{h^2-y^2}{h^2+y^2}}{h}=\lim_{h\to0}y\,\frac{h^2-y^2}{h^2+y^2}=y\cdot\frac{-y^2}{y^2}=-y\quad(y\neq0),$ and $f_x(0,0)=0$. So $f_x(0,y)=-y$ for all $y$ (the value at $y=0$ also fits).

Step 2 — $f_y(x,0)$ for arbitrary $x$.

Similarly, $f_y(x,0)=\lim_{k\to0}\frac{f(x,k)-f(x,0)}{k}=\lim_{k\to0}\frac{xk\frac{x^2-k^2}{x^2+k^2}}{k}=\lim_{k\to0}x\,\frac{x^2-k^2}{x^2+k^2}=x\cdot\frac{x^2}{x^2}=x\quad(x\neq0),$ and $f_y(0,0)=0$. So $f_y(x,0)=x$ for all $x$.

Step 3 — Mixed partials at the origin.

$f_{xy}(0,0)=\frac{\partial}{\partial y}\Big[f_x(0,y)\Big]_{y=0}=\lim_{k\to0}\frac{f_x(0,k)-f_x(0,0)}{k}=\lim_{k\to0}\frac{-k-0}{k}=-1.$ $f_{yx}(0,0)=\frac{\partial}{\partial x}\Big[f_y(x,0)\Big]_{x=0}=\lim_{h\to0}\frac{f_y(h,0)-f_y(0,0)}{h}=\lim_{h\to0}\frac{h-0}{h}=+1.$

Since $f_{xy}(0,0)=-1\neq1=f_{yx}(0,0)$, the order of differentiation matters here (the mixed partials are discontinuous at the origin, so Clairaut’s theorem does not apply).

Answer

$\boxed{\;f_{xy}(0,0)=-1,\qquad f_{yx}(0,0)=+1.\;}$