UPSC 2025 Maths Optional Paper 1 Q4b — Step-by-Step Solution

15 marks · Section A

Partial derivatives · Calculus · asked 8× in 13 yrs · Read the full method →

Question

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.\;}