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UPSC 2021 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

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

Question

$f(x,y)=|x^2-y^2|$. Find $f_{xy}(0,0)$ and $f_{yx}(0,0)$. Hence show $f_{xy}(0,0)=f_{yx}(0,0)$.

Technique

Compute $f_x,f_y$ on each region; evaluate at axes; take partial of partial via limit definition.

Solution

Setup. $|x^2-y^2|=x^2-y^2$ if $|x|\ge|y|$, and $y^2-x^2$ if $|x|<|y|$.

Step 1 — Compute $f_x$ everywhere

For $|x|>|y|$: $f=x^2-y^2$, $f_x=2x$. For $|x|<|y|$: $f=y^2-x^2$, $f_x=-2x$.

At $|x|=|y|\ne 0$: $f=0$, but derivative may not exist (corner). Need to check at origin separately.

$f_x(0,0)=\lim_{h\to 0}\dfrac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\dfrac{|h^2-0|-0}{h}=\lim_{h\to 0}\dfrac{h^2}{h}=\lim_{h\to 0}h=0$.

So $f_x(0,0)=0$.

Similarly $f_y(0,0)=0$.

Step 2 — $f_x(0,y)$ for $y\ne 0$

For $|y|>0$ and $x=0$ (so $|x|<|y|$): $f=y^2-x^2$, so $f_x=-2x$, $f_x(0,y)=0$.

So $f_x(0,y)=0$ for all $y$.

Step 3 — $f_{xy}(0,0)=\partial_y f_x|_{(0,0)}=0$

$f_{xy}(0,0)=\lim_{k\to 0}\dfrac{f_x(0,k)-f_x(0,0)}{k}=\lim_{k\to 0}\dfrac{0-0}{k}=0$.

Step 4 — By symmetry, $f_{yx}(0,0)=0$

$f$ is symmetric in $x,y$ (well, $|x^2-y^2|=|y^2-x^2|$, so $f(x,y)=f(y,x)$). So $f_y(x,0)$ analogously = 0 for all $x$.

$f_{yx}(0,0)=\lim_{h\to 0}\dfrac{f_y(h,0)-f_y(0,0)}{h}=0$.

Step 5 — Conclusion

Answer