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UPSC 2018 Maths Optional Paper 1 Q3b — Step-by-Step Solution

12 marks · Section A

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

Question

Let

Determine which of $\dfrac{\partial f}{\partial x}(0,1)$ and $\dfrac{\partial f}{\partial y}(0,1)$ exists and which does not exist.

Technique

Partial derivatives via one-variable difference quotients; recognise that $(0,1)$ lies interior to a single branch.

Solution

Step 1 — Locate the point relative to the pieces

The point $(0,1)$ has $y=1>0$, which is in the interior of the region $\{y>0\}$. In a whole neighbourhood of $(0,1)$ (any disc of radius $<1$) we have $y>0$, so there $f(x,y)=xy^2$ identically — the piecewise split at $y=0$ is far away and irrelevant for derivatives at $(0,1)$.

Step 2 — $f_x(0,1)$ from the difference quotient

So $f_x(0,1)$ exists and equals $1$. (Consistently, $\partial_x(xy^2)=y^2=1$ at $(0,1)$.)

Step 3 — $f_y(0,1)$ from the difference quotient

For $|k|<1$, $1+k>0$, so $f(0,1+k)=0\cdot(1+k)^2=0$ and $f(0,1)=0$. Hence the quotient is $\frac{0-0}{k}=0$ for all small $k$, giving

So $f_y(0,1)$ also exists and equals $0$. (Consistently, $\partial_y(xy^2)=2xy=0$ at $x=0$.)

Step 4 — Conclusion

Answer