Question

Let the function \( f:\mathbb{R}^2\to \mathbb{R} \) be defined by

Hint:

For mixed partial derivatives at the origin, first compute the first partial derivative along the required axis using the limit definition.

Answer 1

Correct Answer: 0.25

Step 1: Find $\frac{\partial f}{\partial x}$ along $x=0$.
Using the limit definition, $\frac{\partial f}{\partial x}(0,y)=\lim_{h\to 0}\frac{f(h,y)}{h}$, since $f(0,y)=0$.

Step 2: Split the difference quotient.
\[ \frac{f(h,y)}{h}=4h\tan^{-1}\!\Big(\frac{y}{2h}\Big)+\frac{y^3}{h}\tan^{-1}\!\Big(\frac{h}{4y^2}\Big) \]

Step 3: First piece dies.
The factor $\tan^{-1}(\cdot)$ stays bounded, so $4h\tan^{-1}(\cdot)\to 0$ as $h\to 0$.

Step 4: Second piece.
Using $\tan^{-1}u\approx u$ for small $u$, $\frac{y^3}{h}\cdot\frac{h}{4y^2}=\frac{y}{4}$. So $\frac{\partial f}{\partial x}(0,y)=\frac{y}{4}$.

Step 5: Now differentiate in $y$ at the origin.
\[ \frac{\partial^2 f}{\partial y\,\partial x}(0,0)=\lim_{y\to 0}\frac{\frac{y}{4}-0}{y}=\frac14 \]

Step 6: Answer.
\[ \boxed{0.25} \]