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: First find \( \frac{\partial f}{\partial x}(0,y) \).
For \(y\neq 0\),
\[ \frac{\partial f}{\partial x}(0,y) = \lim_{h\to 0}\frac{f(h,y)-f(0,y)}{h} \]

Step 2: Use \(f(0,y)=0\).
Since \(xy=0\) when \(x=0\), we have
\[ f(0,y)=0 \]
Therefore,
\[ \frac{\partial f}{\partial x}(0,y) = \lim_{h\to 0}\frac{f(h,y)}{h} \]

Step 3: Substitute \(f(h,y)\).
\[ \frac{f(h,y)}{h} = 4h\tan^{-1}\left(\frac{y}{2h}\right) + \frac{y^3}{h}\tan^{-1}\left(\frac{h}{4y^2}\right) \]

Step 4: Evaluate the first term.
Since \( \tan^{-1}\left(\frac{y}{2h}\right) \) is bounded,
\[ \lim_{h\to 0}4h\tan^{-1}\left(\frac{y}{2h}\right)=0 \]

Step 5: Evaluate the second term.
Using
\[ \lim_{u\to 0}\frac{\tan^{-1}u}{u}=1 \]
we get
\[ \lim_{h\to 0}\frac{y^3}{h}\tan^{-1}\left(\frac{h}{4y^2}\right) = y^3\cdot \frac{1}{4y^2} = \frac{y}{4} \]

Step 6: Hence find \( \frac{\partial f}{\partial x}(0,y) \).
\[ \frac{\partial f}{\partial x}(0,y)=\frac{y}{4} \]
Also,
\[ \frac{\partial f}{\partial x}(0,0)=0 \]

Step 7: Differentiate with respect to \(y\) at \(0\).
\[ \left(\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\right)_{(0,0)} = \lim_{y\to 0}\frac{\frac{y}{4}-0}{y} = \frac{1}{4} \]
\[ \boxed{0.25} \]