Given $f(x, y) = x^3 y^2 + 2x^2 y + 5$. Calculate the mixed second partial derivative $\frac{\partial^2 f}{\partial x \partial y}$ evaluated at $(x=1, y=2)$.
Answer Key:
16
Step-by-Step Solution & Explanation
First, differentiate with respect to $y$:
$\frac{\partial f}{\partial y} = 2x^3 y + 2x^2$
Next, differentiate with respect to $x$:
$\frac{\partial^2 f}{\partial x \partial y} = 6x^2 y + 4x$
Evaluating at $x = 1, y = 2$:
$\left. \frac{\partial^2 f}{\partial x \partial y} \right|_{(1,2)} = 6(1)^2(2) + 4(1) = 12 + 4 = 16.0$