If $z = x^y$ for $x > 0$, the value of the mixed second partial derivative $\frac{\partial^2 z}{\partial x \partial y}$ evaluated at $x = 1$ and $y = 2$ is ________ (answer in integer).
Answer Key:
1
Step-by-Step Solution & Explanation
First compute the partial derivative with respect to $x$:
$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^y) = y x^{y-1}$
Now differentiate with respect to $y$ using the product rule:
$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(y x^{y-1}) = 1 \cdot x^{y-1} + y \cdot \frac{\partial}{\partial y}(x^{y-1})$
Since $x^{y-1} = e^{(y-1) \ln x}$, its derivative with respect to $y$ is $x^{y-1} \ln x$:
$\frac{\partial^2 z}{\partial x \partial y} = x^{y-1} + y x^{y-1} \ln x$
Evaluating at $x = 1, y = 2$:
$x^{y-1} = 1^{2-1} = 1$
$\ln(1) = 0$
$\frac{\partial^2 z}{\partial x \partial y} = 1 + 2(1)(0) = 1$