Find the four second-order partial derivatives of $z=f(x,y)=x\;\ln\;y+ye^{x}$.

Solution  We begin by finding the first-order partial derivatives $f_{x}(x,y)$ and $f_{y}(x,y)$. $$f_{x}(x,y)=\ln\;y+ye^{x} \qquad f_{y}(x,y)=\dfrac{x}{y}+e^{x}$$

Then the second-order partial derivatives are $$\begin{array}{lll} f_{xx}(x,y)&=&\dfrac{\partial }{\partial x}f_{x}(x,y)=\dfrac{\partial }{\partial x}\left(\ln\;y+ye^{x}\right) =ye^{x} & f_{xy}(x,y)&=&\dfrac{\partial }{\partial y}f_{x}(x,y)=\dfrac{\partial }{\partial y}\left(\ln\;y+ye^{x}\right) =\dfrac{1}{y}+e^{x} \\[5pt] f_{yx}(x,y)&=&\dfrac{\partial }{\partial x}f_{y}(x,y)=\dfrac{\partial }{\partial x}\left(\dfrac{x}{y}+e^{x}\right) =\dfrac{1}{y}+e^{x} & \quad f_{yy}(x,y)&=&\dfrac{\partial }{\partial y}f_{y}(x,y)=\dfrac{\partial }{\partial y}\left( \dfrac{x}{y}+e^{x}\right) =-\dfrac{x}{y^{2}} \end{array}$$