Show that $z=f(x, y)=x^{2}y-1$ is differentiable.

Solution The domain of $f$ is the $xy$-plane and the partial derivatives of $f$ are $$f_{x}(x,y) =2xy\qquad \hbox{and}\qquad f_{y}(x,y) =x^{2}$$

We find the change $\Delta z$ in $z$ and express it in the form of (2). $$\begin{eqnarray*} \Delta z &=& [ (x+\Delta x)^{2}(y+\Delta y)-1 ] - [ x^{2}y-1 ]\\ &=& [ x^{2}+2x\Delta x+ ( \Delta x ) ^{2} ] ( y+\Delta y ) - x^{2}y \\ &=&x^{2}\Delta y+2xy\Delta x+2x\Delta x\Delta y+y(\Delta x)^{2}+(\Delta x)^{2}\Delta y \\ &=&\underset{\color{#0066A7}{\hbox{\(f_{x}(x,y)\)}}}{\underbrace{ ( 2xy ) }}\Delta x+\! \underset{\color{#0066A7}{\hbox{\(f_{y}(x,y)\)}}}{\underbrace{ ( x^{2} ) }} \Delta y+\,\underset{\color{#0066A7}{\hbox{\(\eta _{1}\)}}}{\underbrace{ [ y\Delta x ] }}\Delta x+\,\underset{\color{#0066A7}{\hbox{\(\eta _{2}\)}}}{\underbrace{ [ 2x\Delta x+(\Delta x)^{2} ] }}\Delta y\\ &=&f_{x}(x,y) \Delta x+f_{y}(x,y) \Delta y+\eta _{1}\Delta x+\eta _{2}\Delta y \end{eqnarray*}$$

Equation (2) is satisfied. It remains to show that equation (3) is satisfied. $$\lim_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{1}=\lim_{\Delta x\rightarrow 0}( y\Delta x) =0\quad\!\! \hbox{and}\!\!\quad \lim_{( \Delta x,\Delta y) \rightarrow (0,0) }\eta _{2}=\lim_{\Delta x\rightarrow 0}[ 2x\Delta x+(\Delta x)^{2}] =0$$

So, $z=f(x,y)$ is differentiable on its domain.