Finding the Differential \(dz\) of \(z=f(x,y)\)

Find the differential \(dz\) of each function:

1. \(f(x,y) =e^{x}\cos\;y\)
2. \(f(x,y) =\dfrac{\ln\;x}{y}\)

Solution (a) \(f\) is defined everywhere in the \(xy\)-plane. The partial derivatives of \(f\) are \[ \begin{equation*} f_{x}(x,y) =e^{x}\cos\;y\qquad f_{y}(x,y) =-e^{x}\;\sin\;y \end{equation*} \]

Since \(f_{x}\) and \(f_{y}\) are continuous everywhere in the \(xy\)-plane, the function \(z=f(x,y)\) is differentiable. The differential \(dz\) is \[ dz=e^{x}\cos\;y\,dx-e^{x}\sin\;y\,dy \]

(b) The domain of \(f\) is \(\left\{(x,y) \,|\,x>0,\,y\neq 0\right\} .\) The partial derivatives of \(f\) are \[ f_{x}(x,y) =\dfrac{1}{xy}\qquad f_{y}(x,y) =-\dfrac{\ln\;x}{y^{2}} \]

Since both partial derivatives exist and are continuous at every point \((x_{0},y_{0})\) in the domain of \(f\), the function \(z=f(x,y)\) is differentiable at every point \((x_{0},y_{0})\) in its domain. The differential \(dz\) is \[ dz=\dfrac{1}{xy}\, dx-\dfrac{\ln\;x}{y^{2}}\,dy \]