Find all the second partial derivatives in each of the following cases: \(\begin{aligned}&\text { (a) } z=\ln x(\mathrm{~b}) z=\ln y & (\mathrm{c}) z=\ln x y\end{aligned}\) (d) \(z=x \ln y\) (e) \(z=y \ln x\)

Answer: (a) \(\frac{\partial^2 z}{\partial x^2} = -\frac{1}{x^2}\), \(\frac{\partial^2 z}{\partial x \partial y} = 0\), \(\frac{\partial^2 z}{\partial y^2} = 0\), \(\frac{\partial^2 z}{\partial y \partial x} = 0\). (b) \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial x \partial y} = 0\), \(\frac{\partial^2 z}{\partial y^2} = -\frac{1}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x} = 0\). (c) \(\frac{\partial^2 z}{\partial x^2} = -\frac{1}{x^2}\), \(\frac{\partial^2 z}{\partial x \partial y} = 0\), \(\frac{\partial^2 z}{\partial y^2} = -\frac{1}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x} = 0\). (d) \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial x \partial y} = \frac{1}{y}\), \(\frac{\partial^2 z}{\partial y^2} = -\frac{x}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x} = \frac{1}{y}\). (e) \(\frac{\partial^2 z}{\partial x^2} = -\frac{y}{x^2}\), \(\frac{\partial^2 z}{\partial x \partial y} = \frac{1}{x}\), \(\frac{\partial^2 z}{\partial y^2} = 0\), \(\frac{\partial^2 z}{\partial y \partial x} = \frac{1}{x}\).