Find all the second partial derivatives in each of the following cases: (a) \(z=\frac{1}{x}\) (b) \(z=\frac{y}{x}\) (c) \(z=\frac{x}{y}\) (d) \(z=\frac{1}{x}+\frac{1}{y}\)

Question: Find the second partial derivatives for the following functions: (a) \(z = \frac{1}{x}\), (b) \(z = \frac{y}{x}\), (c) \(z = \frac{x}{y}\), and (d) \(z = \frac{1}{x} + \frac{1}{y}\). Answer: For (a) we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\). For (b) we have: \(\frac{\partial^2 z}{\partial x^2} = \frac{2y}{x^3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{x^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{x^2}\). For (c) we have: \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial y^2}= \frac{2x}{y^3}\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{y^2}\). For (d) we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}= 2y^{-3}\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\).