## Calculus 8th Edition

$$\frac{\partial}{\partial x}f(x,y)=\frac{1}{y};$$ $$\frac{\partial}{\partial y}f(x,y)=-\frac{x}{y^2}.$$
1) Regarding $y$ as constant and differentiating with respect to $x$ we get $$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{x}{y}\right)=\frac{1}{y}\frac{\partial}{\partial x}(x)=\frac{1}{y}\cdot 1=\frac{1}{y}.$$ 2) Regarding $x$ as constant and differentiating with respect to $y$ we get $$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{x}{y}\right)=x\frac{\partial}{\partial y}\left(\frac{1}{y}\right)=x\cdot\left(-\frac{1}{y^2}\right)=-\frac{x}{y^2}.$$