$\begin{align}  & y=\frac{u}{v} \\ & \text{Let }x\text{ receive an increment }\delta x\,\text{and due to this} \\ & \text{let }u\text{ and }v\text{ receive increment }\delta u\text{ and }\delta v\text{ and } \\ & \text{consequently for }y\text{ the increment is }\delta y, \\ & \text{Then } \\ & y+\delta y=\frac{u+\delta u}{v+\delta v} \\ & \delta y=\frac{u+\delta u}{v+\delta v}-\frac{u}{v} \\ & \delta y=\frac{v(u+\delta u)-u(v+\delta v)}{v(v+\delta v)} \\ & \delta y=\frac{v\delta u-u\delta v}{{{v}^{2}}+\delta v} \\ & \delta y=\frac{v\delta u-u\delta v}{{{v}^{2}}+\delta v} \\ & \text{Dividing through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{v\delta u-u\delta v}{(\delta x)({{v}^{2}}+\delta v)} \\ & \frac{\delta y}{\delta x}=\frac{\frac{v\delta u}{\delta x}-\frac{udv}{\delta x}}{({{v}^{2}}+\delta v)} \\ & \text{Taking limits as }\delta x\to 0, \\ & \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{{v}^{2}}} \\ & \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{{v}^{2}}} \\\end{align}$