# Proving The Quotient Rule…

$y=\frac { u }{ v } ,\quad \therefore \quad \left( y+\delta y \right) =\frac { \left( u+\delta u \right) }{ \left( v+\delta v \right) } \\ \\ \left( y+\delta y \right) \left( v+\delta v \right) =u+\delta u\\ \\ yv+y\delta v+v\delta y+\delta y\delta v=u+\delta u\\ \\ but...\quad yv=u,\\ \\ so...\\ \\ y\delta v+v\delta y+\delta y\delta v=\delta u\\ \\ \frac { u }{ v } \delta v+v\delta y+\delta y\delta v=\delta u\\ \\ u\delta v+{ v }^{ 2 }\delta y+v\delta y\delta v=v\delta u\\ \\ { v }^{ 2 }\delta y=v\delta u-u\delta v-v\delta y\delta v\\ \\ { v }^{ 2 }\frac { \delta y }{ \delta x } =v\frac { \delta u }{ \delta x } -u\frac { \delta v }{ \delta x } -v\delta v\frac { \delta y }{ \delta x } \\ \\ But\quad as\quad \delta x\rightarrow 0,\quad \frac { \delta y }{ \delta x } \rightarrow \frac { dy }{ dx } ,\quad \frac { \delta u }{ \delta x } \rightarrow \frac { du }{ dx } ,\quad \frac { \delta v }{ \delta x } \rightarrow \frac { dv }{ dx } \quad and\quad \delta v\rightarrow 0.\\ \\ So\quad you\quad get:\\ \\ { v }^{ 2 }\frac { dy }{ dx } =v\frac { du }{ dx } -u\frac { dv }{ dx } \\ \\ \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } }$