Derivative of y=cotx

y=cotx=\frac { cosx }{ sinx } =\frac { u }{ v } ,\\ \\ If\quad y=\frac { u }{ v } ,\quad \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } } .\\ \\ u=cosx,\quad \therefore \quad \frac { du }{ dx } =-sinx\\ v=sinx,\quad \therefore \quad \frac { dv }{ dx } =cosx,\quad { v }^{ 2 }=\sin ^{ 2 }{ x } \\ \\ So:\\ \\ \frac { dy }{ dx } =\frac { sinx\cdot \left( -sinx \right) -cosxcosx }{ \sin ^{ 2 }{ x } } \\ \\ =\frac { -\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } =\frac { -\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } -\frac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } \\ \\ =-1-\cot ^{ 2 }{ x } =-{ cosec }^{ 2 }x\\ \\ \therefore \quad If\quad y=cotx,\quad \frac { dy }{ dx } =-{ cosec }^{ 2 }x\\ \\ \\

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