y=csc2x .......(1)
Increase from y to y+δy correspondingly x to x+δx in the above equation(1)
⇒y+δy=csc2(x+δx) .....(2)
Eqn(2)-Eqn(1)
⇒y+δy−y=csc2(x+δx)−csc2x
⇒δy=1sin2(x+δx)−1sin2x
⇒δy=sin2x−sin2(x+δx)sin2xsin2(x+δx)
⇒δ=sin(x+x+δx)sin(x−x−δx)sin2xsin2(x+δx) using sin2A−sin2B=sin(A+B)sin(A−B)
⇒δy=sin(x+x+δx)sin(x−x−δx)sin2xsin2(x+δx)
⇒δy=sin(2x+δx)sin(−δx)sin2xsin2(x+δx)
⇒δy=−sin(2x+δx)sin(δx)sin2xsin2(x+δx) since sin(−θ)=−sinθ
⇒limδx→0δyδx=limδx→0−sin(2x+δx)sinδxδxsin2xsin2(x+δx) by applying limits to both sides
⇒dydx=−sin(2x+0)sinδxδxsin2xsin2(x+0)
⇒dydx=−sin2xsin2xsin2x since limθ→0sinθθ=1
⇒dydx=−2sinxcosxsin2xsin2x since sin2x=2sinxcosx
⇒dydx=−2cosxsin3x
∴dydx=−2cotxcsc2x