Increase from y to y+δy correspondingly x to x+δx in the above equation(1)
⇒y+δy=cos2(x+δx) .....(2)
Eqn(2)-Eqn(1)
⇒y+δy−y=cos2(x+δx)−cos2x
⇒δy=cos2(x+δx)−cos2x
Divide both sides by δx we get
⇒δyδx=cos2(x+δx)−cos2xδx
⇒δyδx=−sin(2x+δx)sinδxδx by using cos2B−cos2A=sin(A+B)sin(A−B)
⇒limx→0δyδx=limx→0−sin(2x+δx)sinδxδx
⇒dydx=limx→0−sin(2x+δx)sinδxδx
⇒dydx=−sin(2x+0)×1$since$limx→0sinδxδx=1
∴dydx=−sin2x=−2cosxsinx