If `x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t)` show that `dy/dx =0 at

1.	If `x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t)` show that `dy/dx =0 at t=pi/6`
Answer» `x = ("sin"^(3)t)/(sqrt("cos" 2t))` `rArr (dx)/(dt) = (sqrt("cos"2t) * (d)/(dt)"sin"^(3)t - "sin"^(3)t(d)/(dt) sqrt("cos"2t))/(sqrt("cos" 2t))^(2)` `= (sqrt("cos"2t) * 3"sin"^(2)t "cos"t - "sin"^(3)t * ((-2"sin"2t)/(2sqrt("cos"2t))))/("cos" 2t)` `= ("cos"2t * 3"sin"^(2)t "cos"t + "sin"^(3)t "sin"2t)/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t[3 "cos"t " cos"2t + "sin"t " sin"2t])/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t[3 "cos"t (1-2"sin"^(2)t) + "sin"t * 2"sin"t " cos"t])/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t "cos" t(3-4 "sin"^(2)t))/("cos" 2tsqrt("cos"2t))` `" and "y = ("cos"^(2)t)/(sqrt("cos"2t))` `rArr(dy)/(dt) = (sqrt("cos"2t)(d)/(dt)"cos"^(3)t-"cos"^(3)t(d)/(dt)sqrt("cos"2t))/((sqrt("cos"2t))^(2)` `= (sqrt("cos"2t) * 3"cos"^(2) t (-"sin"t)-"cos"^(3)t * ((-2"sin"2t))/(2sqrt("cos"2t)))/("cos" 2t)` `=-((3"cos"^(2) t "sin"t * "cos" 2t - "sin" 2t "cos"^(3) t))/("cos" 2tsqrt("cos"2t))` `=-("cos"^(2)t[3"sin"t(2 "cos"^(2)t-1)-2 "sin" t "cos"t * "cos"t])/("cos" 2t sqrt("cos"2t))` `=-("cos"^(2)t"sin"t(4"cos"^(2)t-3))/("cos" 2t sqrt("cos"2t))` `Now (dy)/(dx) = (dy//dt)/(dx//dt) = -("cos"^(2)t"sin"t(4"cos"^(2)t-3))/("sin"^(2)t"cos"t(3-4"sin"^(2)t))` `=- ("cos"t(4"cos"^(2)t-3))/("sin"t(3-4"sin"^(2)t))` `=- ((4"cos"^(3)t-3"cos"t))/((3"sin"t-4"sin"^(3)t))` `=-("cos"3t)/("sin"3t) = -"cos" 3t`

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