" Iदि "x=(sin^(3)t)/(sqrt(cos2t))" तथा "y=(cos^(3)t)/(sqrt(cos2t))" तो "(dy)/(dx)" इात कीजिए। "

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x = (sin^3t)/(sqrt(cos 2t)), y = (cos^3 t)/(sqrt(cos 2t)) .

If x=(sin^(3)t)/(sqrt(cos2t)) and y=(cos^(3)t)/(sqrt(cos2t)) , then find (dy)/(dx) .

Find (dy)/(dx) , if x=(sin^3t)/(sqrt(cos2t)) , y=(cos^3t)/(sqrt(cos2t))

If x=(sin^(3)t)/(sqrt(cos 2t)) and y=(cos^(3)t)/( sqrt(cos 2t)) , show that (dy)/(dx)=0 at t=(pi)/(6) .

If x=(sin^(3)t)/(sqrt(cos2t)),y=(cos^(3)t)/(sqrt(cos2t)) show that (dy)/(dx)=0att=(pi)/(6)

Find dy/dx x=cos^(3)t/sqrtcos2t , y = sin^(3)t/sqrt cos2t

If x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t) show that dy/dx =0 at t=pi/6

x= sin^3t/sqrt x cos2t , y= cos^3t/ sqrt x cos2t .

x=sqrt(sin 2t),y=sqrt(cos 2 t)